uva 10594（最小费用最大流）

题意：在一个无向网络中，告诉你边的容量与费用。现在需要传送d个数据问你你否能传送成功，若成功则最小费用是多少。

思路：显然是最小费用最大流问题，这道题的见图比较简单。只需要添加一个原点费用为D指向1就行了。接下来的事情就是套模版了。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <utility>
#include <queue>
#include <stack>
#include <vector>
#define LEN 5010
#define ll long long

using namespace std;

//minCostMaxflow by kuangbin
const ll MAXN = 10000;
const ll MAXM = 100000;
const ll INF = 0x7ffffffffffffffLL;
ll tn, m, vcap, data;

struct Edge
{
    ll to,next,cap,flow,cost;
}edge[MAXM];
ll head[MAXN],tol;
ll pre[MAXN],dis[MAXN];
bool vis[MAXN];
ll N;//节点总个数，节点编号从0~N-1
void init(ll n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(ll u,ll v,ll cap,ll cost)
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(ll s,ll t)
{
    queue<ll>q;
    for(ll 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())
    {
        ll u = q.front();
        q.pop();
        vis[u] = false;
        for(ll i = head[u]; i != -1; i = edge[i].next)
        {
            ll v = edge[i].to;
            if(edge[i].cap > edge[i].flow &&
                    dis[v] > dis[u] + edge[i].cost )
            {
                dis[v] = dis[u] + edge[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
ll minCostMaxflow(ll s,ll t,ll &cost)
{
    ll flow = 0;
    cost = 0;
    while(spfa(s,t))
    {
        ll Min = INF;
        for(ll i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for(ll i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    ll a[LEN], b[LEN], val[LEN];
    while(scanf("%lld%lld", &tn, &m)!=EOF){
        init(tn+1);
        for(ll i=0; i<m; i++){
            scanf("%lld%lld%lld", &a[i], &b[i], &val[i]);
        }
        scanf("%lld%lld", &data, &vcap);
        for(ll i=0; i<m; i++){
            addedge(a[i], b[i], vcap, val[i]);
            addedge(b[i], a[i], vcap, val[i]);
        }
        addedge(0, 1, data, 0);
        ll cost, ans = minCostMaxflow(0, N-1, cost);
        if(ans!=data)printf("Impossible.
");
        else printf("%lld
", cost);
    }
    return 0;
}