POJ2112 Optimal Milking 【最大流+二分】

标签：poj2112

Description

Input

Output

Sample Input

2 3 2
0 3 2 1 1
3 0 3 2 0
2 3 0 1 0
1 2 1 0 2
1 0 0 2 0

Sample Output

2

Source

题意：有k台挤奶器，每台挤奶器最多容纳m头奶牛，该牧场共有c头奶牛，现在给定这k台机器和c头奶牛相互间的直接距离，求让所有奶牛到达挤奶器且满足该条件时奶牛走到挤奶器间的最大距离的最小值。

题解：构图：先用Floyd求出相互间的最短距离，然后设置源点到每头牛的距离为1，每台机器到汇点的距离为m，然后若牛到机器的距离不大于maxdist，那么则将该边加入到新图中，最后对新图求最大流，判断最大流是否等于c，就这样二分枚举maxdist直到找到最小的maxdist为止。

#include <stdio.h>
#include <string.h>
#define inf 0x3fffffff
#define maxn 235

int dist[maxn][maxn], k, c, m, n;
int G[maxn][maxn], Layer[maxn];
int queue[maxn], maxDist;
bool vis[maxn];

void Floyd() {
    int x, i, j;
    maxDist = 200;
    for(x = 1; x <= n; ++x)
        for(i = 1; i <= n; ++i)
            for(j = 1; j <= n; ++j)
                if(dist[i][j] > dist[i][x] + dist[x][j]) {
                    dist[i][j] = dist[i][x] + dist[x][j];
                    if(maxDist < dist[i][j]) maxDist = dist[i][j];
                }
}

void build(int flow) {
    memset(G, 0, sizeof(G));
    int i, j;
    for(i = k + 1; i <= n; ++i) {
        G[0][i] = 1;
        for(j = 1; j <= k; ++j)
            if(dist[i][j] <= flow)
                G[i][j] = 1;
    }
    for(j = 1; j <= k; ++j)
        G[j][n + 1] = m;
}

bool countLayer() {
    int id = 0, front = 0, now, i;
    memset(Layer, 0, sizeof(Layer));
    Layer[0] = 1; queue[id++] = 0;
    while(front < id) {
        now = queue[front++];
        for(i = 0; i <= n + 1; ++i)
            if(G[now][i] && !Layer[i]) {
                Layer[i] = Layer[now] + 1;
                if(i == n + 1) return true;
                else queue[id++] = i;
            }
    }
    return false;
}

bool Dinic() {
    int i, maxFlow = 0, id = 0, now, minCut, pos, u, v;
    while(countLayer()) {
        memset(vis, 0, sizeof(vis));
        vis[0] = 1; queue[id++] = 0;
        while(id) {
            now = queue[id - 1];
            if(now == n + 1) {
                minCut = inf;
                for(i = 1; i < id; ++i) {
                    u = queue[i - 1];
                    v = queue[i];
                    if(G[u][v] < minCut) {
                        minCut = G[u][v];
                        pos = u;
                    } 
                }
                maxFlow += minCut;
                for(i = 1; i < id; ++i) {
                    u = queue[i - 1];
                    v = queue[i];
                    G[u][v] -= minCut;
                    G[v][u] += minCut;
                }
                while(id && queue[id - 1] != pos)
                    vis[queue[--id]] = 0;
            } else {
                for(i = 0; i <= n + 1; ++i) {
                    if(G[now][i] && !vis[i] && Layer[now] + 1 == Layer[i]) {
                        queue[id++] = i;
                        vis[i] = 1; break;
                    }
                }
                if(i > n + 1) --id;
            }
        }
    }
    return maxFlow == c;
}

int binarySolve() {
    int left = 0, right = maxDist, mid;
    while(left < right) {
        mid = (left + right) >> 1;
        build(mid);
        if(Dinic()) right = mid;
        else left = mid + 1;
    }
    return left;
}

int main() {
    //freopen("stdin.txt", "r", stdin);
    int i, j;
    while(scanf("%d%d%d", &k, &c, &m) == 3) {
        for(i = 1, n = k + c; i <= n; ++i)
            for(j = 1; j <= n; ++j) {
                scanf("%d", &dist[i][j]);
                if(!dist[i][j] && i != j)
                    dist[i][j] = inf;
            }
        Floyd();
        printf("%d\n", binarySolve());
    }
    return 0;
}

POJ2112 Optimal Milking 【最大流+二分】

标签：poj2112

原文地址：http://blog.csdn.net/chang_mu/article/details/39680359