hdu 3667 Transportation 费用流

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很明显的费用流的题目，把一条边拆成5条即可。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=1e2+9;int n,m,p;int head[maxn],lon;struct{    int next,to,c,w,from;}e[222222];void edgeini(){    memset(head,-1,sizeof(head));    lon=-1;}void edgemake(int from,int to,int c,int w){    e[++lon].to=to;    e[lon].from=from;    e[lon].c=c;    e[lon].w=w;    e[lon].next=head[from];    head[from]=lon;}void make(int from,int to,int c,int w){    edgemake(from,to,c,w);    edgemake(to,from,0,-w);}bool inque[maxn];int dist[maxn],from[maxn];int que[1111111];bool spfa(int s,int t){    memset(dist,50,sizeof(dist));    int front=1,end=0;    que[++end]=s;    inque[s]=1;    dist[s]=0;    while(front<=end)    {        int u=que[front++];        inque[u]=0;        for(int k=head[u];k!=-1;k=e[k].next)        {            int v=e[k].to;            if(e[k].c==0) continue;            if(dist[v]>dist[u]+e[k].w)            {                dist[v]=dist[u]+e[k].w;                from[v]=k;                if(!inque[v])                {                    inque[v]=1;                    que[++end]=v;                }            }        }    }    return dist[t]<100000000;}int mincostflow(int s,int t){    int ret=0;    while(spfa(s,t)&&(p--))    {        int u=t;        while(u!=s)        {            int k=from[u];            e[k].c--;            e[k^1].c++;            ret+=e[k].w;            u=e[k].from;        }    }    if(p<=0) return ret;    else return -1;}int main(){//    freopen("in.txt","r",stdin);    while(scanf("%d%d%d",&n,&m,&p)!=EOF)    {        edgeini();        for(int i=1,from,to,w,c;i<=m;i++)        {            scanf("%d%d%d%d",&from,&to,&w,&c);            for(int j=1;j<=c;j++)            {                make(from,to,1,w*(2*j-1));            }        }        int ans=mincostflow(1,n);        cout<<ans<<endl;    }    return 0;}