POJ 3189 Steady Cow Assignment (枚举+二分图的多重匹配)

Steady Cow Assignment

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4567		Accepted: 1606

Description

Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. Some cows really like their current barn, and some are not so happy.

FJ would like to rearrange the cows such that the cows are as equally happy as possible, even if that means all the cows hate their assigned barn.

Each cow gives FJ the order in which she prefers the barns. A cow's happiness with a particular assignment is her ranking of her barn. Your job is to find an assignment of cows to barns such that no barn's capacity is exceeded and the size of the range (i.e., one more than the positive difference between the the highest-ranked barn chosen and that lowest-ranked barn chosen) of barn rankings the cows give their assigned barns is as small as possible.

Input

Line 1: Two space-separated integers, N and B

Lines 2..N+1: Each line contains B space-separated integers which are exactly 1..B sorted into some order. The first integer on line i+1 is the number of the cow i's top-choice barn, the second integer on that line is the number of the i'th cow's second-choice barn, and so on.

Line N+2: B space-separated integers, respectively the capacity of the first barn, then the capacity of the second, and so on. The sum of these numbers is guaranteed to be at least N.

Output

Line 1: One integer, the size of the minumum range of barn rankings the cows give their assigned barns, including the endpoints.

Sample Input

6 4
1 2 3 4
2 3 1 4
4 2 3 1
3 1 2 4
1 3 4 2
1 4 2 3
2 1 3 2

Sample Output

2

一开始没看懂这句话：one more than the positive difference between the the highest-ranked barn chosen and that lowest-ranked barn chosen...以为要求的是每一头牛都尽可能高兴，就是把他们尽可能往他们喜欢的barn里住，然后求最少要用到哪个barn...直接用二分答案来做，WA了数次。然后网上搜报告，刚瞄到别人写的大致题意就顿时跪了。题意要求的其实是不管牛高不高兴，而是同个barn里的牛的高兴程度要尽可能接近。所以做法就是从高兴差距为1开始枚举到B，每次求匹配，如果能匹配成功取最小值。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define SIZE 1024

using namespace std;

int N,B,ans,low,high;
int need[SIZE][32];
bool cnt[SIZE][32],vis[32];
int link[32][SIZE],num[32];
int limit[32];

bool dfs(int lt)
{
    for(int r=1; r<=high; r++) 
    {
        if(!vis[need[lt][r]] && cnt[lt][need[lt][r]])
        {
            vis[need[lt][r]] = true;
            if(num[need[lt][r]] < limit[need[lt][r]])
            {
                link[need[lt][r]][++num[need[lt][r]]] = lt;
                return true;
            }
            for(int i=1; i<=limit[need[lt][r]]; i++)
            {
                if(dfs(link[need[lt][r]][i]))
                {
                    link[need[lt][r]][i] = lt;
                    return true;
                }
            }
        }
    }
    return false;
}

bool match()
{
    memset(link,0,sizeof(link));
    memset(num,0,sizeof(num));
    for(int i=1; i<=N; i++)
    {
        memset(vis,0,sizeof(vis));
        if(!dfs(i))
            return false;
    }
    return true;
}

void Search()
{
    low = 1, high = 1; //low和high是喜欢程度,从最喜欢的开始枚举
    while(low <= high && high <= B)
    {
        memset(cnt,0,sizeof(cnt));
        for(int i=1; i<=N; i++)
            for(int j=low; j<=high; j++) //把当前匹配的范围连边
                cnt[i][need[i][j]] = true;
        if(match())
        {
            if(high - low + 1 < ans)
                ans = high - low + 1;
            low ++; //可匹配，将匹配范围缩小，继续看能不能匹配（无视喜欢程度）
        }
        else
            high ++; //不可匹配，扩大当前喜欢的范围
    }
}

int main()
{
    while(~scanf("%d%d",&N,&B))
    {
        for(int i=1; i<=N; i++)
            for(int j=1; j<=B; j++)
                scanf("%d",&need[i][j]);
        for(int i=1; i<=B; i++)
            scanf("%d",&limit[i]);
        ans = 0x7fffffff;
        Search();
        printf("%d\n",ans);
    }
    return 0;
}