HDU 3667 Transportation

LRJ老师白书2上讲网络流的例题，只不过没有给出题号，今天偶然发现了这道题。这里，费用随流量的变化而变化，且正好与流量的平方成正比，可以巧妙的将原本1,4,9,16,25的5种不同情况下的费用拆成5条容量为1的边，他们的费用分别为1,3,5,7,9。因为每次增广的时候，肯定会选择去走费用最小的边，比如说走了cost=1的边之后，如果要再走这条路，肯定会去走cost=3的边（不可能去走cost=5,7,9的边），这样一来，把两条边加起来就是flow=2,cost=4。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define maxn 110
#define maxm 50100
#define INF 1<<30
using namespace std;
struct MCMF{
    int src,sink,e,n;
    int first[maxn];
    int cap[maxm],cost[maxm],v[maxm],next[maxm];
    bool flag;
    void init(){
        e = 0;
        memset(first,-1,sizeof(first));
    }

    void add_edge(int a,int b,int cc,int ww){
        //printf("add:%d to %d,cap = %d,cost = %d
",a,b,cc,ww);
        cap[e] = cc;cost[e] = ww;v[e] = b;
        next[e] = first[a];first[a] = e++;
        cap[e] = 0;cost[e] = -ww;v[e] = a;
        next[e] = first[b];first[b] = e++;
    }

    int d[maxn],pre[maxn],pos[maxn];
    bool vis[maxn];

    bool spfa(int s,int t){
        memset(pre,-1,sizeof(pre));
        memset(vis,0,sizeof(vis));
        queue<int> Q;
        for(int i = 0;i <= n;i++)   d[i] = INF;
        Q.push(s);pre[s] = s;d[s] = 0;vis[s] = 1;
        while(!Q.empty()){
            int u = Q.front();Q.pop();
            vis[u] = 0;
            for(int i = first[u];i != -1;i = next[i]){
                if(cap[i] > 0 && d[u] + cost[i] < d[v[i]]){
                    d[v[i]] = d[u] + cost[i];
                    pre[v[i]] = u;pos[v[i]] = i;
                    if(!vis[v[i]])  vis[v[i]] = 1,Q.push(v[i]);
                }
            }
        }
        return pre[t] != -1;
    }

    int Mincost;
    int Maxflow;

    int MinCostFlow(int s,int t,int nn){
        Mincost = 0,Maxflow = 0,n = nn;
        while(spfa(s,t)){
            int min_f = INF;
            for(int i = t;i != s;i = pre[i])
                if(cap[pos[i]] < min_f) min_f = cap[pos[i]];
            Mincost += d[t] * min_f;
            Maxflow += min_f;
            for(int i = t;i != s;i = pre[i]){
                cap[pos[i]] -= min_f;
                cap[pos[i]^1] += min_f;
            }
        }
        return Mincost;
    }
};
MCMF g;

int N,M,K;

int main(){
    while(scanf("%d%d%d",&N,&M,&K) != EOF){
        g.init();
        int S = 0,T = N;
        for(int i = 0;i < M;i++){
            int from,to,a,cap;
            scanf("%d%d%d%d",&from,&to,&a,&cap);
            for(int j = 1;j <= cap;j++)
                g.add_edge(from,to,1,a*(2*j-1));
        }
        g.add_edge(0,1,K,0);
        int ans = g.MinCostFlow(S,T,T);
        if(g.Maxflow != K)  ans = -1;
        printf("%d
",ans);
    }
    return 0;
}