方格取数 hdu 1565 最小割

来源：互联网发布：怎么做淘宝网店客服编辑：程序博客网时间：2024/06/05 09:18
选取的格子不相邻，即可以在相邻的格子间连一条流量为inf的边，不能被割去
#include<cstdio>#include<iostream>#include<cstring>#include<queue>using namespace std;#define maxn 444#define maxm 33333#define INF 0x3f3f3f3fint head[maxn],cur[maxn],d[maxn],st[maxm],s,e,no,n;struct point{    int u,v,flow,next;    point(){};    point(int x,int y,int z,int w):u(x),v(y),next(z),flow(w){};}p[maxm];void add(int x,int y,int z){    p[no]=point(x,y,head[x],z);    head[x]=no++;    p[no]=point(y,x,head[y],0);    head[y]=no++;}void init(){    memset(head,-1,sizeof(head));    no=0;}bool bfs(){    int i,x,y;    queue<int>q;    memset(d,-1,sizeof(d));    d[s]=0;    q.push(s);    while(!q.empty())    {        x=q.front();        q.pop();        for(i=head[x];i!=-1;i=p[i].next)        {            if(p[i].flow&& d[y = p[i].v]<0)            {                d[y]=d[x]+1;                if(y==e)                    return true;                q.push(y);            }        }    }    return false;}int dinic(){    int i,loc,top,x=s,nowflow,maxflow=0;    while(bfs()){        for(i=s;i<=e;i++)            cur[i]=head[i];        top=0;        while(true)        {            if(x==e)            {                nowflow=INF;                for(i=0;i<top;i++)                {                    if(nowflow>p[st[i]].flow)                    {                        nowflow=p[st[i]].flow;                        loc=i;                    }                }                for(i=0;i<top;i++)                {                    p[st[i]].flow-=nowflow;                    p[st[i]^1].flow+=nowflow;                }                maxflow+=nowflow;                top=loc;                x=p[st[top]].u;            }            for(i=cur[x];i!=-1;i=p[i].next)                if(p[i].flow&&d[p[i].v]==d[x]+1)                    break;            cur[x]=i;            if(i!=-1)            {                st[top++]=i;                x=p[i].v;            }            else            {                if(!top)                    break;                d[x]=-1;                x=p[st[--top]].u;            }        }    }    return maxflow;}int dx[4] = { -1 , 0 , 1 , 0 } ;int dy[4] = { 0 , -1 , 0 , 1 } ;int N,map[22][22];int main(){    while( ~ scanf("%d", &N)){        init();        int sum = 0;        for(int i = 1 ; i <= N ; i ++)            for(int j = 1 ; j <= N ; j ++)                scanf("%d", &map[i][j]) , sum += map[i][j] ;        s = 0;        e = N * N + 1;        n = N * N;        for(int i = 1 ; i <= N ; i ++)            for(int j = 1 ; j <= N ; j ++){                if((i + j) % 2){ /// odd grids                    add(s , (i - 1) * N + j , map[i][j]);                    for(int op = 0 ; op <= 3 ; op ++ ){                        int x = i + dx[op] , y = j + dy[op] ;                        if(x >= 1 && x <= N && y >= 1 && y <= N )                            add((i - 1) * N + j , (x - 1) * N + y , INF) ;                    }                }                else                    add((i - 1) * N + j , e , map[i][j]); /// even grids            }        printf("%d\n" , sum - dinic());    }    return 0;}
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