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;
}