标签：专题 idx int 短路 cin vis add kuangbin dis

【kuangbin】专题四 最短路

https://www.acwing.com/activity/content/90/
（没做的那道是网络流）

先把代码放这...吃完饭回来再总结

1. 青蛙

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
typedef pair<double, int> pdi; //距离 编号
const int N = 205;
int n;
pii g[N];
double d[N];

double count (int a, int b) {
    return sqrt ((g[a].first - g[b].first) * (g[a].first - g[b].first) + (g[a].second - g[b].second) * (g[a].second - g[b].second));
}

void test () {
    for (int i = 1; i <= n; i ++)
        cout << d[i] << ' ';
    cout << endl;
}

void solve () {
    for (int i = 1; i <= n; i ++)   d[i] = 0x3f3f3f3f;
    d[1] = 0;
    priority_queue <pdi, vector <pdi>, greater <pdi>> q;
    q.push ({0, 1});
    //test ();

    while (!q.empty ()) {
        auto t = q.top ();
        q.pop ();
        double dis = t.first;
        int id = t.second;

        if (dis > d[id])    continue;
        //cout << "&";
        for (int i = 2; i <= n; i ++) {
            double cur = max (dis, count (i, id)); //当前最大距离
            if (cur < d[i]) {
                //cout << "&";
                d[i] = cur;
                q.push ({d[i], i});
            }
        }
    }
    //test ();
    cout << fixed << setprecision (3) <<  d[2] << endl << endl;
}
//固定一点，更新距离

int main () {
    int cnt = 0;
    while (cin >> n, n) {
        cnt ++;
        cout << "Scenario #" << cnt << endl;
        cout << "Frog Distance = ";
        for (int i = 1; i <= n; i ++)   cin >> g[i].first >> g[i].second;
        if (n == 2) {
            cout << fixed << setprecision (3) << count (1, 2) << endl << endl;
            continue;
        }

        solve ();
    }
}

2. 货物运输

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 1005;
int n, m;
int e[N][N];
int dis[N];

int dijkstra () {
    priority_queue <pii> q;
    memset (dis, 0, sizeof dis);
    dis[0] = 1e9;
    q.push ({1e9, 1});

    while (!q.empty ()) {
        auto t = q.top ();
        q.pop();
        int dist = t.first, ver = t.second;

        if (dis[ver] > dist)    continue;

        for (int i = 1; i <= n; i ++) {
            int x = min (dist, e[ver][i]);
            if (x > dis[i]) {
                dis[i] = x;
                q.push ({x, i});
            }
        }
    }
    return dis[n];
}

void solve () {
    memset (e, 0, sizeof e);
    cin >> n >> m;
    while (m --) {
        int a, b, c;
        cin >> a >> b >> c;
        e[a][b] = e[b][a] = c;
    }

    cout << dijkstra () << endl << endl;
}

int main () {
    int t;
    cin >> t;
    for (int i = 1; i <= t; i ++) {
        cout << "Scenario #" << i << ":\n";
        solve ();
    }
}

//与青蛙相反，求最小值最大

3. 农场派对

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 1e5 + 5;
int dis[N][2];
int n, m, x;
bool vis[N];

int h[N], e[N], ne[N], w[N], idx;

void add (int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void dijkstra (int id) {
    memset (vis, false, sizeof vis);
    dis[x][id] = 0;
    priority_queue <pii, vector <pii>, greater <pii>> q;
    q.push ({dis[x][id], x});

    while (!q.empty ()) {
        auto t = q.top ();
        q.pop ();
        int dist = t.first, ver = t.second;

        if (vis[ver])   continue;
        vis[ver] = true;

        for (int i = h[ver]; ~i; i = ne[i]) {
            int  j = e[i];
            if (dis[j][id] > dist + w[i]) {
                dis[j][id] = dist + w[i];
                q.push ({dis[j][id], j});
            }

        }

    }

}

int a[N], b[N], c[N];

int main () {
    memset (dis, 0x3f, sizeof dis);
    memset (h, -1, sizeof h);
    cin >> n >> m >> x;
    for (int i = 0; i < m; i ++) {
        cin >> a[i] >> b[i] >> c[i];
        add (a[i], b[i], c[i]);
        
    }

    dijkstra (0);
    
    memset (h, -1, sizeof h), idx = 0;
    for (int i = 0; i < m; i ++)
        add (b[i], a[i], c[i]);

    dijkstra (1);
    int ans = 0;
    for (int i = 1; i <= n; i ++)
        ans = max (ans, dis[i][0] + dis[i][1]);
    cout << ans << endl;

