标签：图论之二分匹配

过山车

6 3 3
1 1
1 2
1 3
2 1
2 3
3 1
0

3

第一道二分匹配题。。。

纯属模板。

参考：http://blog.csdn.net/wellerzhao/article/details/7756956

代码1：

#include <stdio.h>
#include <string.h>
#define M 555

int map[M][M];
int mx[M], my[M];
int vis[M];
int n, m;

int find(int s){
    int i;
    for(i = 1; i<= m; i ++){
        if(!vis[i]&&map[s][i]){
            vis[i] = 1;
            if(my[i] == 0||find(my[i])){
                my[i] = s;
                mx[s] = i;
                return 1;
            }
        }
    }
    return 0;
}

void f(){
    for(int i = 1; i <= n; i ++)
        printf("%d..%d,,%d..%d\n", i, mx[i], i, my[i]);
}
int main(){
    int k;
    while(scanf("%d", &k), k){
        int i;
        memset(map, 0, sizeof(map));
        memset(mx, 0, sizeof(mx));
        memset(my, 0, sizeof(my));
        int a, b;
        scanf("%d%d", &n, &m);
        for(i = 0; i < k; i ++){
            scanf("%d%d", &a, &b);
            map[a][b] = 1;    
        }
        int ans = 0;
        for(i = 1; i<= n; i ++){
            if(!mx[i]){
                memset(vis, 0, sizeof(vis));
                if(find(i)) ++ans;
            }
        }
        printf("%d\n", ans);
    //    f();
    }
    return 0;
}

代码；

#include <stdio.h>
#include <string.h>
#define M 555

int map[M][M];
int mx[M], my[M];
int vis[M];
int n, m;

int find(int s){
    int i;
    for(i = 1; i<= m; i ++){
        if(!vis[i]&&map[s][i]){
            vis[i] = 1;
            if(my[i] == 0||find(my[i])){
                my[i] = s;
                mx[s] = i;
                return 1;
            }
        }
    }
    return 0;
}

void f(){
    for(int i = 1; i <= n; i ++)
        printf("%d..%d,,%d..%d\n", i, mx[i], i, my[i]);
}
int main(){
    int k;
    while(scanf("%d", &k), k){
        int i;
        memset(map, 0, sizeof(map));
        memset(mx, 0, sizeof(mx));
        memset(my, 0, sizeof(my));
        int a, b;
        scanf("%d%d", &n, &m);
        for(i = 0; i < k; i ++){
            scanf("%d%d", &a, &b);
            map[a][b] = 1;    
        }
        int ans = 0;
        for(i = 1; i<= n; i ++){
            if(!mx[i]){
                memset(vis, 0, sizeof(vis));
                if(find(i)) ++ans;
            }
        }
        printf("%d\n", ans);
    //    f();
    }
    return 0;
}

hdoj 2063 过山车 【二分匹配之匈牙利算法】

标签：图论之二分匹配

原文地址：http://blog.csdn.net/shengweisong/article/details/40631763