hdu 3861 The King’s Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3289    Accepted Submission(s): 1165

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

 

 

Tarjan缩点+匈牙利算法

Tarjan写错了。。

屠龙宝刀点击就送

#include <cstring>
#include <cstdio>
#define N 100005

struct Edge
{
    Edge *next;
    int to;
}*head[N],edge[N];
struct EDge
{
    EDge *next;
    int to;
}*newhead[N],newedge[N];
bool instack[N],vis[N];
int cnt,T,n,m,ans,stack[N],top,low[N],dfn[N],tim,col[N],sumcol,f[N];
inline void init()
{
    ans=top=tim=sumcol=cnt=0;
    memset(f,-1,sizeof(f));
    memset(col,0,sizeof(col));
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(head,0,sizeof(head));
    memset(edge,0,sizeof(edge));
    memset(newhead,0,sizeof(newhead));
    memset(newedge,0,sizeof(newedge));
}
struct node
{
    int x,y;
}e[N<<10];
inline int min(int a,int b) {return a>b?b:a;}
void tarjan(int x)
{
    low[x]=dfn[x]=++tim;
    instack[x]=1;
    stack[++top]=x;
    for(Edge * u=head[x];u;u=u->next)
    {
        int v=u->to;
        if(instack[v]) low[x]=min(low[x],dfn[v]);
        else if(!dfn[v])
        {
            tarjan(v);
            low[x]=min(low[x],low[v]);
        }
    }
    if(low[x]==dfn[x])
    {
        int k;
        sumcol++;
        do
        {
            k=stack[top--];
            instack[k]=false;
            col[k]=sumcol;
        }while(k!=x);
    }
}
bool dfs(int x)
{
    for(EDge * u=newhead[x];u;u=u->next)
    {
        int v=u->to;
        if(!vis[v])
        {
            vis[v]=1;
            if(f[v]==-1||dfs(f[v]))
            {
                f[v]=x;
                return 1;
            }
        }
    }
    return 0;
}
inline void ins(int u,int v)
{
    edge[++cnt].next=head[u];
    edge[cnt].to=v;
    head[u]=edge+cnt;
}
inline void insnew(int u,int v)
{
    newedge[++cnt].next=newhead[u];
    newedge[cnt].to=v;
    newhead[u]=newedge+cnt;
}
int Main()
{
    scanf("%d",&T);
    for(;T--;)
    {
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=m;++i)
        {
            scanf("%d%d",&e[i].x,&e[i].y);
            ins(e[i].x,e[i].y);
        }
        for(int i=1;i<=n;++i)
         if(!dfn[i]) tarjan(i);
        cnt=0;
        for(int i=1;i<=m;++i)
        {
            int cx=col[e[i].x],cy=col[e[i].y];
            if(cx!=cy) insnew(cx,cy);
        }
        for(int i=1;i<=sumcol;++i)
        {
            memset(vis,0,sizeof(vis));
            if(dfs(i)) ans++;
        }
        printf("%d
",sumcol-ans);
    }
    return 0;
}
int sb=Main();
int main(int argc,char *argv[]) {;}