[BZOJ 2127]happiness(最小割)

Description

高一一班的座位表是个n*m的矩阵,经过一个学期的相处,每个同学和前后左右相邻的同学互相成为了好朋友。这学期要分文理科了,每个同学对于选择文科与理科有着自己的喜悦值,而一对好朋友如果能同时选文科或者理科,那么他们又将收获一些喜悦值。作为计算机竞赛教练的scp大老板,想知道如何分配可以使得全班的喜悦值总和最大。

Solution

文理分科那道一样啊…1A

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
int n,m,s,t,ans=0;
int level[50005],head[50005],cnt=0;
struct Node
{
    int next,to,cap;
}Edges[500005];
int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')f=-1;c=getchar();
    }
    while(c>='0'&&c<='9'){
        x=x*10+c-'0';c=getchar();
    }
    return x*f;
}
void addedge(int u,int v,int c)
{
    Edges[cnt].next=head[u];
    head[u]=cnt;
    Edges[cnt].to=v;
    Edges[cnt++].cap=c;
}
void insert(int u,int v,int c)
{
    addedge(u,v,c);
    addedge(v,u,0);
}
int pos(int x,int y){return (x-1)*m+y;}
queue<int>q;
bool bfs()
{
    memset(level,-1,sizeof(level));
    q.push(s);level[s]=0;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        for(int i=head[u];~i;i=Edges[i].next)
        {
            int v=Edges[i].to;
            if(level[v]==-1&&Edges[i].cap)
            level[v]=level[u]+1,q.push(v);
        }
    }
    if(level[t]==-1)return false;
    return true;
}
int dfs(int u,int f)
{
    if(u==t)return f;
    int flow=0,d;
    for(int i=head[u];~i&&flow<f;i=Edges[i].next)
    {
        int v=Edges[i].to;
        if(level[v]==level[u]+1&&Edges[i].cap)
        {
            d=dfs(v,min(f-flow,Edges[i].cap));
            flow+=d;
            Edges[i].cap-=d;
            Edges[i^1].cap+=d;
        }
    }
    if(!flow)level[u]=-1;
    return flow;
}
int main()
{
    memset(head,-1,sizeof(head));
    n=read(),m=read();
    s=0,t=n*m+1;
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
        int x=read();
        insert(s,pos(i,j),x);
        ans+=x;
    }
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
        int x=read();
        insert(pos(i,j),t,x);
        ans+=x;
    }
    int num=t+1;
    for(int i=1;i<=n-1;i++)
    for(int j=1;j<=m;j++)
    {
        int x=read();
        insert(s,num,x);
        insert(num,pos(i,j),INF);
        insert(num,pos(i+1,j),INF);
        ans+=x;
        num++;
    }
    for(int i=1;i<=n-1;i++)
    for(int j=1;j<=m;j++)
    {
        int x=read();
        insert(num,t,x);
        insert(pos(i,j),num,INF);
        insert(pos(i+1,j),num,INF);
        ans+=x;
        num++;
    }
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m-1;j++)
    {
        int x=read();
        insert(s,num,x);
        insert(num,pos(i,j),INF);
        insert(num,pos(i,j+1),INF);
        ans+=x;
        num++;
    }
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m-1;j++)
    {
        int x=read();
        insert(num,t,x);
        insert(pos(i,j),num,INF);
        insert(pos(i,j+1),num,INF);
        ans+=x;
        num++;
    }
    int d;
    while(bfs())
    {
        while(d=dfs(s,INF))
        ans-=d;
    }
    printf("%d\n",ans);
    return 0;
} 

 

posted @ 2017-05-17 21:36  Zars19  阅读(191)  评论(0编辑  收藏  举报