The King’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4080    Accepted Submission(s): 1430

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 題目意思：
現在有n個點，m條邊的有向圖，要求劃分的區域最少
規則如下：
1.可以互相到達的點必須屬於一個區域
2.u可以到v或者v可以到v，即一個區域內任意兩點u,v,必須存在路徑從u->v或者從v->u
3.一個點只能屬於一個區域
4.所有點都應該被劃分 分析：
可以互相到達的點肯定是屬於一個強連通分量的，所以利用tarjan將屬於同一個強連通分量的點縮成一個點
得到新圖，現在新圖是一個DAG圖，有向無環圖 最小路徑的定義：在一個有向圖中，找出最少的路徑，使得這些路徑經過了所有的點 對照一下題目：一個區域其實就是一條路徑
最少的區域數目就是最少的路徑數目
所以題目轉換成最小不相交的路徑覆蓋，注意：不相交的路徑，疑問一個點只能屬於一個區域 最小不相交路徑覆蓋=點數-最大二分匹配
所以對得到的新圖，求一遍最大二分匹配就好
最大二分匹配用匈牙利演算法寫

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<math.h>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
#define INF 0x7fffffff
#define mem(a,x) memset(a,x,sizeof(a))
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};

int getval()
{
    int ret(0);
    char c;
    while((c=getchar())==' '||c=='\n'||c=='\r');
    ret=c-'0';
    while((c=getchar())!=' '&&c!='\n'&&c!='\r')
        ret=ret*10+c-'0';
    return ret;
}

#define max_v 5005
int dfn[max_v];
int low[max_v];
int vis[max_v];
int stk[max_v];
int color[max_v];
vector<int> G[max_v];
vector<int> G2[max_v];
int n,m;
int sig,cnt,sp;

int link[max_v];
int match[max_v];
void init()
{
    mem(dfn,0);
    mem(low,0);
    mem(vis,0);
    mem(stk,0);
    mem(color,0);
    for(int i=1;i<=n;i++)
    {
        G[i].clear();
        G2[i].clear();
    }
    sig=0;
    cnt=1;
    sp=-1;
}

int tarjan(int u)
{
    vis[u]=1;
    low[u]=dfn[u]=cnt++;
    stk[++sp]=u;
    for(int j=0;j<G[u].size();j++)
    {
        int v=G[u][j];
        if(vis[v]==0)
            tarjan(v);
        if(vis[v]==1)
            low[u]=min(low[u],low[v]);
    }
    if(low[u]==dfn[u])
    {
        sig++;
        do
        {
            color[stk[sp]]=sig;
            vis[stk[sp]]=-1;
        }while(stk[sp--]!=u);
    }
}

int dfs(int u)
{
    for(int j=0;j<G2[u].size();j++)
    {
        int v=G2[u][j];
        if(vis[v]==0)
        {
            vis[v]=1;
            if(match[v]==-1||dfs(match[v]))
            {
                match[v]=u;
                return 1;
            }
        }
    }
    return 0;
}

int max_match()//匈牙利演算法
{
    mem(match,-1);
    int ans=0;
    for(int i=1;i<=sig;i++)
    {
        mem(vis,0);
        if(dfs(i))
            ans++;
    }
    return ans;
}

int main()
{
    int t;
    cin>>t;
    int x,y;
    while(t--)
    {
        scanf("%d %d",&n,&m);
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&x,&y);
            if(count(G[x].begin(),G[x].end(),y)==0)//重邊
                G[x].push_back(y);
        }
        for(int i=1;i<=n;i++)
        {
            if(vis[i]==0)
                tarjan(i);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<G[i].size();j++)
            {
                if(color[i]!=color[G[i][j]])
                {
                    if(count(G2[color[i]].begin(),G2[color[i]].end(),color[G[i][j]])==0)//重邊
                        G2[color[i]].push_back(color[G[i][j]]);
                }
            }
        }
        printf("%d\n",sig-max_match());//最小不相交路徑覆蓋=新圖點數-最大二分匹配數
    }
    return 0;
}
/*
題目意思：
現在有n個點，m條邊的有向圖，要求劃分的區域最少
規則如下：
1.可以互相到達的點必須屬於一個區域
2.u可以到v或者v可以到v，即一個區域內任意兩點u,v,必須存在路徑從u->v或者從v->u
3.一個點只能屬於一個區域
4.所有點都應該被劃分

分析：
可以互相到達的點肯定是屬於一個強連通分量的，所以利用tarjan將屬於同一個強連通分量的點縮成一個點
得到新圖，現在新圖是一個DAG圖，有向無環圖

最小路徑的定義：在一個有向圖中，找出最少的路徑，使得這些路徑經過了所有的點

對照一下題目：一個區域其實就是一條路徑
最少的區域數目就是最少的路徑數目
所以題目轉換成最小不相交的路徑覆蓋，注意：不相交的路徑，疑問一個點只能屬於一個區域

最小不相交路徑覆蓋=點數-最大二分匹配
所以對得到的新圖，求一遍最大二分匹配就好
最大二分匹配用匈牙利演算法寫

gameover!

*/