POJ 2112 Optimal Milking （floyd+二分+二分图多重匹配）

题意：k个挤奶器，c头奶牛，每个挤奶器可供m头奶牛使用，k+c的矩阵给出挤奶器和奶牛到彼此的距离，前k行是挤奶器，k+1到k+c行是奶牛。求在保证每头牛都能挤到奶的情况下，离得最远的牛的最小距离。

题解：floyd+二分+二分图多重匹配
首先得用floyd求一下最短路，保证奶牛距离挤奶器保存的是最短距离。

奶牛只能去一个挤奶器，每个挤奶器能让多个奶牛挤奶，二分图多重匹配问题。接下来我们要限制最远的牛的最小距离，求最大值的最小值，我们用二分，这里限制距离，将二分距离内的奶牛与挤奶器相连。

吐了，wa了半天居然是floyd写错了。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
using namespace std;
const int MAXN = 1010;
const int MAXM = 1010;
int k, c, m, d[MAXN][MAXN];
int uN, vN;
int g[MAXN][MAXM];
int linker[MAXM][MAXN];
bool used[MAXM];
int num[MAXM];//右边最大的匹配数
bool dfs(int u) {
    for (int v = 1; v <= vN; v++)
        if (g[u][v] && !used[v]) {
            used[v] = true;
            if (linker[v][0] < m) {
                linker[v][++linker[v][0]] = u;
                return true;
            }
            for (int i = 1; i <= m; i++)
                if (dfs(linker[v][i])) {
                    linker[v][i] = u;
                    return true;
                }
        }
    return false;
}
int hungary(int x) {
    memset(g, 0, sizeof(g));
    for (int i = k + 1; i <= k + c; i++) {
        for (int j = 1; j <= k; j++) {
            if (d[i][j] <= x) g[i - k][j] = 1;
        }
    }
    uN = c, vN = k;
    int res = 0;
    for (int i = 1; i <= vN; i++)
        linker[i][0] = 0;
    for (int u = 1; u <= uN; u++) {
        memset(used, false, sizeof(used));
        if (dfs(u))res++;
    }
    return res;
}

int main() {
    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", &d[i][j]);
            if (!d[i][j] && i != j) d[i][j] = 0x3f3f3f3f;
        }
    }
    for (int i = 1; i <= k + c; i++) {
        for (int j = 1; j <= k + c; j++) {
            for (int x = 1; x <= k + c; x++) {
                d[j][x] = min(d[j][i] + d[i][x], d[j][x]);
            }
        }
    }
    int maxd = -1;
    for (int i = k + 1; i <= k + c; i++) {
        for (int j = 1; j <= k; j++) {
            maxd = max(maxd, d[i][j]);
        }
    }
    int l = 1, r = maxd, mid, ans;
    while (l <= r) {
        int mid = (l + r) / 2;
        if (hungary(mid) == c) {
            ans = mid;
            r = mid - 1;
        }
        else
            l = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}