HDU4539:郑厂长系列故事——排兵布阵(状态压缩)

6 6
0 0 0 0 0 0
0 0 0 0 0 0
0 0 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

2
 
 
 
比较简单的状态压缩
 
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int map[105][15];
int dp[1<<10][1<<10],tem[1<<10][1<<10];
int now[1<<10],pre[1<<10],prepre[1<<10],ans[1<<10];
int l_now,l_pre,l_prepre;
int n,m;

void dfs(int id,int k,int p,int sum)
{
    if(k>=m)
    {
        now[++l_now] = p;
        ans[l_now] = sum;
        return;
    }
    if(k>=2 && map[id][k] && !(p&(1<<(k-2))))//这格可以放，而左边第二格不能放，那么在这个位置放下
        dfs(id,k+1,p|(1<<k),sum+1);
    else if(k<2 && map[id][k])//如果是前两列，这格可以放就先放下
        dfs(id,k+1,p|(1<<k),sum+1);
    dfs(id,k+1,p,sum);//这格不放
}

void solve()
{
    int i,j,k,t;
    tem[1][1] = pre[1] = prepre[1] = 0;
    l_now = l_pre = l_prepre = 1;
    for(k = 0; k<n; k++)
    {
        l_now = 0;
        dfs(k,0,0,0);
        for(i = 1; i<=l_now; i++)
            for(j = 1; j<=l_pre; j++)
                dp[i][j] = 0;
        for(i = 1; i<=l_now; i++)
        {
            for(j = 1; j<=l_pre; j++)
            {
                for(t = 1; t<=l_prepre; t++)
                {
                    if(now[i] & prepre[t]) continue;
                    if(now[i] & (pre[j]>>1)) continue;
                    if(now[i] & (pre[j]<<1)) continue;
                    dp[i][j] = max(dp[i][j],tem[j][t]+ans[i]);
                }
            }
        }
        for(i = 1; i<=l_now; i++)
            for(j = 1; j<=l_pre; j++)
                tem[i][j] = dp[i][j];
        for(i = 1; i<=l_pre; i++)
            prepre[i] = pre[i];
        for(i = 1; i<=l_now; i++)
            pre[i] = now[i];
        l_prepre = l_pre;
        l_pre = l_now;
    }
}

int main()
{

    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        for(i = 0; i<n; i++)
            for(j = 0; j<m; j++)
                scanf("%d",&map[i][j]);
        solve();
        int maxn = 0;
        for(i = 1; i<=l_pre; i++)
            for(j = 1; j<=l_prepre; j++)
                maxn = max(maxn,tem[i][j]);
        printf("%d\n",maxn);
    }

    return 0;
}