●BOZJ 2127 happiness

题链：

http://www.lydsy.com/JudgeOnline/problem.php?id=2127

题解：

和 BZOJ 3984 建图类似（最小割模型）。
但是这个建图方法效率有点低。
另外这里这个解法比较高效，写得很好很清晰：
http://blog.youkuaiyun.com/hzj1054689699/article/details/53038620

代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXN 65000
#define MAXM 300000
#define INF 0x3f3f3f3f
using namespace std;
struct Edge{
	int to[MAXM],cap[MAXM],nxt[MAXM],head[MAXN],ent;
	void Init(){
		ent=2; memset(head,0,sizeof(head));
	}
	void Adde(int u,int v,int w){
		to[ent]=v; cap[ent]=w; nxt[ent]=head[u]; head[u]=ent++;
		to[ent]=u; cap[ent]=0; nxt[ent]=head[v]; head[v]=ent++;
	}
	int Next(int i,bool type){
		return type?head[i]:nxt[i];
	}
}E;
int cur[MAXN],d[MAXN];
int N,M,S,T,ans;
int idx(int i,int j,int k){
	return (i-1)*M+j+k*N*M;
}
bool bfs(){
	queue<int>q; int u,v;
	memset(d,0,sizeof(d));
	d[S]=1; q.push(S);
	while(!q.empty()){
		u=q.front(); q.pop();
		for(int i=E.Next(u,1);i;i=E.Next(i,0)){
			v=E.to[i];
			if(d[v]||!E.cap[i]) continue;
			d[v]=d[u]+1; q.push(v);
		}
	}
	return d[T];
}
int dfs(int u,int reflow){
	if(u==T||!reflow) return reflow;
	int flowout=0,f,v;
	for(int i=E.Next(u,1);i;i=E.Next(i,0)){
		v=E.to[i];
		if(d[v]!=d[u]+1) continue;
		f=dfs(v,min(reflow,E.cap[i]));
		flowout+=f; E.cap[i^1]+=f;
		reflow-=f;	E.cap[i]-=f;
		if(!reflow) break;
	}
	if(!flowout) d[u]=0;
	return flowout;
}
int Dinic(){
	int flow=0;
	while(bfs()){
		memcpy(cur,E.head,sizeof(E.head));
		flow+=dfs(S,INF);
	}
	return flow;
}
int main()
{
	E.Init();
	scanf("%d%d",&N,&M);
	S=N*M*5+1; T=N*M*5+2;
	for(int i=1,x;i<=N;i++)
		for(int j=1;j<=M;j++)
			scanf("%d",&x),ans+=x,E.Adde(S,idx(i,j,0),x);
	for(int i=1,x;i<=N;i++)
		for(int j=1;j<=M;j++)
			scanf("%d",&x),ans+=x,E.Adde(idx(i,j,0),T,x);
	for(int i=1,_i,_j,x;i<N;i++)
		for(int j=1;j<=M;j++){
			scanf("%d",&x); ans+=x;
			E.Adde(S,idx(i,j,1),x);
			E.Adde(idx(i,j,1),idx(i,j,0),INF);
			E.Adde(idx(i,j,1),idx(i+1,j,0),INF);
		}
	for(int i=1,_i,_j,x;i<N;i++)
		for(int j=1;j<=M;j++){
			scanf("%d",&x); ans+=x;
			E.Adde(idx(i,j,2),T,x);
			E.Adde(idx(i,j,0),idx(i,j,2),INF);
			E.Adde(idx(i+1,j,0),idx(i,j,2),INF);
		}
	for(int i=1,_i,_j,x;i<=N;i++)
		for(int j=1;j<M;j++){
			scanf("%d",&x); ans+=x;
			E.Adde(S,idx(i,j,3),x);
			E.Adde(idx(i,j,3),idx(i,j,0),INF);
			E.Adde(idx(i,j,3),idx(i,j+1,0),INF);
		}
	for(int i=1,_i,_j,x;i<=N;i++)
		for(int j=1;j<M;j++){
			scanf("%d",&x); ans+=x;
			E.Adde(idx(i,j,4),T,x);
			E.Adde(idx(i,j,0),idx(i,j,4),INF);
			E.Adde(idx(i,j+1,0),idx(i,j,4),INF);
		}
	int der=Dinic();
	ans-=der;
	printf("%d",ans);
	return 0;
}

转载于:https://www.cnblogs.com/zj75211/p/7931331.html