洛谷 P1540 乌龟棋

第一感觉是定义状态f[n][i][j][k][kk]，但这样空间和时间都承受不下。我们可以设状态为f[i][j][k][kk]，这样可以省掉一个n，因为我们依据行走步数可以直接算出行走距离.

Code:

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 42;
long long f[maxn][maxn][maxn][maxn];
int val[400], steps[10];
inline int dist(int a,int b,int c,int d)
{
    return a + 2*b + 3*c + 4*d;
}
int main()
{
   // freopen("in.txt","r",stdin);
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 0;i < n;++i)scanf("%d",&val[i]);
    for(int i = 1;i <= m;++i)
    {
        int a;
        scanf("%d",&a);
        ++ steps[a];
    }
    f[0][0][0][0] = val[0];
    for(int i = 0;i <= steps[1]; ++i)
        for(int j = 0;j <= steps[2];++j)
            for(int k = 0;k <= steps[3]; ++k)
                for(int kk = 0; kk <= steps[4]; ++kk)
                {
                    if(i + j + k + kk == 0) continue;
                    int h = dist(i,j,k,kk);
                    if(i >= 1)f[i][j][k][kk] = max(f[i][j][k][kk], f[i-1][j][k][kk] + val[h]);
                    if(j >= 1)f[i][j][k][kk] = max(f[i][j][k][kk], f[i][j-1][k][kk] + val[h]);
                    if(k >= 1)f[i][j][k][kk] = max(f[i][j][k][kk], f[i][j][k-1][kk] + val[h]);
                    if(kk >= 1)f[i][j][k][kk] = max(f[i][j][k][kk], f[i][j][k][kk-1] + val[h]);
                }
    printf("%lld",f[steps[1]][steps[2]][steps[3]][steps[4]]);
    return 0;
}