HDU 2063 过山车（匈牙利算法模板）

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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)