题意:
给了一个图无向联通图..国王还要划分州..若在划分州前两点v,u有路径(v,u)、(u,v)那么他们必须要在同一个州中...并且建立了州以后...一个区域内的任意两点至少要有单向路径..问最少建立多少个州可以满足要求...
题解:
路径覆盖所有的点且任何的点只于一条选取的路径有关联....有向图最小路径覆盖=缩点后的顶点数-二分图最大匹配....
Program:
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<stack>
#include<string.h>
#include<queue>
#define ll long long
#define esp 1e-5
#define MAXN 5006
#define MAXM 100005
#define oo 100000007
using namespace std;
struct node
{
int x,y,next;
}line[MAXM];
int Lnum,_next[MAXN],dfn[MAXN],low[MAXN],tp[MAXN],tpnum,DfsIndex,match[MAXN];
bool instack[MAXN],used[MAXN];
stack<int> mystack;
void addline(int x,int y)
{
line[++Lnum].next=_next[x],_next[x]=Lnum;
line[Lnum].x=x,line[Lnum].y=y;
}
void tarjan(int x)
{
mystack.push(x),instack[x]=true;
dfn[x]=low[x]=++DfsIndex;
for (int k=_next[x];k;k=line[k].next)
{
int y=line[k].y;
if (!dfn[y])
{
tarjan(y);
low[x]=min(low[x],low[y]);
}else
if (instack[y])
low[x]=min(low[x],dfn[y]);
}
if (low[x]==dfn[x])
{
tpnum++;
do
{
x=mystack.top();
mystack.pop();
instack[x]=false;
tp[x]=tpnum;
}while (low[x]!=dfn[x]);
}
}
bool dfs(int x)
{
for (int k=_next[x];k;k=line[k].next)
{
int y=line[k].y;
if (used[y]) continue;
used[y]=true;
if (!match[y] || dfs(match[y]))
{
match[y]=x;
return true;
}
}
return false;
}
int getmax(int n)
{
int sum=0;
memset(match,0,sizeof(match));
for (int i=1;i<=n;i++)
{
memset(used,false,sizeof(used));
sum+=dfs(i);
}
return sum;
}
int main()
{
int i,n,m,cases;
scanf("%d",&cases);
while (cases--)
{
scanf("%d%d",&n,&m);
Lnum=0,memset(_next,0,sizeof(_next));
while (m--)
{
int x,y;
scanf("%d%d",&x,&y);
addline(x,y);
}
memset(dfn,0,sizeof(dfn));
while (!mystack.empty()) mystack.pop();
memset(instack,false,sizeof(instack));
DfsIndex=tpnum=0;
for (i=1;i<=n;i++)
if (!dfn[i]) tarjan(i);
int temp=Lnum;
Lnum=0,memset(_next,0,sizeof(_next));
for (i=1;i<=temp;i++)
{
int x=tp[line[i].x],y=tp[line[i].y];
if (x==y) continue;
addline(x,y);
}
n=tpnum;
printf("%d\n",n-getmax(n));
}
return 0;
}
