HDU 2063 过山车（匈牙利算法模板）

链接：

http://acm.hdu.edu.cn/showproblem.php?pid=2063

分析与总结：

这题是裸的二分匹配，用来验证模板的

代码：

1. DFS

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 505;
int Nx,Ny;  
int G[MAXN][MAXN];
int Mx[MAXN], My[MAXN];  
bool mark[MAXN];

bool FindPath(int u){
    for(int v=0; v<Ny; ++v){
        if(G[u][v] && !mark[v]){
            mark[v] = true;
            if(My[v]==-1 || FindPath(My[v])){
                My[v] = u;
                Mx[u] = v;
                return true;
            }
        }
    }
    return false;
}
int MaxMatch(){
    int ret=0;
    memset(Mx, -1, sizeof(Mx));
    memset(My, -1, sizeof(My));
    for(int u=0; u<Nx; ++u){
        if(Mx[u] == -1) {
            memset(mark, 0, sizeof(mark));
            if(FindPath(u)) ++ret;
        }
    }
    return ret;
}

int main(){
    int k,a,b;
    while(~scanf("%d%d%d",&k,&Nx,&Ny) && k){
        memset(G,0,sizeof(G));
        for(int i=0; i<k; ++i) {
            scanf("%d%d",&a,&b);
            G[a-1][b-1]=1;
        }
        printf("%d\n",MaxMatch());
    }
    return 0;
}

2. BFS

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 505;
int Nx,Ny;  
int G[MAXN][MAXN];
int Mx[MAXN], My[MAXN], prev[MAXN], Q[MAXN];  
int mark[MAXN];

int MaxMatch(){
    int res = 0;
    int front, rear;
    
    memset(Mx, -1, sizeof(Mx));
    memset(My, -1, sizeof(My));
    memset(mark, -1, sizeof(mark));
    
    for (int i = 0; i < Nx; i++){
        if (Mx[i] == -1){
            front = rear = 0;
            Q[rear++] = i;
            prev[i] = -1;
            
            bool flag = 0;
            while (front < rear && !flag){
                int u = Q[front];
                
                for (int v = 0; v < Ny && !flag; v++){
                    if (G[u][v] && mark[v] != i){
                        mark[v] = i;
                        Q[rear++] = My[v];
                        if (My[v] >= 0)
                            prev[My[v]] = u;
                        else{
                            flag = 1;
                            int d = u, e = v;
                            while (d != -1){
                                int t = Mx[d];
                                Mx[d] = e;
                                My[e] = d;
                                d = prev[d];
                                e = t;
                            }
                        }
                    }
                }
                front++;
            }
            if (Mx[i] != -1)
                res++;
        }
    }
    return res;
}
int main(){
    int k,a,b;
    while(~scanf("%d%d%d",&k,&Nx,&Ny) && k){
        memset(G,0,sizeof(G));
        for(int i=0; i<k; ++i) {
            scanf("%d%d",&a,&b);
            G[a-1][b-1]=1;
        }
        printf("%d\n",MaxMatch());
    }
    return 0;
}

—— 生命的意义，在于赋予它意义士。

原创http://blog.csdn.net/shuangde800，By D_Double (转载请标明)