3189 Steady Cow Assignment//MaxMatch

Steady Cow Assignment
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 2134
 
Accepted: 730
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

Hint
Explanation of the sample:
 
Each cow can be assigned to her first or second choice: barn 1 gets cows 1 and 5, barn 2 gets cow 2, barn 3 gets cow 4, and barn 4 gets cows 3 and 6.
Source
USACO 2006 February Gold
아니면 이분도의 다중 일치
선호치는 구간[1,b]이기 때문에 이 차치를 일일이 열거할 수 있다. 최소 선호치는low이고 최대 선호치는high이다. 처음에 low=high=1이다. 만약에 선호치가 [low,high]의 범위 내에서 이 n마리소를 주제를 만족시키는 방식으로 안배할 수 있다면 low값을 증가시켜 차치를 줄이는 데 사용하고 안배 방안이 없으면 높은 값을 증가시켜 새로운 안배 방안을 찾는다.
 
#include#includeint mat[1010][25];int c[25];bool usedif[25];int link[25][1010];int n,b,low,high;bool can(int t){    for(int i=1;i<=b;i++)    if(usedif[i]==0&&mat[t][i]<=high&&mat[t][i]>=low)    {        usedif[i]=true;        if(link[i][0]