POJ3041 - Asteroids

来源：互联网发布：关于数据爆炸编辑：程序博客网时间：2024/05/22 16:22

此题说难也难说简单也简单。关键就看有没有理解这样的一条定理：

最小点覆盖数=最大匹配数

解题思路如下：

1、以行和列分别建立两个顶点集V1，V2

2、将障碍物（x1，y1）视为两个顶点集中的两点V1(x1)，V2(y1)的边

3、求V1和V2的最大二分匹配，由最小点覆盖数=最大匹配数可知，此即为答案(the minimum number of times Bessie must shoot)

//Memory Time//956K0MS#include<stdio.h>int edge[501][501];int  edgeNum[501];int couple[501];bool isVisited[501];bool isHoped[501];int search(int N);bool dfs(int src);int main(){int N, K;scanf("%d %d", &N, &K);for(int i = 1; i <= N; i++){edgeNum[i] = 0;couple[i] = -1;isVisited[i] = 0;isHoped[i] = 1;}for(int i = 0; i < K; i++){int src,dst;scanf("%d %d", &src, &dst);edge[src][edgeNum[src]] = dst;edgeNum[src]++;}printf("%d\n", search(N));return 0;}int search(int N){int temp = 0;for(int i = 1; i <= N; i++){for(int j = 1; j <= N; j++){isVisited[j] = 0;}if(dfs(i))temp++;}return temp;}bool dfs(int src){int tempNum = edgeNum[src];for(int i = 0; i < tempNum; i++){int tempDst = edge[src][i];if(isVisited[tempDst] == 0 && isHoped[tempDst]){isVisited[tempDst] = 1;if(couple[tempDst] == -1){couple[tempDst] = src;return 1;}else{if(dfs(couple[tempDst])){couple[tempDst] = src;return 1;}else{isHoped[tempDst] = 0;}}}}return 0;}

0 0