[BZOJ 2055]80人環遊世界 有上下界網絡流

由於每一個節點都具體規定了人數,所以上下界一樣,然後就像可行流一樣建圖後跑最小費用最大流就好了.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#define inf 0x3fffffff
#define maxn 205
#define maxm 500021
using namespace std;
int head[maxn],n,m,in[maxn],tot,ans,vis[maxn],fr[maxn],dis[maxn];
int S,T,s;
queue<int>q;
struct edge{int fr,v,next,w,c;}e[maxm];
void adde(int a,int b,int x,int y){
	e[tot]=(edge){a,b,head[a],x,y};head[a]=tot++;
	e[tot]=(edge){b,a,head[b],0,-y};head[b]=tot++;
}
bool spfa(){
	for(int i=S;i<=T;i++)fr[i]=0,dis[i]=1e9,vis[i]=0;
	q.push(S);dis[S]=0;
	while(!q.empty()){
		int u=q.front();q.pop();
		vis[u]=0;
		for(int v,i=head[u];i!=-1;i=e[i].next){
			if(e[i].w&&dis[v=e[i].v]>dis[u]+e[i].c){
				fr[v]=i;
				dis[v]=dis[u]+e[i].c;
				if(!vis[v]){
					vis[v]=1;
					q.push(v);
				}
			}
		}
	}
	return dis[T]!=1e9;
}
void solve(){
	int Min=1e9;
	for(int x=T;x!=S;x=e[fr[x]].fr){
		Min=min(Min,e[fr[x]].w);
	}
	for(int x=T;x!=S;x=e[fr[x]].fr ){
		ans+=e[fr[x]].c*Min;
		e[fr[x]].w-=Min,e[fr[x]^1].w+=Min;
	}
}
int main(){
	memset(head,-1,sizeof(head));
	scanf("%d%d",&n,&m);
	S=0,s=2*n+1,T=2*n+2;
	for(int x,i=1;i<=n;i++){
		scanf("%d",&x);
		adde(i,i+n,0,0);
		in[i]-=x,in[i+n]+=x;
	}
	adde(S,s,m,0);
	for(int i=1;i<=n;i++)adde(s,i,inf,0);
	for(int i=1;i<=2*n;i++)
		if(in[i]>0)adde(S,i,in[i],0);
		else adde(i,T,-in[i],0);
	for(int i=1;i<=n;i++){
		for(int j=i+1,x;j<=n;j++){
			scanf("%d",&x);
			if(x!=-1)adde(i+n,j,inf,x);
		}
	}
	while(spfa())
		solve();
	printf("%d",ans);
	return 0;
}