HDU 3861 Tarjan + 缩点 + 最小路径覆盖

The King’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2685    Accepted Submission(s): 978

Problem Description

In the Kingdom of Silence, the king has a new problem. There are N cities in the kingdom and there are M directional roads between the cities. That means that if there is a road from u to v, you can only go from city u to city v, but can’t go from city v to city u. In order to rule his kingdom more effectively, the king want to divide his kingdom into several states, and each city must belong to exactly one state. What’s more, for each pair of city (u, v), if there is one way to go from u to v and go from v to u, (u, v) have to belong to a same state. And the king must insure that in each state we can ether go from u to v or go from v to u between every pair of cities (u, v) without passing any city which belongs to other state.
  Now the king asks for your help, he wants to know the least number of states he have to divide the kingdom into.

 

Input

The first line contains a single integer T, the number of test cases. And then followed T cases. 

The first line for each case contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the number of cities and roads in the kingdom. The next m lines each contains two integers u and v (1 <= u, v <= n), indicating that there is a road going from city u to city v.

 

Output

The output should contain T lines. For each test case you should just output an integer which is the least number of states the king have to divide into.

 

Sample Input

13 21 21 3

 

Sample Output

2

 

题意是将一些点划分区域，同时有两个规定：

1.若有u,v两个点，u->v且v->u 即n,v两点可以互相到达形成环，则一定分在同一区域

思路：Tarjan求强连通分量然后缩点。

2.在同一区域的任意两点至少存在一条路径可以相互到达，即（设同一区域两点u,v）有u->v或 v->u。

思路：二分图，很明显是最小路径覆盖，缩点后建新图，跑个匈牙利得到最大匹配 ans，结果就为： 缩点后的点数 num 减去 ans。

代码：

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int INF = 1e8;const int maxn = 5010;vector<int> G[maxn],G2[maxn];int low[maxn],dfn[maxn]; int vis[maxn],instack[maxn],point[maxn],match[maxn];int n,tot,num;stack<int> S;void init(void){    tot = num = 0;    for(int i=0 ;i<=n ;i++){        G[i].clear();        G2[i].clear();        match[i] = -1;        low[i] = dfn[i] = 0;        vis[i] = instack[i] = point[i] = 0;    }    while(S.size())    S.pop();}void Tarjan(int x){    low[x] = dfn[x] = tot++;    vis[x] = instack[x] = 1;    S.push(x);    for(int i=0 ;i<G[x].size();i++){        int v = G[x][i];        if(!vis[v]){            Tarjan(v);            low[x] = min(low[x],low[v]);        }        else if(instack[v]){            low[x] = min(low[x],dfn[v]);        }    }    if(low[x] == dfn[x]){        while(1){            int t = S.top();            S.pop();            instack[t] = 0;            point[t] = num;             if(t == x)    break;        }        num++;    }}bool find(int x){for(int i=0 ;i<G2[x].size() ;i++){int t = G2[x][i];if(!vis[t]){vis[t] = 1;if(match[t] == -1 || find(match[t])){match[t] = x;return true;}}}return false;}int main(){   int T;   scanf("%d",&T);   while(T--){int m;scanf("%d%d",&n,&m);init();   while(m--){   int x,y;scanf("%d%d",&x,&y);G[x].push_back(y);}for(int i=1 ;i<=n ;i++){if(!vis[i]){Tarjan(i);}}for(int i=1 ;i<=n ;i++){for(int j=0 ;j<G[i].size() ;j++){if(point[i] != point[G[i][j]]){G2[point[i]].push_back(point[G[i][j]]);}}}int ans = 0;for(int i=0 ;i<num ;i++){memset(vis,0,sizeof(vis));if(find(i))ans++;}cout << num-ans << endl;}    return 0;}