}
//最短路中最长的一条
//正反做两次dij

4. 货币兑换

#include <bits/stdc++.h>

using namespace std;
const int N = 1005;
int n, m, s;
double v;
double r[N], c[N], dis[N];
int h[N], e[N], ne[N], idx;
int cnt[N];
bool vis[N]; //是否在队列中

void add (int a, int b, double rr, double cc) {
    e[idx] = b, ne[idx] = h[a], r[idx] = rr, c[idx] = cc, h[a] = idx ++;
}

bool spfa () {
    queue<int> q;
    q.push (s), vis[s] = true;
    dis[s] = v;
    // for (int i = 1; i <= n; i ++) {
    //     q.push (i);
    //     vis[i] = true;
    // }

    while (!q.empty ()) {
        auto t = q.front ();
        q.pop ();
        vis[t] = false;

        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            double k = (dis[t] - c[i]) * r[i];

            if (dis[j] < k) { //注意反向，因为是看正回路
                //if (j == s)     return true;
                dis[j] = k;
                cnt[j] = cnt[t] + 1;

                if (cnt[j] >= n)    return true;
                if (!vis[j])    vis[j] = true, q.push (j);
            }
        }
    }
    return false;
}

int main () {
    memset (h, -1, sizeof h);
    cin >> n >> m >> s >> v;
    while (m --) {
        int a, b;
        double r1, r2, c1, c2;
        cin >> a >> b >> r1 >> c1 >> r2 >> c2;
        add (a, b, r1, c1);
        add (b, a, r2, c2);
    }
    if (spfa ())    cout << "YES\n";
    else    cout << "NO\n";

}
//求是否存在正权回路
//spfa判正环

5. 虫洞

#include <bits/stdc++.h>

using namespace std;
const int N = 505, M = 5205; //注意边的数量是5000+200
int t, n, m1, m2;
int h[N], e[M], ne[M], w[M], idx;
int dis[N], cnt[N];
bool vis[N];

void add (int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}

bool spfa () {
    queue <int> q;
    for (int i = 1; i <= n; i ++) {
        q.push (i);
        vis[i] = true;
        dis[i] = cnt[i] = 0;
    }

    while (!q.empty ()) {
        int t = q.front ();
        q.pop ();
        vis[t] = false; //出队

        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dis[t] + w[i]) {
                dis[j] = dis[t] + w[i];
                cnt[j] = cnt[t] + 1;

                if (cnt[j] >= n)    return true;

                if (!vis[j])    vis[j] = true, q.push (j); //入队
            }
        }
    }
    return false;
}

int main () {
    
    cin >> t;
    while (t --) {
        memset (h, -1, sizeof h), idx = 0; //初始化要做全套啊
        cin >> n >> m1 >> m2;
        while (m1 --) {
            int a, b, c;
            cin >> a >> b >> c;
            add (a, b, c), add (b, a, c);
        }

        while (m2 --) {
            int a, b, c;
            cin >> a >> b >> c;
            add (a, b, -c);
        }

        if (spfa())
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
}
//时间变小了也就相当于存在负环，把值变小
//转化为spfa判断是否存在负环

6. 传递信息

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio (0);cin.tie(0);

using namespace std;
typedef pair <int, int> pii;
const int N = 20005;
int n, dis[N];
int h[N], e[N], ne[N], w[N], idx;
bool vis[N];

void add (int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

int to_int(string s){
    int res = 0;
    for(int i = 0,l = s.size();i < l; ++i) res = res * 10 + s[i] - '0';
    return res;
}

void dijkstra () {
    memset (dis, 0x3f, sizeof dis);
    dis[1] = 0;
    priority_queue <pii, vector <pii>, greater<pii>> q;
    q.push ({0, 1});

    while (!q.empty ()) {
        auto t = q.top ();
        q.pop();
        int dist = t.first, ver = t.second;

        if (vis[ver])   continue;
        vis[ver] = true;

        for (int i = h[ver]; ~i; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dist + w[i]) {
                dis[j] = dist + w[i];
                q.push ({dis[j], j});
            }
        }

