Optimal Milking poj2112 二分+最大流

Description

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K个产奶机，C头奶牛，且每个产奶机最多可供M头奶牛使用；已知奶牛之间的两两距离，求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？

Solution

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看到最大值最小就要想到二分答案了

先floyd求两两之间最短路，二分的时候上界直接粗暴地用INF，或者别的什么也行

机器和牛棚分别连向源点和汇点，其中机器的限制m作为源点连向机器的容量

Code

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#include <stdio.h>
#include <string.h>
#include <queue>
#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)
#define erg(i, now) for (int i = ls[now]; i; i = e[i].next)
#define fill(x, t) memset(x, t, sizeof(x))
#define INF 0x3f3f3f3f
#define N 531
#define E N * N / 2 + 1
struct edge{int x, y, w, next;}e[E];
int rc[N][N], dis[N], ls[N];
inline void addEdge(int &cnt, int x, int y, int w){
    e[++ cnt] = (edge){x, y, w, ls[x]}; ls[x] = cnt;
    e[++ cnt] = (edge){y, x, 0, ls[y]}; ls[y] = cnt;
}
using std:: queue;
inline int bfs(int st, int ed){
    fill(dis, -1);
    dis[st] = 1;
    queue<int> q;
    q.push(st);
    while (!q.empty()){
        int now = q.front(); q.pop();
        erg(i, now){
            if (e[i].w > 0 && dis[e[i].y] == -1){
                q.push(e[i].y);
                dis[e[i].y] = dis[now] + 1;
                if (e[i].y == ed){
                    return 1;
                }
            }
        }
    }
    return 0;
}
inline int min(int x, int y){
    return x<y?x:y;
}
inline int find(int now, int ed, int mn){
    if (now == ed || !mn){
        return mn;
    }
    int ret = 0;
    erg(i, now){
        if (e[i].w > 0 && dis[now] + 1 == dis[e[i].y]){
            int d = find(e[i].y, ed, min(mn - ret, e[i].w));
            e[i].w -= d;
            e[i ^ 1].w += d;
            ret += d;
            if (ret == mn){
                break;
            }
        }
    }
    return ret;
}
inline int dinic(int st, int ed){
    int mxFlow = 0;
    while (bfs(st, ed)){
        mxFlow += find(st, ed, INF);
    }
    return mxFlow;
}
inline int cheque(int lim, int k, int c, int m){
    int st = 0, ed = N - 1;
    int edgeCnt = 1;
    fill(ls, 0);
    rep(i, 1, k){
        rep(j, k + 1, k + c){
            if (rc[i][j] <= lim){
                addEdge(edgeCnt, i, j, 1);
            }
        }
    }
    rep(i, 1, k){
        addEdge(edgeCnt, st, i, m);
    }
    rep(i, k + 1, k + c){
        addEdge(edgeCnt, i, ed, 1);
    }
    int mxFlow = dinic(st, ed);
    return mxFlow;
}
int main(void){
    int p, c, m;
    scanf("%d%d%d", &p, &c, &m);
    rep(i, 1, p + c){
        rep(j, 1, p + c){
            scanf("%d", &rc[i][j]);
            if (!rc[i][j]){
                rc[i][j] = INF;
            }
        }
    }
    rep(k, 1, p + c){
        rep(i, 1, p + c){
            rep(j, 1, p + c){
                if (i != j && j != k && i != k && rc[i][k] + rc[k][j] < rc[i][j]){
                    rc[i][j] = rc[i][k] + rc[k][j];
                }
            }
        }
    }
    int l = 0, r = INF - 1;
    int ans = 0;
    while (l <= r){
        int mid = (l + r) >> 1;
        int mxFlow = cheque(mid, p, c, m);
        if (mxFlow >= c){
            ans = mid;
            r = mid - 1;
        }else{
            l = mid + 1;
        }
    }
    printf("%d\n", ans);
    return 0;
}