【最小路径覆盖+注意】POJ 2594

这题要注意一点，题中说了You should notice that the roads of two different robots may contain some same point. 就是要注意这个细节/article/6002635.html ，要用一次传递闭包floyd找出所有通路

#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <string>
#include <cstdio>
#include <math.h>
#include <iomanip>
#include <cstdlib>
#include <limits.h>
#include <string.h>
#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;

#define LL long long
#define MIN -99999999
#define MAX 99999999
#define pii pair<int ,int>

#define bug cout<<"here!!"<<endl
#define PI acos(-1.0)
#define FRE freopen("input.txt","r",stdin)
#define FF freopen("output.txt","w",stdout)
#define eps 1e-8
#define N 505
bool g

;
int match
;
bool vis
;
int n;
void floyd(){
int i,j,k;
for(k=1;k<=n;k++){
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(g[i][k] && g[k][j])
g[i][j] = true;
}
}
}
}
bool sear(int s){
int i;
for(i=1;i<=n;i++){
if(g[i][s] && !vis[i]){
vis[i] = 1;
if(match[i]==0 || sear(match[i])){
match[i] = s;
return true;
}
}
}
return false;
}

int main(){
int m;
while(scanf("%d%d",&n,&m) && (n+m)){
int i,j;
memset(g,0,sizeof(g));
memset(match,0,sizeof(match));
while(m--){
int a,b;
scanf("%d%d",&a,&b);
g[a][b] = true;
}
int cnt = 0;
floyd();
for(i=1;i<=n;i++){
memset(vis,0,sizeof(vis));
if(sear(i))cnt++;
}
printf("%d\n",n-cnt);
}
return 0;
}