HDU1556 方格取数(1) 状态压缩+dp

这题相对于前面的little kings 来说简单了一些，没有了步数的限制，dp方程也简洁了不少。
只需要考虑当前状态是否与上一个状态冲突即可。

代码如下：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#define MAXN 20000
using namespace std;

long long f[25][MAXN];
int N, M, map[25][25], s[MAXN];

inline long long max(long long &x, long long &y)
{
    return x < y ? y : x;
}

void dfs(int l, int state)
{  // 枚举出每一行的所有状态
    if (l == N)
    {
        s[++M] = state;
        return;
    }
    dfs(l+1, state<<1);
    if (!(state&1))
        dfs(l+1, state<<1|1);
}

inline bool judge(int s1, int s2)
{
    if (s[s1] & s[s2])
        return false;
    else
        return true;
}

inline int get(int x, int k)
{
    int ans = 0;
    for (int i = N, a = s[k]; a; --i, a >>= 1)
    {
        if (a & 1)
            ans += map[x][i];
    }
    return ans;
}

void dp()
{
    for( int i = 1; i <= M; ++i )
        f[1][i] = get(1,i);
    for (int i = 2; i <= N; ++i)
    {
        for (int j = 1; j <= M; ++j)
        { 
            int res = get(i, j);
            for (int k = 1; k <= M; ++k)
            { 
                if (judge(j, k))
                {
                    f[i][j] = max(f[i][j], f[i-1][k]+res);
                }
            }
        }
    }
}

int main()
{
    long long ans;
    while (scanf("%d", &N) == 1)
    {
        M = 0; 
        ans = 0;
        dfs(0, 0);
        for (int i = 0; i <= N; ++i)
        {
            for (int j = 0; j <= M; ++j)
                f[i][j] = 0;
        }    
        for (int i = 1; i <= N; ++i)
        {
            for (int j = 1; j <= N; ++j)
            {
                scanf("%d", &map[i][j]);
            }
        }
        dp();
        for (int i = 1; i <= M; ++i)
        {
            ans = max(ans, f[N][i]);
        }
        cout << ans << endl;
    }
    return 0;
}