    }


}

int main () {
    IOS;
    memset (h, -1, sizeof h);
    cin >> n;
    for (int i = 2; i <= n; i ++)
        for (int j = 1; j < i; j ++) {
            string s;
            cin >> s;
            if (s != "x") {
                int x = to_int(s);
                add (i, j, x);
                add (j, i, x);
            }
        }

    dijkstra ();
    int ans = 0;
    for (int i = 2; i <= n; i ++)   ans = max (ans, dis[i]);
    cout << ans << endl;
}

//以1为源点的最小生成树
//烦死了。。读错题了，不是最小生成树
//传递是同时进行的，所以要看的是1到所有点的距离当中的最大值
//直接跑dijkstra

7. 牛的比赛

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 105;
int n, m;
int f[N][N];

int main () {
    cin >> n >> m;
    while (m --) {
        int a, b;
        cin >> a >> b;
        f[b][a] = 1;
    }

    //test ();

    for (int k = 1; k <= n; k ++)
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= n; j ++)
                f[i][j] |= (f[i][k] && f[k][j]); //直接有值或间接更新得到

    int ans = 0;
    for (int i = 1; i <= n; i ++) {
        int cnt = 0;
        for (int j = 1; j <= n; j ++)
            if (f[i][j] | f[j][i])   cnt ++;
        if (cnt == n - 1)   ans ++; //和我分析的一致
    }

    cout << ans << endl;
    
}

//度为n-1的节点一定能确定
//再根据已经确定的推未知的
//迭代更新的思想还是挺像最短路的

8. 最短路径和

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 1e6 + 5;
int h[N], hh[N], e[N<<1], ne[N<<1], w[N<<1], idx;
int dis1[N], dis2[N], n, m;
bool vis[N];

void add (int a, int b, int c, int *h) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void dijkstra (int *h, int *dis) {
    memset (dis, 0x3f, sizeof dis);
    memset (vis, false, sizeof vis);
    dis[1] = 0;
    priority_queue <pii, vector <pii>, greater <pii>>q;
    q.push ({0, 1});

    while (!q.empty ()) {
        auto t = q.top();
        q.pop();
        int ver = t.second, dist = t.first;

        if (vis[ver])   continue;
        vis[ver] = true;

        for (int i = h[ver]; ~i; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dist + w[i]) {
                dis[j] = dist + w[i];
                q.push ({dis[j], j});
            }
        }
    }
}

void solve () {
    memset (h, -1, sizeof h);
    memset (hh, -1, sizeof hh);
    //记得dis要初始化
    memset (dis1, 0x3f, sizeof dis1);
    memset (dis2, 0x3f, sizeof dis2);

    cin >> n >> m;
    while (m --) {
        int a, b, c;
        cin >> a >> b >> c;
        add (a, b, c, h);
        add (b, a, c, hh);
    }
    dijkstra (h, dis1);
    dijkstra (hh, dis2);

    long long ans = 0;
    for (int i = 1; i <= n; i ++)
        ans += dis1[i] + dis2[i];
    cout << ans << endl;
    
}

int main () {
    int t;
    cin >> t;
    while (t --) {
        solve ();
    }
}

//正反两次dijkstra

9. 糖果

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 30005, M = 150005;
int n, m;
int h[M], e[M], ne[M], w[M], idx;
int dis[N];
bool vis[N];

void add (int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

int dijkstra () {
    memset (dis, 0x3f, sizeof dis);
    memset (vis, false, sizeof vis);
    dis[1] = 0;
    priority_queue <pii, vector<pii>, greater <pii>>q;
    q.push ({0, 1});

    while (!q.empty ()) {
        auto t  = q.top ();
        q.pop();
        int ver = t.second, dist = t.first;

        if (vis[ver])   continue;
        vis[ver] = true;

        for (int i = h[ver]; ~i; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dist + w[i]) {
                dis[j] = dist + w[i];
                q.push ({dis[j], j});
            }
        }
    }
    return dis[n];
}

int main () {
    cin >> n >> m;
    memset (h, -1, sizeof h);
    while (m --) {
        int a, b, c;
        cin >> a >> b >> c;
        add (a, b, c);
    }
    cout << dijkstra () << endl;
}
// n 最多比小朋友 1 多分到的糖果数量的最大可能
//dijkstra板子

