模板[最小费用最大流]

事实上这个问题是最大费用最大流，为了用最小费用最大流存，把费用改为负去加边。

#include <iostream>#include <algorithm>#include <cstdio>#include <queue>#include<memory.h>using namespace std;const int sz = 75000;const int INF = 200000000;struct Edge{int v,w,capacity;int next;}adj[sz * 8];int head[sz],pre[sz],dist[sz],road[sz],src,dest,k;bool inq[sz];void init(const int &n){src = 1;dest = n;k = 0;fill(head,head+dest+1,-1);}void add(const int &v1,const int &v2,const int &w,const int &capacity) {adj[k].v = v2;adj[k].w = w;adj[k].capacity = capacity;adj[k].next = head[v1];head[v1] = k ++;}void addEdge(const int &v1,const int &v2,const int &w,const int &capacity){add(v1,v2,w,capacity);add(v2,v1,-w,0);}bool SPFA() {fill(dist,dist+dest+1,INF);fill(inq,inq+dest+1,false);inq[src] = true;dist[src] = 0;queue < int > q;q.push(src);while(!q.empty()) {int t = q.front();q.pop();inq[t] = false;int index = head[t];while(index != -1){if(adj[index].capacity > 0 && dist[t] + adj[index].w < dist[adj[index].v]) {dist[adj[index].v] = dist[t] + adj[index].w;pre[adj[index].v] = t;road[adj[index].v] = index;if(!inq[adj[index].v]){q.push(adj[index].v);inq[adj[index].v] = true;}} index = adj[index].next;}}return (dist[dest] != INF);}int minCostMaxFlow(){int minCost = 0,t,nowCost; while(SPFA()){int flow = INF;for(t = dest ; t != src ; t = pre[t] ) {if(adj[road[t]].capacity < flow)flow = adj[road[t]].capacity;}nowCost=0;for(t = dest ; t != src ; t = pre[t] ) {adj[road[t]].capacity -= flow;//positive edge capacity -= flowadj[road[t]^1].capacity += flow;//nagative edge capacity += flow//cout<<flow<<endl;//cout<<adj[road[t]].w<<endl;nowCost += flow * adj[road[t]].w;}if (nowCost<0) minCost+=nowCost;}return minCost;}int a[200];int main(){int n,m,v,i,v1,v2,w,c;while(~scanf("%d%d",&n,&m)){init(n+1);for(i=2;i<=n;i++){scanf("%d",&a[i]);addEdge(i,n+1,-a[i],INF);}for(i=1;i<=m;i++){scanf("%d%d%d%d",&v1,&v2,&c,&w);addEdge(v1,v2,w,c);addEdge(v2,v1,w,c);}printf("%d/n",-minCostMaxFlow());}return 0;}

 

（ZOJ 3362 Beer Problem）