【最小费用可行流模板】

 可能再也用不到了吧,今天整理电脑文件看到的,作为图论选手,留个纪念,

//原图: 对于pi,拆点xi,yi s->S,[m,m],0 S->xi,[0,inf],0 yi->t,[0,inf],0 xi->yi,[vi,vi],0 对于有航线的pi和pj,yi->xj,[0,inf],cost

//这样就建好了原图 那么有源汇有上下界的费用流的改造方法: 首先建立附加源汇ss,tt 对于原图里有的一条边x->y,[l,r],cost,变成x->y,r-l,cost 每一个点的权di定义为所有流入这个点的边的下界和-所有流出这个点的边的下界和 对于一个点i,若di>0,ss->i,di,0;若di<0,i->tt,-di,0 连边t->s,inf,0 然后对ss,tt做最小费用最大流 最终的费用为(网络流中计算的费用+原图中有费用的边的下界*这条边的费用)

//代码
#include<algorithm> 
#include<iostream> 
#include<cstring> 
#include<cstdio>
#include<cmath> 
#include<queue> 
using namespace std;
#define N 40005 
#define inf 2000000000
int n,m,x,mincost,s,t,S,ss,tt;
int tot,point[N],nxt[N],v[N],remain[N],c[N];
int dis[N],last[N],d[N];
bool vis[N];
queue <int> q;
void addedge(int x,int y,int cap,int z) {
	++tot;
	nxt[tot]=point[x];
	point[x]=tot;
	v[tot]=y;
	remain[tot]=cap;
	c[tot]=z;
	++tot;
	nxt[tot]=point[y];
	point[y]=tot;
	v[tot]=x;
	remain[tot]=0;
	c[tot]=-z;
}
int addflow(int s,int t) {
	int now=t,ans=inf;
	while (now!=s) {
		ans=min(ans,remain[last[now]]);
		now=v[last[now]^1];
	}
	now=t;
	while (now!=s) {
		remain[last[now]]-=ans;
		remain[last[now]^1]+=ans;
		now=v[last[now]^1];
	}
	return ans;
}
bool spfa(int s,int t) {
	memset(dis,127,sizeof(dis));
	dis[s]=0;
	memset(vis,0,sizeof(vis));
	vis[s]=1;
	while (!q.empty()) q.pop();
	q.push(s);
	while (!q.empty()) {
		int now=q.front();
		q.pop();
		vis[now]=0;
		for (int i=point[now]; i!=-1; i=nxt[i]) if (dis[v[i]]>dis[now]+c[i]&&remain[i]) {
				dis[v[i]]=dis[now]+c[i];
				last[v[i]]=i;
				if (!vis[v[i]]) {
					vis[v[i]]=1;
					q.push(v[i]);
				}
			}
	}
	if (dis[t]>inf) return 0;
	int flow=addflow(s,t);
	mincost+=flow*dis[t];
	return 1;
}
int main() {
	tot=-1;
	memset(point,-1,sizeof(point));
	scanf("%d%d",&n,&m);
	S=n+n+1,s=S+1,t=s+1;
	ss=t+1,tt=ss+1;
	d[s]-=m,d[S]+=m;
	for (int i=1; i<=n; ++i) {
		scanf("%d",&x);
		addedge(S,i,inf,0);
		addedge(n+i,t,inf,0);
		d[i]-=x,d[n+i]+=x;
	}
	for (int i=1; i<n; ++i) for (int j=i+1; j<=n; ++j) {
			scanf("%d",&x);
			if (x==-1) continue;
			addedge(n+i,j,inf,x);
		}
	for (int i=1; i<=t; ++i) {
		if (d[i]>0) addedge(ss,i,d[i],0);
		if (d[i]<0) addedge(i,tt,-d[i],0);
	}
	addedge(t,s,inf,0);
	while (spfa(ss,tt));
	PRintf("%d\n",mincost);
}


















#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define INF 1e9
const int N=30000;
int tot=-1,nxt[N],point[N],remind[N],c[N],last[N],dis[N],v[N],mincost,d[N];
bool vis[N];
void addline(int x,int y,int cap,int cc) {
	++tot;
	nxt[tot]=point[x];
	point[x]=tot;
	v[tot]=y;
	remind[tot]=cap;
	c[tot]=cc;
	++tot;
	nxt[tot]=point[y];
	point[y]=tot;
	v[tot]=x;
	remind[tot]=0;
	c[tot]=-cc;
}
int addflow(int s,int t) {
	int now=t,ans=INF;
	while (now!=s) {
		ans=min(ans,remind[last[now]]);
		now=v[last[now]^1];
	}
	now=t;
	while (now!=s) {
		remind[last[now]]-=ans;
		remind[last[now]^1]+=ans;
		now=v[last[now]^1];
	}
	return ans;
}
bool spfa(int s,int t) {
	queue<int>q;
	q.push(s);
	memset(dis,0x7f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	dis[s]=0;
	while (!q.empty()) {
		int x=q.front();
		q.pop();
		vis[x]=0;
		for (int i=point[x]; i!=-1; i=nxt[i])
			if (dis[v[i]]>dis[x]+c[i] && remind[i]) {
				last[v[i]]=i;
				dis[v[i]]=dis[x]+c[i];
				if (!vis[v[i]]) vis[v[i]]=1,q.push(v[i]);
			}
	}
	if (dis[t]>INF) return 0;
	int flow=addflow(s,t);
	mincost+=flow*dis[t];
	return 1;
}
int main() {
	memset(point,-1,sizeof(point));
	int x,n,m;
	scanf("%d%d",&n,&m);

	int s=n*2+1,S=s+1,t=S+1,ss=t+1,tt=ss+1;
	addline(s,S,m,0);
//   d[s]-=m; d[S]+=m;
	for (int i=1; i<=n; i++) addline(S,i,INF,0),addline(i+n,t,INF,0);
	for (int i=1; i<=n; i++) scanf("%d",&x),d[i]-=x,d[i+n]+=x;
	for (int i=1; i<n; i++)
		for (int j=i+1; j<=n; j++) {
			scanf("%d",&x);
			if (x!=-1) addline(i+n,j,INF,x);
		}
	for (int i=1; i<=n; i++) addline(i+n,t,INF,0);
	for (int i=1; i<=t; i++) {
		if (d[i]>0) addline(ss,i,d[i],0);
		if (d[i]<0) addline(i,tt,-d[i],0);
	}
	addline(t,s,INF,0);
	while (spfa(ss,tt));
	printf("%d",mincost);
}
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