标签：24 可重 P3357 int define vis maxn include dis
费用流好题，k重区间plus版本
https://www.luogu.com.cn/problem/P3357
从 \(k\) 重区间变成 \(k\) 重线段，仍旧只看 \(x\) 轴的重叠。
这样就多出来一个问题，对于平行于 \(x\) 轴的线段，还向上一题那样连边会变成自环，于是T了两个点。考虑拆点，每个点拆成 \(2n\) 和 \(2n+1\). 剩下的正常连就行了~
点击查看代码
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <vector>
typedef long long ll;
#define endl '\n'
#define IOS \
ios::sync_with_stdio(0); \
cin.tie(0); \
cout.tie(0)
#define P pair<int, int>
#define endl '\n'
using namespace std;
// #define int long long
const int maxn = 500 * 2 + 10;
const int inf = 0x3f3f3f3f;
int n1, n2, cnt_edge = 1, S, T;
int head[maxn];
int dis[maxn];
bool vis[maxn];
struct edge {
int to, nxt;
int flow, cost;
} e[(maxn * maxn) << 2];
inline void add(int u, int v, int w, int c) {
e[++cnt_edge].nxt = head[u];
head[u] = cnt_edge;
e[cnt_edge].to = v;
e[cnt_edge].flow = w;
e[cnt_edge].cost = c;
}
inline void addflow(int u, int v, int w, int c) {
add(u, v, w, c);
add(v, u, 0, -c);
}
inline bool spfa(int on) {
memset(vis, 0, sizeof(vis));
if (on == 1)
for (int i = 0; i <= T; i++) dis[i] = inf;
else
for (int i = 0; i <= T; i++) dis[i] = -inf;
queue<int> q;
q.push(S);
dis[S] = 0;
vis[S] = 1;
while (!q.empty()) {
int x = q.front();
q.pop();
vis[x] = 0;
for (int i = head[x]; i; i = e[i].nxt) {
int y = e[i].to;
// cout << "->" << y << endl;
if ((on == 1 && e[i].flow && dis[y] > dis[x] + e[i].cost) ||
(on == -1 && e[i].flow && dis[y] < dis[x] + e[i].cost)) {
dis[y] = dis[x] + e[i].cost;
if (!vis[y]) q.push(y), vis[y] = 1;
}
}
}
if (on == 1)
return dis[T] != inf;
else
return dis[T] != -inf;
}
ll dfs(int x, ll lim) {
vis[x] = 1;
if (x == T || lim <= 0) return lim;
int res = lim;
for (int i = head[x]; i; i = e[i].nxt) {
int y = e[i].to;
if (dis[y] != dis[x] + e[i].cost || e[i].flow <= 0 || vis[y]) continue;
ll tmp = dfs(y, min(res, e[i].flow));
res -= tmp;
e[i].flow -= tmp;
e[i ^ 1].flow += tmp;
if (res <= 0) break;
}
return lim - res;
}
inline ll Dinic(int on) {
ll res = 0, cost = 0;
while (spfa(on)) {
ll flow = dfs(S, inf);
res += flow, cost += flow * dis[T];
}
return cost;
}
int nn, k;
int lx[maxn], rx[maxn], ly[maxn], ry[maxn], a[maxn << 1], len[maxn];
int cdis(int i) {
return int(sqrt(1ll * (lx[i] - rx[i]) * (lx[i] - rx[i]) +
1ll * (ly[i] - ry[i]) * (ly[i] - ry[i])));
}
signed main() {
// IOS;
cin >> nn >> k;
for (int i = 1; i <= nn; i++) {
cin >> lx[i] >> ly[i] >> rx[i] >> ry[i];
if (lx[i] > rx[i])
swap(lx[i], rx[i]), swap(ly[i], ry[i]);
len[i] = cdis(i);
lx[i] <<= 1, rx[i] <<= 1;
if (lx[i] == rx[i])
rx[i]++;
else lx[i]++;
a[i] = lx[i], a[i + nn] = rx[i];
}
sort(a + 1, a + nn * 2 + 1);
int n = unique(a + 1, 1 + a + nn * 2) - a - 1;
S = n + 1, T = S + 1;
for (int i = 1; i <= nn; i++) {
int L = lower_bound(a + 1, a + 1 + n, lx[i]) - a;
int R = lower_bound(a + 1, a + 1 + n, rx[i]) - a;
addflow(L, R, 1, len[i]);
}
for (int i = 1; i < n; i++) {
addflow(i, i + 1, inf, 0);
}
addflow(S, 1, k, 0);
addflow(n, T, k, 0);
// cout << "build c\n";
cout << Dinic(-1) << endl;
return 0;
}
标签：24,可重,P3357,int,define,vis,maxn,include,dis 来源： https://www.cnblogs.com/FushimiYuki/p/15435250.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。