Optimal Milking

//二分枚举最大距离的最小值+Floyd找到最短路+Dinic算法
//参考图论算法书，并对BFS构建层次网络算法进行改进
//Time:157Ms    Memory:652K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

#define MAX 250
#define INF 0x3f3f3f3f

int K, C, M;
int s, t;
int d[MAX][MAX];	//各点间最短距离
int res[MAX][MAX];	//残留网络
int lev[MAX];

void build_map(int limit)
{
    memset(res,0,sizeof(res));
	for (int i = K + 1; i <= K + C; i++)
		res[s][i] = 1;
	for (int i = 1; i <= K; i++)
		res[i][t] = M;
	for (int i = K + 1; i <= K + C; i++)
		for (int j = 1; j <= K; j++)
			if (d[i][j] <= limit)	res[i][j] = 1;
}

bool bfs()  //BFS标记层次网络
{
    memset(lev, -1, sizeof(lev));
	queue<int> q;
	q.push(s);
	lev[s] = 0;
	while (!q.empty()) {    //构建层次网络
		int cur = q.front();
		q.pop();
		for(int i = 1; i <= t; i++)
        {
            if(lev[i] == -1 && res[cur][i])    //未访问且正向有流量
            {
                q.push(i);
                lev[i] = lev[cur] + 1;
            }
        }
	}
	return lev[t] != -1;
}

int dfs(int v, int alpha)  //DFS进行多次增广
{
    if(v == t || alpha == 0)  return alpha;
    int src = alpha;    //原可改进量
    for(int i = 1; i <= t; i++)
    {
        if(res[v][i] && lev[i] == lev[v] + 1){  //识别下一层次
            int tmp = dfs(i, min(alpha, res[v][i]));
            res[v][i] -= tmp;
            res[i][v] += tmp;
            alpha -= tmp;   //可改进量减少
        }
    }
    return src - alpha; //总改进量
}

int main()
{
	//freopen("in.txt", "r", stdin);

	scanf("%d%d%d", &K,&C,&M);
	s = 0;  t = K + C + 1;  //源点-汇点
	for (int i = 1; i < t; i++)
		for (int j = 1; j < t; j++)
		{
			scanf("%d", &d[i][j]);
			if (d[i][j] == 0)	d[i][j] = INF;
		}

	for (int k = 1; k < t; k++)
		for (int i = 1; i < t; i++)
		{
			if (d[i][k] != INF) {
				for (int j = 1; j < t; j++)
					d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
			}
		}

	s = 0;	t = K + C + 1;	//源点 汇点
	int l = 0, r = 9000;
	while (l < r)
	{
		int ans = 0;    //到达目的地的奶牛数量
		int mid = (l + r) / 2;
		build_map(mid);
		while (bfs())
			ans += dfs(0,INF); //第二参数指定该点可改进量
        ans == C ? r = mid: l = mid+1;
	}
	printf("%d\n", r);

	return 0;
}