HDU 4067 Random Maze 最小费用最大流

最新推荐文章于 2018-10-22 00:51:23 发布

原创最新推荐文章于 2018-10-22 00:51:23 发布 · 421 阅读

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#最小费用最大流 #网络流

图论——网络流_费用流同时被 2 个专栏收录

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A了再说然而并没理解

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本文详细介绍了如何使用MCMF算法解决最大流量问题，并通过实例演示了求解过程，包括流量网络的构建、SPFA算法的应用以及求解步骤的实现。

完全参考自：

http://www.cnblogs.com/wally/archive/2013/08/29/3288898.html

建好图，套个模板就OK了。

code：

#include <bits/stdc++.h>
using namespace std;

const int N = 105;
const int M = 1005;
const int INF = 0x3f3f3f3f;
typedef long long LL;

struct Edge {
    int from, to, cap, flow, cost ,next;
}edge[512345];

struct MCMF {
    int s, t, cnt;
    int head[N], a[N], d[N], p[N];
    bool inq[N];
    int init(int ss, int tt) {
        s = ss, t = tt;
        cnt = 0;
        memset(head, -1, sizeof(head));
    }
    inline void addEdge(int u, int v, int cap, int cost) {
        edge[cnt] = (Edge){u, v, cap, 0, cost, head[u]};
        head[u] = cnt++;
        edge[cnt] = (Edge){v, u, 0, 0, -cost, head[v]};
        head[v] = cnt++;
    }
    bool SPFA(int &flow, int &cost) {
        memset(d, INF, sizeof(d));
        memset(inq, false, sizeof(inq));
        queue <int> q;
        q.push(s), inq[s] = true;
        a[s] = INF, d[s] = 0, p[s] = -1;
        while(!q.empty()) {
            int u = q.front();
            q.pop(), inq[u] = false;
            for(int i = head[u];i != -1; i = edge[i].next) {
                Edge &e = edge[i];
                if(e.cap > e.flow && d[e.to] > d[u]+e.cost) {
                    d[e.to] = d[u]+e.cost;
                    a[e.to] = min(a[u], e.cap-e.flow);
                    p[e.to] = i;
                    if(!inq[e.to]) {
                        inq[e.to] = true;
                        q.push(e.to);
                    }
                }
            }
        }
        if(d[t] == INF) return false;
        cost += d[t]*a[t];
        flow += a[t];

        int u = t;
        while(u != s) {
            edge[p[u]].flow += a[t];
            edge[p[u]^1].flow -= a[t];
            u = edge[p[u]].from;
        }
        return true;
    }
    void solve(int& flow, int& cost) {
        while(SPFA(flow, cost));
    }
}mcmf;

int n, m, s, t;
int in[N], out[N];
int main() {
    int T, cas = 0;
    scanf("%d" ,&T);
    while(T--) {
        scanf("%d%d%d%d", &n, &m, &s, &t);
        memset(in, 0, sizeof(in));
        memset(out, 0, sizeof(out));
        mcmf.init(0, n+1);

        in[s]++, out[t]++;
        int sum = 0;
        {
            int u, v, a, b;
            for(int i = 0;i < m; i++) {
                scanf("%d%d%d%d", &u, &v, &a, &b);
                if(a > b){
                    mcmf.addEdge(u, v, 1, a-b);
                    sum += b;
                }
                else{
                    mcmf.addEdge(v, u, 1, b-a);
                    in[v]++, out[u]++;
                    sum += a;
                }
            }
        }
        int tolflow = 0;
        for(int i = 1;i <= n; i++) {
            if(in[i] == out[i]) continue;
            if(in[i] > out[i]) {
                mcmf.addEdge(0, i, in[i]-out[i], 0);
                tolflow += in[i]-out[i];
            }
            else 
                mcmf.addEdge(i, n+1, out[i]-in[i], 0);
        }

        printf("Case %d: ", ++cas);
        int flow = 0, cost = 0;
        mcmf.solve(flow, cost);
        if(flow == tolflow) 
            printf("%d\n", sum+cost);
        else 
            puts("impossible");
    }
    return 0;
}