
题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=677
题意转化:将点0与所有的有间谍的点相连,则题意变为求点0到点n的最小割,直接套最大流EK算法~
下面代码顶点是从1~n+1
代码入下:
#include "stdio.h"
#include "string.h"
#include "queue"
using namespace std; #define N 205
#define INF 0x3fffffff int start,end;
int route[N],dis[N];
int map[N][N]; int MIN(int x,int y){ return x<y?x:y; } int BFS()
{
int i;
int x,y;
memset(route,-1,sizeof(route));
dis[start] = INF;
queue<int> q;
q.push(start);
while(!q.empty())
{
x = q.front();
q.pop();
for(y=1; y<=end; ++y)
{
if(route[y]==-1 && map[x][y]!=0)
{
route[y] = x;
dis[y] = MIN(dis[x],map[x][y]);
q.push(y);
}
}
}
if(route[end]==-1) return 0;
return dis[end];
} int EK(int n)
{
int x,y;
int ans=0;
int kejia;
while(kejia = BFS())
{
ans += kejia;
y = end;
while(y!=start)
{
x = route[y];
map[x][y] -= kejia;
map[y][x] += kejia;
y = x;
}
}
return ans;
} int main()
{
int T,Case;
int i,k;
int n,m,p;
int x,y;
scanf("%d",&T);
for(Case=1; Case<=T; ++Case)
{
memset(map,0,sizeof(map));
scanf("%d%d%d",&n,&m,&p);
for(i=0; i<p; ++i)
{
scanf("%d",&k);
k++;
map[1][k] = map[k][1] = INF;
}
while(m--)
{
scanf("%d %d",&x,&y);
x++; y++;
map[x][y] = map[y][x] = 1;
}
start = 1; end = n+1;
printf("Case #%d: %d\n",Case,EK(end));
}
}