BZOJ 3894 文理分科

题解：最小割建模

如果某些元素在一起会得到收益考虑最小割

答案 总收益-最小割

注意：对0点的处理

一开始模型是错的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<cassert>
#include<algorithm>
using namespace std;
const int maxn=100000;
const int oo=1000000000;

int n,m; 
int ans=0;
int totn=0;
int p[200][200];

struct Edge{
	int from,to,cap,flow;
	Edge(){
		from=to=cap=flow=0;
	}
};
vector<int>G[maxn];
vector<Edge>edges;
void Addedge(int x,int y,int z){
	if((x==0)||(y==0))return; 
	Edge e;
	e.from=x;e.to=y;e.cap=z;e.flow=0;
	edges.push_back(e);
	e.from=y;e.to=x;e.cap=0;e.flow=0;
	edges.push_back(e);
	int c=edges.size();
	G[x].push_back(c-2);
	G[y].push_back(c-1);
}

int s,t;
queue<int>q;
int vis[maxn];
int d[maxn];
int Bfs(){
	memset(vis,0,sizeof(vis));
	vis[s]=1;d[s]=0;q.push(s);
	while(!q.empty()){
		int x=q.front();q.pop();
		for(int i=0;i<G[x].size();++i){
			Edge e=edges[G[x][i]];
			if((!vis[e.to])&&(e.cap>e.flow)){
				vis[e.to]=1;
				d[e.to]=d[x]+1;
				q.push(e.to);
			}
		}
	}
	return vis[t];
}

int Dfs(int x,int a){
	if((x==t)||(a==0))return a;
	
	int nowflow=0,f=0;
	for(int i=0;i<G[x].size();++i){
		Edge e=edges[G[x][i]];
		if((d[e.to]==d[x]+1)&&((f=Dfs(e.to,min(a,e.cap-e.flow)))>0)){
			nowflow+=f;
			a-=f;
			edges[G[x][i]].flow+=f;
			edges[G[x][i]^1].flow-=f;
			if(a==0)break;
		}
	}
	return nowflow;
}

int Maxflow(){
	int flow=0;
	while(Bfs())flow+=Dfs(s,oo);
	return flow;
}


//s->Ñ¡ÎÄ
//t->Ñ¡Àí
//hashmapÅжϱßÊÇ·ñ´æÔÚ 
int main(){
	scanf("%d%d",&n,&m);
	
	s=n*m+1;t=n*m+2;totn=t;
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			p[i][j]=(i-1)*m+j;
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			int x;scanf("%d",&x);
			ans+=x;
			Addedge(s,p[i][j],x);
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			int x;scanf("%d",&x);
			ans+=x;
			Addedge(p[i][j],t,x);
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			int x;scanf("%d",&x);
			ans+=x;
			++totn;
			Addedge(s,totn,x);
			Addedge(totn,p[i][j],oo);
			Addedge(totn,p[i][j-1],oo);
			Addedge(totn,p[i][j+1],oo);
			Addedge(totn,p[i-1][j],oo);
			Addedge(totn,p[i+1][j],oo);
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=1;j<=m;++j){
			int x;scanf("%d",&x);
			ans+=x;
			++totn;
			Addedge(totn,t,x);
			Addedge(p[i][j],totn,oo);
			Addedge(p[i][j-1],totn,oo);
			Addedge(p[i][j+1],totn,oo);
			Addedge(p[i-1][j],totn,oo);
			Addedge(p[i+1][j],totn,oo);
		}
	}
	
	printf("%d
",ans-Maxflow());
	return 0;
}