10. 地铁

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 205;
double dis[N][N];
int n;
vector <pii> v(2);

double dist (int i, int j) {
    return sqrt ((v[i].first - v[j].first) *(v[i].first - v[j].first) + (v[i].second - v[j].second) *(v[i].second - v[j].second)) * 0.006;
    //km/h -> m/min
}

int main () {
    for (int i = 0; i < N; i ++)
        for (int j = 0; j < N; j ++) {
            if (i == j)     dis[i][j] = 0;
            else    dis[i][j] = 0x3f3f3f3f;
        }

    cin >> v[0].first >> v[0].second >> v[1].first >> v[1].second;

    int x, y, cnt = 2;
    while (cin >> x >> y) {
        //统计更新路径
        if (x == -1 && y == -1) {
            for (int i = cnt; i < v.size (); i ++)
                for (int j = i + 1; j < v.size (); j ++)
                    dis[j][i] = dis[i][j] = dis[i][j-1] + dist (j, j-1) / 4;
            cnt = v.size(); //下一条路线的起点
        }
        else    v.push_back ({x, y});
    }

    n = v.size ();
    for (int i = 0; i < n; i ++)
        for (int j = 0; j < n; j ++)
            dis[i][j] = min (dis[i][j], dist (i, j)); //坐地铁or走路

    //floyd
    for (int k = 0; k < n; k ++)
        for (int i = 0; i < n; i ++)
            for (int j = 0; j < n; j ++)
                dis[i][j] = min (dis[i][j], dis[i][k] + dis[k][j]);

    cout << (int) (dis[0][1] + 0.50) << endl;

}

//n只有200，可以floyd

11. 昂贵的聘礼

#include <bits/stdc++.h>

using namespace std;
const int N = 105;
int n, m;
bool vis[N];
int dis[N], a[N][N], level[N];

int dijkstra (int st, int ed) {
    memset (dis, 0x3f, sizeof dis);
    memset (vis, false, sizeof vis);
    dis[0] = 0;

    //普通dijkstra
    for (int i = 1; i <= n; i  ++) {
        int t = -1;
        for (int j = 0; j <= n; j ++) {  //从0开始！！虚拟源点
            if (!vis[j] && (t == -1 || dis[t] > dis[j]))
                t = j;
        }
        vis[t] = true;

        for (int j = 1; j <= n; j ++)
            if (level[j] >= st && level[j] <= ed)
                dis[j] = min (dis[j], dis[t] + a[t][j]);
    }
    return dis[1];
}

int main () {
    memset (a, 0x3f, sizeof a);
    cin >> m >> n;
    for (int i = 1; i <= n; i ++)
        a[i][i] = 0;  

    for (int i = 1; i <= n; i ++) {
        int p, x;
        cin >> p >> level[i] >> x;
        a[0][i] = min (p, a[0][i]);
        while (x --) {
            int t, v;
            cin >> t >> v;
            a[t][i] = min (a[t][i], v);
        }
    }  

    int ans = 0x7fffffff;
    for (int i = level[1] - m; i <= level[1]; i ++) //从酋长开始交易的
        ans = min (ans, dijkstra(i, i + m));
    cout << ans << endl;
}

12. 电车

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 2e4 + 5;
int n, st, ed;
int h[N], e[N], ne[N], w[N], idx;
bool vis[N];
int dis[N];

void add (int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

int dijkstra () {
    memset (dis, 0x3f, sizeof dis);
    memset (vis, false, sizeof vis);
    dis[st] = 0;
    priority_queue <pii, vector <pii>, greater <pii>> q;
    q.push ({0, st});

    while (!q.empty ()) {
        auto t = q.top ();
        q.pop();
        int ver = t.second, dist = t.first;

        if (vis[ver])   continue;
        vis[ver] = true;

        for (int i = h[ver]; ~i; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dist + w[i]) {
                dis[j] = dist + w[i];
                q.push ({dis[j], j});
            }
        }
    }
    if (dis[ed] == 0x3f3f3f3f)  return -1;
    return dis[ed];
}

int main () {
    memset (h, -1, sizeof h);
    cin >> n >> st >> ed;
    for (int i = 1; i <= n; i ++) {
        int m;  cin >> m;
        for (int j = 0; j < m; j ++) {
            int x;  cin >> x;
            if (j == 0) add (i, x, 0);
            else    add (i, x, 1);
        }
    }
    cout << dijkstra () << endl;
}

//最短路
//建边,i到d[1]的权值为0

13. Nya图最短路

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 2e5 + 5, M = N << 2;
int h[N], e[M], ne[M], w[M], idx;
int dis[N];
bool vis[N];
int n, m, c;

void add (int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

int spfa () {
    memset (vis, false, sizeof vis);
    memset (dis, 0x3f, sizeof dis);
    dis[1] = 0, vis[1] = true;
    queue <int> q;
    q.push (1);

    while (!q.empty ()) {
        auto t = q.front ();
        q.pop();
        vis[t] = false;

