题目大意：有n张幻灯片和n个数字，幻灯片放置有重叠，每个数字隶属于一个幻灯片，现在问你能够确定多少数字一定属于某个幻灯片

思路：上次刷过二分图的必须点后这题思路就显然了 做一次二分匹配后将当前匹配的边删除，再做一次二分匹配，如果能匹配到那么说明这条边不是必须边

顺便说下 删边以后不用从头开始二分匹配，而是在原来二分匹配的基础上对这个点进行增广就行，另外这题格式需要注意，很容易PE

#include<cstdio>

#include<string.h>

#include<iostream>

#include<algorithm>

#include<math.h>

#define maxn 10000

#define MOD 1000000007

using namespace std;

int head[maxn],next[maxn],point[maxn],now;

int match[maxn],x1[maxn],x2[maxn],y11[maxn];

int y2[maxn],an[maxn],cop[maxn],ann[maxn];

int rematch[maxn],ans[maxn];

bool visit[maxn];

void add(int x,int y)

{

next[++now]=head[x];

head[x]=now;

point[now]=y;

}

int dfs(int k,int forbid)

{

for(int i=head[k];i;i=next[i])if(!visit[point[i]])

if(i!=forbid)

{

int u=point[i];

visit[u]=1;

if(match[u]==-1||dfs(match[u],forbid))

{

match[u]=k;

an[k]=i;

return 1;

}

}

return 0;

}

int main()

{

int n,x,y,cas=0;

while(1)

{

now=0;

memset(head,0,sizeof(head));

memset(ans,0,sizeof(ans));

scanf("%d",&n);

if(n==0)break;

for(int i=1;i<=n;i++)

scanf("%d%d%d%d",&x1[i],&x2[i],&y11[i],&y2[i]);

for(int i=1;i<=n;i++)

{

scanf("%d%d",&x,&y);

for(int j=1;j<=n;j++)

if(x1[j]<=x&&x<=x2[j]&&y11[j]<=y&&y<=y2[j])

{

add(i,j);

//printf("%d %d\n",i,j);

}

}

memset(match,-1,sizeof(match));

memset(an,-1,sizeof(an));

for(int i=1;i<=n;i++)

{

memset(visit,0,sizeof(visit));

dfs(i,-1);

}

//for(int i=1;i<=n;i++)printf("%d  ",match[i]);

for(int i=1;i<=n;i++)rematch[match[i]]=i;

memcpy(cop,match,sizeof(match));

memcpy(ann,an,sizeof(an));

for(int i=1;i<=n;i++)

{

memset(visit,0,sizeof(visit));

match[rematch[i]]=-1;

if(!dfs(i,an[i]))ans[rematch[i]]=i;else ans[rematch[i]]=-1;

memcpy(an,ann,sizeof(an));

memcpy(match,cop,sizeof(match));

}

printf("Heap %d\n",++cas);

int flag=0,last=-1;

for(int i=1;i<=n;i++)

{

if(ans[i]!=-1)last=i;

}

for(int i=1;i<=n;i++)if(ans[i]!=-1 && i!=last)

{

flag=1;

printf("(%c,%d) ",(char)(i-1+'A'),ans[i]);

}

if(last!=-1)printf("(%c,%d)\n",(char)(last-1+'A'),ans[last]);

else printf("none\n");

printf("\n");

}

return 0;

}