HDU 3861 The King’s Problem

http://acm.hdu.edu.cn/showproblem.php?pid=3861

The King’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1828    Accepted Submission(s): 664

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

 

Source

2011 Multi-University Training Contest 3 - Host by BIT

 

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题目大意：将n个点组成的有向图划分成最少的集合数，要求任意两个相互到达的点在一个集合中，且属于同一个集合的任意两个点对(u,v)至少存在一条路径使得v对于u可达，或者u对于v可达。

解题思路：首先tarjan缩点，然后重建图求最小路径覆盖即为答案

最小路径覆盖=顶点数-最大匹配数。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn 5000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b)
{
    return a>b;
}
vector<int>G[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
stack<int>S;
int n;
void dfs(int u)
{
pre[u]=lowlink[u]=++dfs_clock;
S.push(u);
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(!pre[v])
{
dfs(v);
lowlink[u]=min(lowlink[u],lowlink[v]);
}
else if(!sccno[v])lowlink[u]=min(lowlink[u],pre[v]);
}
if(lowlink[u]==pre[u])
{
scc_cnt++;
for(;;)
{
int x=S.top();S.pop();
sccno[x]=scc_cnt;
if(x==u)break;
}
}
}
void find_scc(int n)
{
dfs_clock=scc_cnt=0;
cle(sccno),cle(pre);
for(int i=1;i<=n;i++)
if(!pre[i])dfs(i);
}

vector<int>g[maxn];
int match[maxn];
int vis[maxn];
int tot;
int dfs2(int x)
{
for(int i=0;i<g[x].size();i++)
if(!vis[g[x][i]])
{
vis[g[x][i]]=1;
if(match[g[x][i]]==-1||dfs2(match[g[x][i]]))
{
match[g[x][i]]=x;
return 1;
}
}
return 0;
}
int hungary()
{
tot=0;
memset(match,-1,sizeof match);
for(int i=1;i<=n;i++)
{
cle(vis);
if(dfs2(i))tot++;
}
return scc_cnt-tot;
}
void solve()
{
find_scc(n);
for(int i=1;i<maxn;i++)
g[i].clear();
//用scc构建图
for(int i=1;i<=n;i++)
for(int j=0;j<G[i].size();j++)
{
int k=G[i][j];
if(sccno[i]==sccno[k])continue;
g[sccno[i]].push_back(sccno[k]);
}
printf("%d\n",hungary());
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int m,t;
    int a,b;
    scanf("%d",&t);
    while(t--)
{
scanf("%d%d",&n,&m);
for(int i=1;i<maxn;i++)
G[i].clear();
for(int i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
G[a].push_back(b);
}
solve();
}
    return 0;
}