POJ_1274_The Perfect Stall(二分图匹配)

时间：2015-05-07 22:16:10 阅读：111 评论：0 收藏：0 [点我收藏+]

The Perfect Stall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19908		Accepted: 9009

Description

Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall.
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.

Input

The input includes several cases. For each case, the first line contains two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. Each of the following N lines corresponds to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow.

Output

For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made.

Sample Input

Sample Output

题意：农夫新建了牛棚，一个牛栏只能容纳一只牛，而每只牛都有自己喜欢的牛栏，给出每只牛喜欢的牛棚，问最多能够匹配多少头牛。

分析：二分图匹配问题。求出最大分配方案。

题目链接：http://poj.org/problem?id=1274

代码清单：

#include<map>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;

const int maxv = 200 + 5;

int N,M;
int n,m;
bool vis[maxv];
int match[maxv];
bool graph[maxv][maxv];

void init(){
    memset(match,-1,sizeof(match));
    memset(graph,false,sizeof(graph));
}

bool dfs(int u){
    for(int v=1;v<=M;v++){
        if(!vis[v]&&graph[u][v]){
            vis[v]=true;
            if(match[v]==-1 || dfs(match[v])){
                match[v]=u;
                return true;
            }
        }
    }return false;
}

int max_match(){
    int sum=0;
    for(int i=1;i<=N;i++){
        memset(vis,false,sizeof(vis));
        if(dfs(i)) sum++;
    }return sum;
}

int main(){
    while(scanf("%d%d",&N,&M)!=EOF){
        init();
        for(int i=1;i<=N;i++){
            scanf("%d",&n);
            while(n--){
                scanf("%d",&m);
                graph[i][m]=true;
            }
        }
        printf("%d\n",max_match());
    }return 0;
}

静态邻接表：

#include<map>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;

const int maxv = 200 + 5;

struct Edge{
    int to,next;
}edge[maxv*maxv];

int N,M;
int n,m,index;
bool vis[maxv];
int match[maxv];
int head[maxv*maxv];

void init(){
    index=1;
    memset(head,0,sizeof(head));
    memset(match,-1,sizeof(match));
}

void add(int u,int v){
    edge[index].to=v;
    edge[index].next=head[u];
    head[u]=index++;
}

bool dfs(int u){
    for(int i=head[u];i!=0;i=edge[i].next){
        int v=edge[i].to;
        if(!vis[v]){
            vis[v]=true;
            if(match[v]==-1 || dfs(match[v])){
                match[v]=u;
                return true;
            }
        }
    }return false;
}

int max_match(){
    int sum=0;
    for(int i=1;i<=N;i++){
        memset(vis,false,sizeof(vis));
        if(dfs(i)) sum++;
    }return sum;
}

int main(){
    while(scanf("%d%d",&N,&M)!=EOF){
        init();
        for(int i=1;i<=N;i++){
            scanf("%d",&n);
            while(n--){
                scanf("%d",&m);
                add(i,m);
            }
        }
        printf("%d\n",max_match());
    }return 0;
}

POJ_1274_The Perfect Stall(二分图匹配)

标签：acm algorithm poj 二分图匈牙利算法

原文地址：http://blog.csdn.net/jhgkjhg_ugtdk77/article/details/45564853

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