费用流 hdu3667 Transportation

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传送门：点击打开链接

题意：n个节点m条有向边，每条有向边的容量是C,且费用是a*x^2，x是流量，问从1运送k流量到n的最小费用

一般做的费用流边的费用都是固定的，而这题并不是固定的。

但是，看到了C<=5，其实就是在提示可以拆边..

假如C是3，我们就可以把一条边拆成3条边。

假如不拆，如果通过的流量是1，2，3，那么费用分别是a，4a，9a

如果拆成3条边，那么3条边的费用分别是a，3a，5a，容量都是1

这样就完美的解决了边的费用问题了~

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cctype>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int MX = 1e2 + 5;
const int MM = 2e5 + 5;
const int INF = 0x3f3f3f3f;

struct Edge {
    int to, next, cap, flow, cost;
    Edge() {}
    Edge(int _to, int _next, int _cap, int _flow, int _cost) {
        to = _to; next = _next; cap = _cap; flow = _flow; cost = _cost;
    }
} E[MM];

int Head[MX], tol;
int pre[MX]; //储存前驱顶点
int dis[MX]; //储存到源点s的距离
bool vis[MX];
int N;//节点总个数，节点编号从0~N-1

void init(int n) {
    tol = 0;
    N = n + 2;
    memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v, int cap, int cost) {
    E[tol] = Edge(v, Head[u], cap, 0, cost);
    Head[u] = tol++;

    E[tol] = Edge(u, Head[v], 0, 0, -cost);
    Head[v] = tol++;
}
bool spfa(int s, int t) {
    queue<int>q;
    for (int i = 0; i < N; i++) {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = Head[u]; i != -1; i = E[i].next) {
            int v = E[i].to;
            if (E[i].cap > E[i].flow && dis[v] > dis[u] + E[i].cost) {
                dis[v] = dis[u] + E[i].cost;
                pre[v] = i;
                if (!vis[v]) {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if (pre[t] == -1) return false;
    else return true;
}

//返回的是最大流， cost存的是最小费用
int minCostMaxflow(int s, int t, int &cost) {
    int flow = 0;
    cost = 0;
    while (spfa(s, t)) {
        int Min = INF;
        for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {
            if (Min > E[i].cap - E[i].flow)
                Min = E[i].cap - E[i].flow;
        }
        for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {
            E[i].flow += Min;
            E[i ^ 1].flow -= Min;
            cost += E[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}

inline int read() {
    char c = getchar();
    while(!isdigit(c)) c = getchar();

    int x = 0;
    while(isdigit(c)) {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

int dist[] = {0, 1, 3, 5, 7, 9};

int main() {
    int n, m, k; //FIN;
    while(~scanf("%d%d%d", &n, &m, &k)) {
        int s = 0, t = n + 1;
        init(t);

        for(int i = 1; i <= m; i++) {
            int u, v, a, c;
            scanf("%d%d%d%d", &u, &v, &a, &c);
            for(int j = 1; j <= c; j++) {
                edge_add(u, v, 1, a * dist[j]);
            }
        }
        edge_add(s, 1, k, 0);
        edge_add(n, t, k, 0);

        int ans = 0;
        if(minCostMaxflow(s, t, ans) != k) {
            printf("-1\n");
        } else printf("%d\n", ans);
    }
    return 0;
}

费用流 hdu3667 Transportation

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原文地址：http://blog.csdn.net/qwb492859377/article/details/47760761