HDU 3667 Transportation 费用流（拆边）

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Transportation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1722    Accepted Submission(s): 684

Problem Description

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 road i, there is a coefficient a_i. If you want to carry x units of goods along this road, you should pay a_i * x² dollars to hire guards to protect your goods. And what’s worse, for each road i, there is an upper bound C_i, which means that you cannot transport more than C_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 <= N <= 100, 1 <= M <= 5000, 0 <= K <= 100). Then M lines followed, each contains four integers (u_i, v_i, a_i, C_i), indicating there is a directed road from city u_i to v_i, whose coefficient is a_i and upper bound is C_i. (1 <= u_i, v_i <= N, 0 < a_i <= 100, C_i <= 5)

 

 

Output

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

 

 

Sample Input

2 1 21 2 1 22 1 21 2 1 12 2 21 2 1 21 2 2 2

 

 

Sample Output

4-13

 

 

Source

2010 Asia Regional Harbin

 

 

Recommend

lcy

 

 

 

题意是说N个村庄，M条路,你需要运送K单位的物品从1号村庄到第N号村庄。路与路之间的费用为a*x*x其中x为流量，a为参数，求最小费用。

要用拆变法，因为求的是最小费用流，如果这条弧的流量是1，走的肯定是cost=1的那条弧，如果流量是2，肯定走的是cost=1和cost=3的那两条弧；如果流量是3，走的肯定是cost为1,3,5那三条。如图所示：

 

 

 

#include<iostream>#include<algorithm>#include<cstring>#include<queue>#include<cstdio>using namespace std;const int MAXN=105;const int inf=1<<29;int pre[MAXN];          // pre[v] = k：在增广路上，到达点v的边的编号为kint dis[MAXN];          // dis[u] = d：从起点s到点u的路径长为dint vis[MAXN];         // inq[u]：点u是否在队列中int path[MAXN];int head[MAXN];int NE,tot,ans,max_flow,map[666][666];int n,m,k;struct node{    int u,v,cap,cost,next;} Edge[100007];void addEdge(int u,int v,int cap,int cost){    Edge[NE].u=u;    Edge[NE].v=v;    Edge[NE].cap=cap;    Edge[NE].cost=cost;    Edge[NE].next=head[u];    head[u]=NE++;    Edge[NE].v=u;    Edge[NE].u=v;    Edge[NE].cap=0;    Edge[NE].cost=-cost;    Edge[NE].next=head[v];    head[v]=NE++;}int SPFA(int s,int t)                   //  源点为0，汇点为sink。{    int i;    for(i=s;i<=t;i++) dis[i]=inf;    memset(vis,0,sizeof(vis));    memset(pre,-1,sizeof(pre));    dis[s] = 0;    queue<int>q;    q.push(s);    vis[s] =1; while(!q.empty())        //  这里最好用队列，有广搜的意思，堆栈像深搜。    {        int u =q.front();        q.pop();        for(i=head[u]; i!=-1;i=Edge[i].next)        {            int v=Edge[i].v;            if(Edge[i].cap >0&& dis[v]>dis[u]+Edge[i].cost)            {                dis[v] = dis[u] + Edge[i].cost;                pre[v] = u;                path[v]=i;                if(!vis[v])                {                    vis[v] =1;                    q.push(v);                }            }        }        vis[u] =0;    }    if(pre[t]==-1)        return 0;    return 1;}void end(int s,int t){    int u, sum = inf;    for(u=t; u!=s; u=pre[u])    {        sum = min(sum,Edge[path[u]].cap);    }    max_flow+=sum;                          //记录最大流    for(u = t; u != s; u=pre[u])    {        Edge[path[u]].cap -= sum;        Edge[path[u]^1].cap += sum;        ans += sum*Edge[path[u]].cost;     //  cost记录的为单位流量费用，必须得乘以流量。    }}int main(){    int i,j,s,t;    while(scanf("%d%d%d",&n,&m,&k)!=EOF)    {        memset(head,-1,sizeof(head));        NE=ans=max_flow=s=0;        int a,b,c,d;        while(m--)        {            scanf("%d%d%d%d",&a,&b,&c,&d);            for(int i=1;i<=d;i++)            {                addEdge(a,b,1,c*(2*i-1));            }        }        addEdge(0,1,k,0);        s=0;t=n;        while(SPFA(s,t))        {            end(s,t);        }        if(max_flow<k)printf("-1\n");        else        printf("%d\n",ans);    }    return 0;}