POJ 2112--Optimal Milking【二分找最大距离的最小值 && 最大流】

标签：

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

上面图片有详细的题意和解析，来自图论算法理论，实现及应用这本书。

先用floyd求解任意两点之间的最短距离，再构建网络，将牛与源点相连容量为1（则从原点出去的中流量为C），将挤奶器与汇点相连容量为m。然后用dinic算法求最大流，二分枚举答案，将牛与挤奶器之间的距离小于枚举值得边 加入网络，大于枚举值得边去掉。找到可以使汇点的流入量为C的且符合条件的最小值就是答案。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#define maxn 1100
#define maxm 550000
#define INF 1000000
using namespace std;

int dist[maxn], vis[maxn];
int head[maxn], cur[maxn], cnt;
int map[maxn][maxn];
struct node{
    int u, v, cap, flow, next;
};

node edge[maxm];
int K, C, M, n;

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w){
    edge[cnt] = {u, v, w, 0, head[u]};
    head[u] = cnt++;
    edge[cnt] = {v, u, 0, 0, head[v]};
    head[v] = cnt++;
}

void getmap(int min_max){
    int i, j;
    for(i = K + 1; i <= n; ++i)
        add(0, i, 1);
    for(i = 1; i <= K; ++i)
        add(i, n + 1, M);
    for(i = K + 1; i <= n; ++i)
        for(j = 1; j <= K; ++j)
        if(map[i][j] <= min_max) add(i, j, 1);
}

bool BFS(int st, int ed){
    queue<int>q;
    memset(vis, 0, sizeof(vis));
    memset(dist, -1, sizeof(dist));
    q.push(st);
    vis[st] = 1;
    dist[st] = 0;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next){
            node E = edge[i];
            if(!vis[E.v] && E.cap > E.flow){
                vis[E.v] = 1;
                dist[E.v] = dist[u] + 1;
                if(E.v == ed) return true;
                q.push(E.v);
            }
        }
    }
    return false;
}

int DFS(int x, int ed, int a){
    if(a == 0 || x == ed)
        return a;
    int flow = 0, f;
    for(int &i = cur[x]; i != -1; i =edge[i].next){
        node &E = edge[i];
        if(dist[E.v] == dist[x] + 1 && (f = DFS(E.v, ed, min(a, E.cap - E.flow))) > 0){
            E.flow += f;
            edge[i ^ 1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}

int maxflow (int st, int ed){
    int flowsum  = 0;
    while(BFS(st, ed)){
        memcpy(cur, head, sizeof(head));
        flowsum += DFS(st, ed, INF);
    }
    return flowsum;
}

int main (){
    while(scanf("%d%d%d", &K,&C,&M) != EOF){
        n = C + K;
        int i, j , k;
        for(i = 1; i <= n; ++i)
        for(j = 1; j <= n; ++j){
            scanf("%d", &map[i][j]);
            if(map[i][j] == 0) map[i][j] = INF;
        }
        for(k = 1; k <= n; ++k)
            for(i = 1; i <= n; ++i)
                if(map[i][k] != INF){
                    for(j = 1; j <= n; ++j)
                        map[i][j] = min(map[i][k] + map[k][j], map[i][j]);
            }
        int L = 0, R = 60000, mid, ans;
        while( R > L){
            mid = (L + R) / 2;
            ans = 0;
            init();
            getmap(mid);
            ans = maxflow(0, n + 1);
            if(ans >= C) R = mid;
            else L = mid + 1;
        }
        printf("%d\n", R);
    }
    return 0;
}

POJ 2112--Optimal Milking【二分找最大距离的最小值 && 最大流】

标签：

原文地址：http://blog.csdn.net/hpuhjh/article/details/47300911