HDU 1569(二分图+最大流)

 

 

根据题目意思 很容易得到一个二分的概念 即只要确定了一个格子 就有能再取的格子和不能再取的格子 2种情况 可以把图化作二分图 相当于在能取的部分再取然后求取的和最大  引入最大流的概念 首先对图进行黑白染色(我们现在设i+j为奇数为黑) 然后连接所有黑的格子旁的白格子 将所有白格子连入虚拟的汇点 将所有黑格子连入虚拟的源点 用数的总和扣去求出的结果（即总和-最大流）

 

代码如下:

#include <iostream>
using namespace std;

const long maxn=3000;
const long maxm=1000000;
const long inf=0x7fffffff;

struct node
{
    long v,next;
    long val;
}s[maxm*2];

long level[maxn],p[maxn],que[maxn],out[maxn],ind;

void init()
{
    ind=0;
    memset(p,-1,sizeof(p));
}

inline void insert(long x,long y,long z)
{
    s[ind].v=y;
    s[ind].val=z;
    s[ind].next=p[x];
    p[x]=ind++;
    s[ind].v=x;
    s[ind].val=0;
    s[ind].next=p[y];
    p[y]=ind++;
}


long max_flow(long n,long source,long sink)
{
    long ret=0;
    long h=0,r=0;
    while(1)
    {
        long i;
        for(i=0;i<n;++i)
            level[i]=0;
        h=0,r=0;
        level[source]=1;
        que[0]=source;
        while(h<=r)
        {
            long t=que[h++];
            for(i=p[t];i!=-1;i=s[i].next)
            {
                if(s[i].val&&level[s[i].v]==0)
                {
                    level[s[i].v]=level[t]+1;
                    que[++r]=s[i].v;
                }
            }
        }
        if(level[sink]==0)break;
        for(i=0;i<n;++i)out[i]=p[i];
        long q=-1;
        while(1)
        {
            if(q<0)
            {
                long cur=out[source];
                for(;cur!=-1;cur=s[cur].next)
                {
                    if(s[cur].val&&out[s[cur].v]!=-1&&level[s[cur].v]==2)
                    {
                        break;
                    }
                }
                if(cur>=0)
                {
                    que[++q]=cur;
                    out[source]=s[cur].next;
                }
                else
                {
                    break;
                }
            }
            long u=s[que[q]].v;
            if(u==sink)
            {
                long dd=inf;
                long index=-1;
                for(i=0;i<=q;i++)
                {
                    if(dd>s[que[i]].val)
                    {
                        dd=s[que[i]].val;
                        index=i;
                    }
                }
                ret+=dd;
                for(i=0;i<=q;i++)
                {
                    s[que[i]].val-=dd;
                    s[que[i]^1].val+=dd;    
                }
                for(i=0;i<=q;i++)
                {
                    if(s[que[i]].val==0)
                    {
                        q=index-1;
                        break;
                    }
                }
            }
            else
            {
                long cur=out[u];
                for(;cur!=-1;cur=s[cur].next)
                {
                    if(s[cur].val&&out[s[cur].v]!=-1&&level[u]+1==level[s[cur].v])
                    {
                        break;
                    }
                }
                if(cur!=-1)
                {
                    que[++q]=cur;
                    out[u]=s[cur].next;
                }
                else
                {
                    out[u]=-1;
                    q--;
                }
            }
        }
    }
    return ret;
}


long Map[60][60];

int main()
{
    long m,n;
    while (scanf("%ld%ld",&m,&n)!=EOF)
    {
        init();
        long i,j;
        long Start=0;
        long End=1;
        long sum=0;

        for (i=0;i<m;++i)
        {
            for (j=0;j<n;++j)
            {
                scanf("%ld",&Map[i][j]);

                sum+=Map[i][j];
                
                long from,to;

                from=i*n+j+2;

                if ((i+j)&1)
                {
                    insert(Start,from,Map[i][j]);

                    if (i>0)
                    {
                        to=(i-1)*n+j+2;
                        insert(from,to,Map[i][j]);
                    }

                    if (j>0)
                    {
                        to=i*n+j-1+2;
                        insert(from,to,Map[i][j]);
                    }

                    if (i<m-1)
                    {
                        to=(i+1)*n+j+2;
                        insert(from,to,Map[i][j]);
                    }

                    if (j<n-1)
                    {
                        to=i*n+j+1+2;
                        insert(from,to,Map[i][j]);
                    }
                }
                else
                {
                    insert(from,End,Map[i][j]);
                }

            }
        }

        printf("%ld\n",sum-max_flow(m*n+2,Start,End));


    }
    return 0;
}