ZOJ3362 Beer Problem（最小费用任意流）
题目大概说有n个城市，由m条无向边相连，每条边每天最多运送cap桶酒且其运送一桶的花费是cost。现在从1号城市开始出发运酒，供应到2到n号城市，这些城市的收购单价是price，问最大的盈利是多少。
。。。顺路AC
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<algorithm> 5 using namespace std; 6 #define INF (1<<30) 7 #define MAXN 111 8 #define MAXM 111*222 9 struct Edge{ 10 int u,v,cap,cost,next; 11 }edge[MAXM]; 12 int head[MAXN]; 13 int NV,NE,vs,vt; 14 15 void addEdge(int u,int v,int cap,int cost){ 16 edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost; 17 edge[NE].next=head[u]; head[u]=NE++; 18 edge[NE].u=v; edge[NE].v=u; edge[NE].cap=0; edge[NE].cost=-cost; 19 edge[NE].next=head[v]; head[v]=NE++; 20 } 21 bool vis[MAXN]; 22 int d[MAXN],pre[MAXN]; 23 bool SPFA(){ 24 for(int i=0;i<NV;++i){ 25 vis[i]=0; 26 d[i]=INF; 27 } 28 vis[vs]=1; 29 d[vs]=0; 30 queue<int> que; 31 que.push(vs); 32 while(!que.empty()){ 33 int u=que.front(); que.pop(); 34 for(int i=head[u]; i!=-1; i=edge[i].next){ 35 int v=edge[i].v; 36 if(edge[i].cap && d[v]>d[u]+edge[i].cost){ 37 d[v]=d[u]+edge[i].cost; 38 pre[v]=i; 39 if(!vis[v]){ 40 vis[v]=1; 41 que.push(v); 42 } 43 } 44 } 45 vis[u]=0; 46 } 47 return d[vt]!=INF; 48 } 49 int MCMF(){ 50 int res=0; 51 while(SPFA()){ 52 int flow=INF,cost=0; 53 for(int u=vt; u!=vs; u=edge[pre[u]].u){ 54 flow=min(flow,edge[pre[u]].cap); 55 } 56 for(int u=vt; u!=vs; u=edge[pre[u]].u){ 57 edge[pre[u]].cap-=flow; 58 edge[pre[u]^1].cap+=flow; 59 cost+=flow*edge[pre[u]].cost; 60 } 61 if(cost>=0) break; 62 res+=cost; 63 } 64 return res; 65 } 66 int main(){ 67 int n,m,a,b,c,d; 68 while(~scanf("%d%d",&n,&m)){ 69 vs=1; vt=n+1; NV=vt+1; NE=0; 70 memset(head,-1,sizeof(head)); 71 for(int i=2; i<=n; ++i){ 72 scanf("%d",&a); 73 addEdge(i,vt,INF,-a); 74 } 75 while(m--){ 76 scanf("%d%d%d%d",&a,&b,&c,&d); 77 addEdge(a,b,c,d); 78 addEdge(b,a,c,d); 79 } 80 printf("%d\n",-MCMF()); 81 } 82 return 0; 83 }
