HDU 1565 方格取数(1)(状态压缩DP）

题目链接

水过的，就是普普通通的枚举上一层和这一层，应该可以加上队列优化，降复杂度。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define eps 1e-9
#define INF -10000000
#define ll __int64
using namespace std;
int dp[21][18000],p[21][21],o[18000];
int main()
{
    int n,i,j,k,u,tem,num;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(i = 1; i <= n; i ++)
        {
            for(j = 1; j <= n; j ++)
                scanf("%d",&p[i][j]);
        }
        num = 1;
        for(i = 1; i < 1<<n; i ++)
        {
            if(i&(i<<1)||i&(i>>1))
            {
                continue;
            }
            dp[1][num] = 0;
            o[num] = i;
            for(j = 0; j <= n-1; j ++)
            {
                if(i&(1<<j))
                {
                    dp[1][num] += p[1][j+1];
                }
            }
            num ++;
        }
        for(i = 2; i <= n; i ++)
        {
            for(k = 1; k < num; k ++)
            {
                for(j = 1; j < num; j ++)
                {
                    if(o[j]&o[k])
                        continue;
                    tem = 0;
                    for(u = 0; u <= n-1; u ++)
                    {
                        if(o[k]&(1<<u))
                            tem += p[i][u+1];
                    }
                    if(dp[i][k] < dp[i-1][j]+tem)
                        dp[i][k] = dp[i-1][j]+tem;
                }
            }
        }
        /*for(i = 1; i <= n; i ++)
        {
            for(j = 1; j < num; j ++)
            {
                printf("%d ",dp[i][j]);
            }
            printf("\n");
        }*/
        tem = 0;
        for(i = 1; i < num; i ++)
        {
            if(tem < dp[n][i])
                tem = dp[n][i];
        }
        printf("%d\n",tem);
    }
    return 0;
}