Optimal Milking (二分图多重匹配+最短路)

FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow locations are named by ID numbers K+1..K+C. 

Each milking point can "process" at most M (1 <= M <= 15) cows each day. 

Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse severa

l paths on the way to their milking machine. 
Input
* Line 1: A single line with three space-separated integers: K, C, and M. 

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line. 
Output
A single line with a single integer that is the minimum possible total distance for the furthest walking cow. 
Sample Input
2 3 2
0 3 2 1 1
3 0 3 2 0
2 3 0 1 0
1 2 1 0 2
1 0 0 2 0
Sample Output
2


#include
#include
#include
#include
#include
#include
#include
const int INF=100000;
using namespace std;
int kk,c,m,n,mid;
int dis[1300][1300];
int cnt[1300];
int a[1300][1300];
int v[1300];
void floyd()
{
    int i,j,k;
    for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
            for(k=1;k<=n;k++)
                dis[j][k]=min(dis[j][i]+dis[i][k],dis[j][k]);
}
int dfs(int u)
{
    for(int i=1;i<=kk;i++)
    {
        if(v[i]) continue;
        if(dis[u][i]>mid) continue;
        v[i]=1;
        if(cnt[i]




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