标签：Real cnt ch int System ++ dfn Traffic low
Problem Description City C is really a nightmare of all drivers for its traffic jams. To solve the traffic problem, the mayor plans to build a RTQS (Real Time Query System) to monitor all traffic situations. City C is made up of N crossings and M roads, and each road connects two crossings. All roads are bidirectional. One of the important tasks of RTQS is to answer some queries about route-choice problem. Specifically, the task is to find the crossings which a driver MUST pass when he is driving from one given road to another given road.Input There are multiple test cases.
For each test case:
The first line contains two integers N and M, representing the number of the crossings and roads.
The next M lines describe the roads. In those M lines, the ith line (i starts from 1)contains two integers Xi and Yi, representing that roadi connects crossing Xi and Yi (Xi≠Yi).
The following line contains a single integer Q, representing the number of RTQs.
Then Q lines follows, each describing a RTQ by two integers S and T(S≠T) meaning that a driver is now driving on the roads and he wants to reach roadt . It will be always at least one way from roads to roadt.
The input ends with a line of “0 0”.
Please note that: 0<N<=10000, 0<M<=100000, 0<Q<=10000, 0<Xi,Yi<=N, 0<S,T<=M
Output For each RTQ prints a line containing a single integer representing the number of crossings which the driver MUST pass. Sample Input 5 6 1 2 1 3 2 3 3 4 4 5 3 5 2 2 3 2 4 0 0 Sample Output 0 1 ——————————————————————————————————————————————————————————————————
给你一个无向图，询问边a和边b，问从边a到边b的路径中几个点是必须要经过的。
题解
求无向图的必经点，跟上题类似，求出点双后缩点，然后转化成查询路径上割点的个数。由于点双缩点的特殊性，所以算距离的时候写的不一样。
后面就是代码了：
1 #include<iostream>
2 #include<cstring>
3 #include<queue>
4 #define rg register
5 #define il inline
6 #define co const
7 template<class T>il T read(){
8 rg T data=0,w=1;rg char ch=getchar();
9 for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
10 for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
11 return data*w;
12 }
13 template<class T>il T read(rg T&x) {return x=read<T>();}
14 typedef long long ll;
15 using namespace std;
16
17 co int N=2e4+1,M=2e5+2;
18 int n,m,t;
19 int head[N],ver[M],next[M],tot;
20 int dfn[N],low[N],stack[N],new_id[N],c[N],belong[M],num,root,top,cnt;
21 int d[N],dist[N],f[N][16];
22 bool cut[N];
23 vector<int> dcc[N];
24 int hc[N],vc[M],nc[M],tc;
25 queue<int> q;
26
27 il void add(int x,int y){
28 ver[++tot]=y,::next[tot]=head[x],head[x]=tot;
29 }
30 il void add_c(int x,int y){
31 vc[++tc]=y,nc[tc]=hc[x],hc[x]=tc;
32 }
33 void tarjan(int x){
34 dfn[x]=low[x]=++num;
35 stack[++top]=x;
36 if(x==root&&head[x]==0){
37 dcc[++cnt].push_back(x);
38 return;
39 }
40 int flag=0;
41 for(int i=head[x];i;i=::next[i]){
42 int y=ver[i];
43 if(!dfn[y]){
44 tarjan(y);
45 low[x]=min(low[x],low[y]);
46 if(low[y]>=dfn[x]){
47 ++flag;
48 if(x!=root||flag>1) cut[x]=1;
49 ++cnt;
50 int z;
51 do{
52 z=stack[top--];
53 dcc[cnt].push_back(z);
54 }while(z!=y);
55 dcc[cnt].push_back(x);
56 }
57 }
58 else low[x]=min(low[x],dfn[y]);
59 }
60 }
61 void bfs(int s){
62 d[s]=1,dist[s]=s>cnt,q.push(s);
63 for(int i=0;i<16;++i) f[s][i]=0;
64 while(q.size()){
65 int x=q.front();q.pop();
66 for(int i=hc[x];i;i=nc[i]){
67 int y=vc[i];
68 if(d[y]) continue;
69 d[y]=d[x]+1,dist[y]=dist[x]+(y>cnt),f[y][0]=x;
70 for(int j=1;j<16;++j) f[y][j]=f[f[y][j-1]][j-1];
71 q.push(y);
72 }
73 }
74 }
75 int lca(int x,int y){
76 if(d[x]<d[y]) swap(x,y);
77 for(int i=15;i>=0;--i)
78 if(d[f[x][i]]>=d[y]) x=f[x][i];
79 if(x==y) return x;
80 for(int i=15;i>=0;--i)
81 if(f[x][i]!=f[y][i]) x=f[x][i],y=f[y][i];
82 return f[x][0];
83 }
84 int main(){
85 while(read(n)|read(m)){
86 memset(head,0,sizeof head);
87 memset(hc,0,sizeof hc);
88 memset(dfn,0,sizeof dfn);
89 memset(d,0,sizeof d);
90 memset(cut,0,sizeof cut);
91 memset(c,0,sizeof c);
92 for(int i=1;i<=n;++i) dcc[i].clear();
93 tot=1,num=cnt=top=0;
94 for(int x,y;m--;){
95 read(x),read(y);
96 add(x,y),add(y,x);
97 }
98 for(int i=1;i<=n;++i)
99 if(!dfn[i]) root=i,tarjan(i);
100 num=cnt;
101 for(int i=1;i<=n;++i)
102 if(cut[i]) new_id[i]=++num;
103 tc=1;
104 for(int i=1;i<=cnt;++i){
105 for(int j=0;j<dcc[i].size();++j){
106 int x=dcc[i][j];
107 if(cut[x]) add_c(i,new_id[x]),add_c(new_id[x],i);
108 c[x]=i;
109 }
110 for(int j=0;j<dcc[i].size()-1;++j){
111 int x=dcc[i][j];
112 for(int k=head[x];k;k=::next[k]){
113 int y=ver[k];
114 if(c[y]==i) belong[k/2]=i; // 输入的第k/2条边(x,y)处于第i个v-DCC内
115 }
116 }
117 }
118 // 编号 1~cnt 的为原图的v-DCC，编号 cnt+1~num 的为原图割点
119 for(int i=1;i<=num;++i)
120 if(!d[i]) bfs(i);
121 read(t);
122 while(t--){
123 int es=read<int>(),et=read<int>();
124 int x=belong[es],y=belong[et],z=lca(x,y);
125 printf("%d\n",dist[x]+dist[y]-2*dist[z]+(z>cnt));
126 }
127 }
128 return 0
标签：Real,cnt,ch,int,System,++,dfn,Traffic,low 来源： https://www.cnblogs.com/DZN2004/p/12842540.html
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