Optimal Milking（POJ2112+二分+Dinic）

题目链接：http://poj.org/problem?id=2112

题目：

题意：有k台挤奶机，c头奶牛，每台挤奶机每天最多生产m的奶，给你每个物品到其他物品的距离（除了物品到自己本省的距离为0外，两者之间没有路线直接到达也为0，此时需要将距离处理为inf），问跑最远距离的奶牛要跑多远。

思路：先用floyd将各物品间的最短距离求出来，再二分跑最远距离的奶牛走的距离x，将小于x的两者之间进行连边，流量为1，引入一个超级源点和超级汇点，我们采用方向建图的方式，让超级源点与挤奶机连边，且流量为m，奶牛与超级汇点连边，流量为1，跑一边Dinic，如果最大流为c那么ub变成mid-1，否则lb为mid+1。

代码实现如下：

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#define lson i<<1
#define rson i<<1|1
#define bug printf("*********
");
#define FIN freopen("D://code//in.txt", "r", stdin);
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define IO ios::sync_with_stdio(false),cin.tie(0);

const double eps = 1e-8;
const int mod = 10007;
const int maxn = 250 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f;

int k, c, m, s, e, num, tot, maxflow;
int mp[maxn][maxn], head[maxn], d[maxn];

queue<int> q;

struct edge {
    int v, w, next;
}ed[maxn*maxn];

void addedge(int u, int v, int w) {
    ed[tot].v = v;
    ed[tot].w = w;
    ed[tot].next = head[u];
    head[u] = tot++;
    ed[tot].v = u;
    ed[tot].w = 0;
    ed[tot].next = head[v];
    head[v] = tot++;
}

bool bfs() {
    memset(d, 0, sizeof(d));
    while(!q.empty()) q.pop();
    q.push(s);
    d[s] = 1;
    while(!q.empty()) {
        int x = q.front(); q.pop();
        for(int i = head[x]; ~i; i = ed[i].next) {
            int y = ed[i].v;
            if(ed[i].w && !d[y]) {
                q.push(y);
                d[y] = d[x] + 1;
                if(y == e) return 1;
            }
        }
    }
    return 0;
}

int dinic(int x, int flow) {
    if(x == e) return flow;
    int rest = flow, k;
    for(int i = head[x]; ~i && rest; i = ed[i].next) {
        int y = ed[i].v;
        if(ed[i].w && d[y] == d[x] + 1) {
            k = dinic(y, min(rest, ed[i].w));
            if(!k) d[y] = 0;
            ed[i].w -= k;
            ed[i^1].w += k;
            rest -= k;
        }
    }
    return flow - rest;
}

bool check(int x) {
    tot = 0;
    memset(head, -1, sizeof(head));
    for(int i = 1; i <= k; i++) {
        for(int j = k + 1; j <= num; j++) {
            if(mp[i][j] <= x) {
                addedge(i, j, 1);
            }
        }
    }
    for(int i = 1; i <= k; i++) {
        addedge(s, i, m);
    }
    for(int i = k + 1; i <= num; i++) {
        addedge(i, e, 1);
    }
    int flow = 0;
    maxflow = 0;
    while(bfs()) {
        while((flow = dinic(s, inf))) maxflow += flow;
    }
    return maxflow == c;
}

int main() {
    //FIN;
    scanf("%d%d%d", &k, &c, &m);
    num = k + c;
    for(int i = 1; i <= num; i++) {
        for(int j = 1; j <= num; j++) {
            scanf("%d", &mp[i][j]);
            if(i != j && mp[i][j] == 0) {
                mp[i][j] = inf;
            }
        }
    }
    for(int k = 1; k <= num; k++) {
        for(int i = 1; i <= num; i++) {
            for(int j = 1; j <= num; j++) {
                if(mp[i][j] > mp[i][k] + mp[k][j]) {
                    mp[i][j] = mp[i][k] + mp[k][j];
                }
            }
        }
    }
    s = 0, e = num + 1;
    int ub = inf, lb = 0, mid;
    while(ub >= lb) {
        mid = (ub + lb) >> 1;
        if(check(mid)) ub = mid - 1;
        else lb = mid + 1;
    }
    printf("%d
", lb);
    return 0;
}