The King’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

1
3 2
1 2
1 3

Sample Output

2

Source

2011 Multi-University Training Contest 3 - Host
by BIT

Recommend

lcy

题意：n个城市m条有向边，把这些城市分成若干个州，分的原则是（1）u和v可以互相到达的话他们两个必须在同一个州（2）同一个州里任意两个城市u和v要满足u可以到达v或者v可以到达u。问州的最小个数是多少。

思路：先用Tarjan算法进行缩点，在缩点后的图上进行二分图匹配，最后求得最小路径覆盖=强连通个数-最大匹配数。

可以看一下：

http://blog.csdn.net/hellobabygogo3/article/details/7900812

http://www.cnblogs.com/ka200812/archive/2011/07/31/2122641.html

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 5005;
const int MAXM = 100010;

int n,m;

struct Edge
{
    int to,next;
}edge[MAXM],e[MAXM];

int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Belong[MAXN],Stack[MAXN];
int Index,scc,top;
bool Inq[MAXN];

int uN,vN;
int linker[MAXN];
bool used[MAXN];
int hed[MAXN],num;

void init()
{
    tot=0;num=0;
    memset(hed,-1,sizeof(hed));
    memset(head,-1,sizeof(head));
}

void add(int u,int v)
{
    e[num].to=v;
    e[num].next=hed[u];
    hed[u]=num++;
}

void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Inq[u]=true;
    for (int i=head[u];i+1;i=edge[i].next)
    {
        v=edge[i].to;
        if (!DFN[v])
        {
            Tarjan(v);
            if (Low[u]>Low[v]) Low[u]=Low[v];
        }
        else if (Inq[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if (Low[u]==DFN[u])
    {
        scc++;
        do{
            v=Stack[--top];
            Inq[v]=false;
            Belong[v]=scc;
        }while (v!=u);
    }
}
void solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Inq,false,sizeof(Inq));
    Index=scc=top=0;
    for (int i=1;i<=N;i++)
        if (!DFN[i])
            Tarjan(i);
    for (int u=1;u<=N;u++)
    {
        for (int i=head[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if (Belong[u]==Belong[v]) continue;
            add(Belong[u],Belong[v]);
        }
    }
    uN=vN=scc;
}

bool dfs(int u)
{
    for (int i=hed[u];~i;i=e[i].next)
    {
        int v=e[i].to;
        if (!used[v])
        {
            used[v]=true;
            if (linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int ans=0;
    memset(linker,-1,sizeof(linker));
    for (int i=1;i<=uN;i++)
    {
        memset(used,false,sizeof(used));
        if (dfs(i)) ans++;
        if (ans==uN) break;
    }
    return ans;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,t,u,v;
    sf(t);
    while (t--)
    {
        sff(n,m);
        init();
        for (i=0;i<m;i++)
        {
            sff(u,v);
            addedge(u,v);
        }
        solve(n);
        int ans=hungary();
        printf("scc=%d \n",scc);
        printf("%d\n",scc-ans);
    }
    return 0;
}

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