HDU 3667.Transportation 最小费用流

Transportation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3045    Accepted Submission(s): 1318

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 2

1 2 1 2

2 1 2

1 2 1 1

2 2 2

1 2 1 2

1 2 2 2

 

Sample Output

4

-1

3

 

Source

2010 Asia Regional Harbin

 

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3667

题意：求从1运送K个货物到N最少花费，每条边有一个运送上限，运送费用为所有边x*x*w的和，其中x为这条边的运送货物量，w为价格。

思路：最小费用流。但是费用不是与货物成正比，而是与货物的平方成正比，所以不能直接跑最小费用最大流，最后用费用*流量*流量，所以需要建立新的模型。当流量为1是，费用为w，流量为2是，费用是4w，但流量为3时，费用是9w......。所以，可以建立一个这样的模型，跑1流量的花费为w，跑2流量的花费为4w，跑3流量的花费为9w，可以这样建立，直接将容量c的边拆成c条容量为1的边，每条边的费用不一样才能满足要求，费用分别为为w，3w，5w.....，这样的话就会使得满足费用与流量的平方成正比。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define PI acos(-1.0)
const int maxn=1e3+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e13+7;
struct edge
{
    int from,to;
    ll c,w;
};
int n;
vector<edge>es;
vector<int>G[maxn];
ll dist[maxn];
int pre[maxn];
inline void addedge(int u,int v,ll c,ll w)
{
    es.push_back((edge)
    {
        u,v,c,w
    });
    es.push_back((edge)
    {
        v,u,0,-w
    });
    int x=es.size();
    G[u].push_back(x-2);
    G[v].push_back(x-1);
}

bool spfa(int s,int t)
{
    static std::queue<int> q;
    static bool inq[maxn];
    for(int i=0; i<=n+10; i++) dist[i]=INF,inq[i]=false;
    pre[s]=-1;
    dist[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        inq[u]=false;
        for(int i=0; i<G[u].size(); i++)
        {
            edge e=es[G[u][i]];
            if(e.c&&dist[e.to]>dist[u]+e.w)
            {
                pre[e.to]=G[u][i];
                dist[e.to]=dist[u]+e.w;
                if(!inq[e.to]) q.push(e.to),inq[e.to]=true;
            }
        }
    }
    return dist[t]<inf;
}

void dinic(int s,int t,ll f)
{
    ll flow=0,cost=0;
    while(spfa(s,t))
    {
        ll d=f;
        for(int i=t; i!=s; i=es[pre[i]].from)
            d=min(d,es[pre[i]].c);
        f-=d;
        flow+=d;
        cost+=d*dist[t];
        for(int i=t; i!=s; i=es[pre[i]].from)
        {
            es[pre[i]].c-=d;
            es[pre[i]^1].c+=d;
        }
        if(f<=0) break;
    }
    if(f) puts("-1");
    else printf("%lld
",cost);
}

int main()
{
    int m;
    ll k;
    while(~scanf("%d%d%lld",&n,&m,&k))
    {
        for(int i=1; i<=m; i++)
        {
            int u,v;
            ll c,w;
            scanf("%d%d%lld%lld",&u,&v,&w,&c);
            for(ll t=1; t<=c; t++)
                addedge(u,v,1LL,(t*t-(t-1)*(t-1))*w);
        }
        dinic(1,n,k);
        es.clear();
        for(int i=0; i<=n+10; i++) G[i].clear();
    }
    return 0;
}