【完全二分匹配必须边】POJ 1486

刚开始没看清楚题目，直接匈牙利算法，后来看到一句话Then print a series of all the slides whose numbers can be uniquely determined from the input

输出要唯一！就是问每条边是否唯一，做法：先用一次匈牙利求出一个匹配，再逐条边拆，当出现另一个完全匹配时就是none，否则可以输出结果。注意match[i] = j的i是x，j是y。

#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 -9999999#define MAX INT_MAX#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 1005int min(int a,int b){return a>b?b:a;}int max(int a,int b){return a>b?a:b;}bool g[N][N],vis[N];int match[N];int n;int ans[N];struct re{    int xmin,xmax,ymin,ymax;}p[N];struct node{    int x,y;}q[N];bool sear(int x){//匈牙利算法    for(int i=0;i<n;i++){        if(g[x][i] && !vis[i]){            vis[i] = 1;            if(match[i]== -1 || sear(match[i])){                match[i] = x;                return true;            }        }    }    return false;}bool chk(int i,int j){    if(q[j].x>=p[i].xmin && q[j].x<=p[i].xmax && q[j].y>=p[i].ymin && q[j].y<=p[i].ymax)    return true;    return false;}void gao(){    int i,j;    bool ok = 0;    for(i=0;i<n;i++){        int tmp = match[i];        g[tmp][i] = 0;        match[i] = -1;        memset(vis,0,sizeof(vis));        if(!sear(tmp)) {            if(ok)printf(" ");            printf("(%c,%d)",i+'A',tmp+1);            ok = 1;match[i] = tmp;        }        g[tmp][i] = 1;    }    if(!ok){        puts("none\n");    } else {        printf("\n\n");    }}int main(){int t=1;    while(scanf("%d",&n) && n){        int i,j;        memset(g,0,sizeof(g));        for(i=0;i<n;i++){            scanf("%d%d%d%d",&p[i].xmin,&p[i].xmax,&p[i].ymin,&p[i].ymax);            match[i] = -1;        }        for(i=0;i<n;i++){            scanf("%d%d",&q[i].x,&q[i].y);        }        for(i=0;i<n;i++){            for(j=0;j<n;j++){                if(chk(i,j)){                    g[j][i] = 1;                }            }        }        int cnt=0;        for(i=0;i<n;i++){            memset(vis,0,sizeof(vis));            if(sear(i))cnt++;        }        printf("Heap %d\n",t++);        if(cnt!=n){            puts("none");            puts("");        } else {            gao();        }    }    return 0;}