uva 1486 - Transportation（最小费用流）

最新推荐文章于 2018-01-21 23:42:37 发布

原创最新推荐文章于 2018-01-21 23:42:37 发布 · 1.2k 阅读

0 ·

CC 4.0 BY-SA版权

图论专栏收录该内容

114 篇文章

订阅专栏

本文介绍了一种解决特定运输问题的方法：通过最小费用最大流算法来计算从城市1到城市N运输K单位货物所需的最低成本。该算法考虑了道路的安全系数及货物运输上限，并通过拆边法将二次费用转化为线性费用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

There are N cities, and M directed roads connecting them. Now you want to transport K units of goods from city 1 to city N. There are many robbers on the road, so you must be very careful. The more goods you carry, the more dangerous it is. To be more specific, for each roadi, there is a coefficient a_i. If you want to carryx units of goods along this road, you should paya_i* x² dollars to hire guards to protect your goods. And what's worse, for each roadi, there is an upper boundC_i, which means that you cannot transport more thanC_i units of goods along this road. Please note you can only carry integral unit of goods along each road.

You should find out the minimum cost to transport all the goods safely.

Input

There are several test cases. The first line of each case contains three integers,N,M and K. (1 $\le$ N $\le$ 100, 1 $\le$ M $\le$ 5000, 0 $\le$ K $\le$ 100). Then M lines followed, each contains four integers(u_i,v_i, a_i,C_i), indicating there is a directed road from cityu_i to v_i, whose coefficient isa_i and upper bound isC_i. (1 $\le$ u_i,v_i $\le$ N, 0 <a_i $\le$ 100,C_i $\le$ 5)

Output

Output one line for each test case, indicating the minimum cost. If it is impossible to transport all theK units of goods, output `-1'.

Sample Input

Sample Output

4 
-1 
3

首先，构图上要用到拆边法，把ax2这样的边，拆为a，3a，5a，7a这样的容量为1的边。然后此题是固定流量的，所以在求最小费用，增广时要检查一下，在flow+a>=k的时候只增广k - flow单位的流量，然后终止程序。

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;

const int maxn = 100+10;
const int INF = 1000000000;

struct Edge{
    int from,to,cap,flow,cost;
};

struct MCMF {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  int inq[maxn];         // 是否在队列中
  int d[maxn];           // Bellman-Ford
  int p[maxn];           // 上一条弧
  int a[maxn];           // 可改进量

  void init(int n) {
    this->n = n;
    for(int i = 0; i < n; i++) G[i].clear();
    edges.clear();
  }

  void AddEdge(int from, int to, int cap, int cost) {
    edges.push_back((Edge){from, to, cap, 0, cost});
    edges.push_back((Edge){to, from, 0, 0, -cost});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
  }

  bool BellmanFord(int k,int s, int t, int &flow,int &cost) {
    for(int i = 0; i < n; i++) d[i] = INF;
    memset(inq, 0, sizeof(inq));
    d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;

    queue<int> Q;
    Q.push(s);
    while(!Q.empty()) {
      int u = Q.front(); Q.pop();
      inq[u] = 0;
      for(int i = 0; i < G[u].size(); i++) {
        Edge& e = edges[G[u][i]];
        if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
          d[e.to] = d[u] + e.cost;
          p[e.to] = G[u][i];
          a[e.to] = min(a[u], e.cap - e.flow);
          if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
        }
      }
    }
    if(d[t] == INF) return false;//s-t不连通，失败退出
    
    if(flow + a[t] >= k){//判断流量是否达到了k
        cost += d[t] * (k-flow);
        flow = k;
        return false;
    }
    
    flow += a[t];
    cost += d[t] * a[t];
    int u = t;
    while(u != s) {
      edges[p[u]].flow += a[t];
      edges[p[u]^1].flow -= a[t];
      u = edges[p[u]].from;
    }
    return true;
  }

  // 需要保证初始网络中没有负权圈
  int Mincost(int &can,int s, int t) {
    int flow = 0,cost = 0;
    while(BellmanFord(can,s, t,flow, cost));
    can = flow;
    return cost;
  }

};

MCMF g;

int main(){
    int n,m,k;
    while(scanf("%d%d%d",&n,&m,&k) != EOF){
        g.init(n);
        int sorce = 0,sink = n-1;
        for(int i = 0;i < m;i++){
            int from,to,a,cap;
            scanf("%d%d%d%d",&from,&to,&a,&cap);
            for(int i = 1;i <= cap;i++){
                g.AddEdge(from-1,to-1,1,a*(2*i-1));
            }
        }
        int can = k;
        int ans = g.Mincost(can,sorce,sink);
        if(can == k)    printf("%d\n",ans);
        else printf("-1\n");
    }
    return 0;
}