poj 2112 最大流

题意：K个产奶机，C头奶牛，每个产奶机最多可供M头奶牛使用；并告诉了产奶机、奶牛之间的两两距离Dij(0<=i,j<K+C)。

问题：如何安排使得在任何一头奶牛都有自己产奶机的条件下，奶牛到产奶机的最远距离最短？最短是多少？

二分答案。

注意先floyd求得两两之间最小距离，二分的时候上界不是200。200只是直接相连的最大距离。

// File Name: 2112.cpp
// Author: Missa
// Created Time: 2013/4/11 星期四 14:20:52

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<cstdlib>
#include<map>
#include<set>
using namespace std;
#define CL(x,v) memset(x,v,sizeof(x));
#define R(i,st,en) for(int i=st;i<en;++i)
#define LL long long
#define inf 0x3f3f3f3f
int a[240][240];
int K,C,M;

const int maxn = 240;
int cap[maxn][maxn];
int lev[maxn];
int n, m;
int st, en;

bool bfs()
{
    queue <int> q;
    while (!q.empty()) q.pop();
    memset(lev, -1, sizeof(lev));
    lev[st] = 0;
    q.push(st);
    while (!q.empty())
    {
        int u = q.front();q.pop();
        for (int v = 0; v <= n+1; ++v)
            if (lev[v] == -1 && cap[u][v] != 0)
            {
                lev[v] = lev[u] + 1;
                q.push(v);
            }
    }
    return lev[en] != -1;
}
int dfs(int u, int cur_flow)
{
    int dt = cur_flow;
    if (u == en) return dt;
    for (int v = 0; v <= n + 1; ++v)
    {
        if (cap[u][v] > 0 && lev[u] + 1 == lev[v])
        {
            int flow = dfs(v, min(dt, cap[u][v]));
            cap[u][v] -= flow;
            cap[v][u] += flow;
            dt -= flow;
        }
    }
    return cur_flow - dt;
}

int dinic()
{
    int cur_flow, ans = 0;
    while(bfs())
        while(cur_flow = dfs(st, inf))
            ans += cur_flow;
    return ans;
}
void build(int dis)
{
    memset(cap, 0, sizeof(cap));
    n = K + C;
    st = 0;
    en = n + 1;
    for (int i = 1; i <= K; ++i)
        cap[0][i] += M;
    for (int i = K + 1; i <= K + C; ++i)
        cap[i][en] += 1;
    for (int i = 1; i <= K; ++i)
        for (int j = K + 1; j <= K + C; ++j)
            if (a[i][j] <= dis)
                cap[i][j] += 1;
}


int main()
{
    while(~scanf("%d%d%d",&K, &C, &M))
    {
        for (int i = 1; i <= K + C; ++i)
            for (int j = 1; j <= K + C; ++j)
            {
                scanf("%d", &a[i][j]);
                if (a[i][j] == 0) a[i][j] = inf;
            }
        for (int k = 1; k <= K + C; ++k)
            for (int i = 1; i <= K + C; ++i)
                for (int j = 1; j <= K + C; ++j)
                    a[i][j]= min(a[i][j],a[i][k]+a[k][j]);
        /*for (int i = 1; i <= K + C; ++i)
        {
            for (int j = 1; j <= K + C; ++j)
                cout<<a[i][j]<<" ";
            cout<<endl;
        }*/
        int ans = 1, low = 1, high = inf;
        while(low <= high)
        {
            int mid = (low + high) >> 1;
            build(mid);
            if (dinic() == C)
            {
                ans = mid;
                high = mid - 1;
            }
            else
                low = mid + 1;
        }
        printf("%d\n", ans);

    }
    return 0;
}