        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dis[t] + w[i]) {
                dis[j] = dis[t] + w[i];
                if (!vis[j])    q.push (j), vis[j] = true;
            }
        }
    }
    if (dis[n] == 0x3f3f3f3f)   return -1;
    return dis[n];
}

int main () {
    int t;
    cin >> t;
    for (int _ = 1; _ <= t; _ ++) {
        cout<<"Case #"<<_<<": ";
        cin >> n >> m >> c;
        memset (h, -1, sizeof h), idx = 0;
        for (int i = 1; i <= n; i ++) {
            int x;  cin >> x;
            //更新阶梯
            add (x + n, i, 0);
            if (x < n)  add (i, x + n + 1, c);
            if (x > 1)  add (i, x + n - 1, c);
        }

        while (m --) {
            //更新额外边
            int a, b, c;
            cin >> a >> b >> c;
            add (a, b, c), add (b, a, c);
        }
        cout << spfa () << endl;
    }
}

14. 规划最短路

网络流,pass

15. 0或1

#include <bits/stdc++.h>

using namespace std;
const int N = 305;
int a[N][N], dis[N];
bool vis[N];
int n;

void spfa (int x) {
    queue <int> q;
    for (int i  =1; i <= n; i ++)
        dis[i] = a[x][i], vis[i] = true, q.push (i);
    dis[x] = 0x3f3f3f3f, vis[x] = false; //环,先把自己放飞，再通过后续更新回来

    while (!q.empty ()) {
        auto t = q.front ();
        q.pop ();
        vis[t] = false;

        for (int i = 1; i <= n; i ++) {
            if (dis[i] > dis[t] + a[t][i]) {
                dis[i] = dis[t] + a[t][i];
                if (!vis[i])    vis[i] = true, q.push (i);
            }
        }

    }
}

int main () {
    while (cin >> n) {
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= n; j ++)
                cin >> a[i][j];

        spfa (1);
        int d1 = dis[1], d2 = dis[n]; //1环,1-n之间的最短路
        spfa (n);
        int d3 = dis[n];
        cout << min (d1 + d3, d2) << endl;
    }
}

// 1. 是说1只有一个出度
// 2. n只有一个入度
// 3. 2~n-1之间,每一个数的出入度都相等

//关键在于抽象题意

//ans(取min):
//1出发的一个环和从n出发的环之和
//1-n之间的最短路

16. 排队布局

#include <bits/stdc++.h>

using namespace std;
const int N = 1005, M = N + 2e4;
int n, L, D;
int dis[N], cnt[N];
bool vis[N];
int h[N], e[M], ne[M], w[M], idx;

void add (int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}

bool spfa (int x) {
    queue <int> q;
    memset (dis, 0x3f, sizeof dis);
    memset (cnt, 0, sizeof cnt);
    memset (vis, false, sizeof vis);

    for (int i = 1; i <= x; i ++) //x
        q.push (i), vis[i] = true, dis[i] = 0;

    while (!q.empty()) {
        int t = q.front();
        q.pop();
        vis[t] = false;

        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dis[t] +w[i]) {
                dis[j] = dis[t] + w[i];
                cnt[j] = cnt[t] + 1;

                if (cnt[j] >= n)
                    return false;

                if (!vis[j])
                    q.push (j), vis[j] = true;
            }
        }
    }
    return true;
}

int main () {
    memset (h, -1, sizeof h);
    cin >> n >> L >> D;
    for (int i = 1; i < n; i ++)
        add (i + 1, i, 0);
    while (L --) {
        int a, b, c;
        cin >> a >> b >> c;
        if (a > b)
            swap (a, b);
        add (a, b, c); //注意最短路的符号是相反的
    }
    while (D --) {
        int a, b, c;
        cin >> a >> b >> c;
        if (a > b)
            swap (a, b);
        add (b, a, -c);
    }
    if (!spfa(n)) 
        puts("-1");
    else {
        spfa(1);
        if (dis[n] == 0x3f3f3f3f )   
            puts("-2");
        else 
            cout << dis[n] << endl;
    }
}

标签：专题,idx,int,短路,cin,vis,add,kuangbin,dis
来源： https://www.cnblogs.com/CTing/p/16457806.html

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