poj1274 The Perfect Stall（二分圖匹配 / 最大流）

The Perfect Stall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20547		Accepted: 9263

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

5 5
2 2 5
3 2 3 4
2 1 5
3 1 2 5
1 2 

Sample Output

4

題意：n頭牛，m個位置，每頭牛可以待在s個位置上，給每個牛分配一個位置，求問最多幾頭牛可以分配到位置。

這是二分圖匹配的基礎題，也可以用最大流來做，直接貼代碼。。

二分圖匹配：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define maxn 205

using namespace std;

int maps[maxn][maxn], vis[maxn], link[maxn];
int n, m;

bool Find(int x)
{
    for(int i = 1; i <= m; i++)
    {
        if(vis[i] == 0 && maps[x][i] == 1)
        {
            vis[i] = 1;
            if(link[i] == 0 || Find(link[i]))
            {
                link[i] = x;
                return true;
            }
        }
    }
    return false;
}

int main()
{
    int x, y;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        memset(maps, 0, sizeof(maps));
        memset(link, 0, sizeof(link));
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &x);
            for(int j = 0; j < x; j++)
            {
                scanf("%d", &y);
                maps[i][y] = 1;
            }
        }
        int ans = 0;
        for(int i = 1; i <= n; i++)
        {
            memset(vis, 0, sizeof(vis));
            if(Find(i))
                ans++;
        }
        printf("%d\n", ans);
    }
}

網絡流最大流：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

#define maxn 505

using namespace std;

int n, m;
int maps[maxn][maxn], q[maxn * maxn];
int d[maxn];//記錄每個點所在的層

int BFS()//BFS分層，判斷是否存在增廣路
{
    memset(d, -1, sizeof(d));
    d[0] = 0;
    int f, r;
    f = 1;
    r = 1;
    q[r++] = 0;
    while(f < r)
    {
        int x = q[f++];
        for(int i = 0; i <= n + m + 1; i++)
        {
            if(d[i] < 0 && maps[x][i] > 0)
            {
                d[i] = d[x] + 1;
                q[r++] = i;
            }
        }
    }
    if(d[n + m + 1] > 0)
        return 1;
    else
        return 0;
}

int Find(int x, int v)//找出某一條增廣路的最大流
{
    int a;
    if(x == n + m + 1)
        return v;
    for(int i = 0; i <= n + m + 1; i++)
    {
        if(maps[x][i] > 0 && d[i] == d[x] + 1 && (a = Find(i, min(v, maps[x][i]))))//x到i有流量 且 i是x的下一層 且 i到匯點存在最大流
        {
            maps[x][i] -= a;//增廣路
            maps[i][x] += a;//回退邊
            return a;
        }
    }
    return 0;
}

int main()
{
    int x, y, t;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        memset(maps, 0, sizeof(maps));
        for(int i = 1; i <= n; i ++)
        {
            scanf("%d", &x);
            for(int j = 1; j <= x; j ++)
            {
                scanf("%d", &y);
                maps[i][n + y] = 1;
            }
        }
        for(int i = 1; i <= n; i++)
            maps[0][i] = 1;
        for(int i = n + 1; i <= n + m; i++)
            maps[i][n + m + 1] = 1;
//        for(int i = 0; i <= n + m + 1; i++)
//        {
//            for(int j = 0; j <= n + m + 1; j++)
//            {
//                printf("%d ",maps[i][j]);
//            }
//            printf("\n");
//        }
        int ans = 0;
        while(BFS())//BFS搜索判斷是否有從源點到匯點的通路
        {
            while(t = Find(0, 2000000005))//查找增廣路，求其最大流
                ans += t;
        }
        printf("%d\n", ans);
    }
}