题意：给你n张照片的范围，n个点的坐标，问你能唯一确定那几个点属于那几张照片，例如样例中4唯一属于A，2唯一属于C，1唯一属于B，3唯一属于C
思路：进行二分图完全匹配，怎么判断唯一属于？匹配完之后删掉某一条匹配边再跑一次二分图匹配，如果还能完全匹配，那么就不是唯一，反之唯一。
代码：
#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = + ;
const int MOD = 1e9 + ;
const int INF = 0x3f3f3f3f;
struct node{
int x1, x2, y1, y2;
}p[maxn];
struct Node{
int x, y;
}a[maxn];
struct Edge{
int to, next;
}edge[maxn * ];
int g[maxn][maxn], linker[maxn], ans[maxn], n;
bool used[maxn];
bool dfs(int u){
for(int v = ; v < * n; v++){
if(g[u][v] && !used[v]){
used[v] = true;
if(linker[v]== - || dfs(linker[v])){
linker[v] = u;
return true;
}
}
}
return false;
}
int hungary(){
int res = ;
memset(linker, -, sizeof(linker));
for(int u = ; u < * n; u++){
memset(used, false, sizeof(used));
if(dfs(u)) res++;
}
return res;
}
bool inside(node m, Node n){
if(n.x > m.x1 && n.x < m.x2 && n.y > m.y1 && n.y < m.y2) return true;
return false;
}
int main(){
int ca = ;
while(~scanf("%d", &n) && n){
for(int i = ; i < n; i++){
scanf("%d%d%d%d", &p[i].x1, &p[i].x2, &p[i].y1, &p[i].y2);
}
memset(g, , sizeof(g));
for(int i = ; i < n; i++){
scanf("%d%d", &a[i].x, &a[i].y);
for(int j = ; j < n; j++){
if(inside(p[j], a[i])){
g[i][n + j] = g[n + j][i] = ;
}
}
}
printf("Heap %d\n", ca++);
int ret = hungary();
memcpy(ans, linker, sizeof(ans));
if(ret != * n) printf("none\n");
else{
bool ok = false;
for(int i = n; i < * n; i++){
g[i][ans[i]] = g[ans[i]][i] = ;
ret = hungary();
if(ret == * n) continue;
printf("%s(%c,%d)", ok == true? " " : "",'A' + (i - n), ans[i] + );
ok = true;
g[i][ans[i]] = g[ans[i]][i] = ;
}
if(!ok) printf("none\n");
else printf("\n");
}
printf("\n");
}
return ;
}