codevs 1068 乌龟棋(记忆化搜索)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 55;
int dp[maxn][maxn][maxn][maxn];
int n, m, a[505];
int cnt1, cnt2, cnt3, cnt4;

int dfs(int c1, int c2, int c3, int c4)
{
if(!(c1+c2+c3+c4)) return a[1];
if(dp[c1][c2][c3][c4]) return dp[c1][c2][c3][c4];
if(c1) dp[c1][c2][c3][c4] = max(dp[c1][c2][c3][c4], dfs(c1-1, c2, c3, c4)+a[c1*1+c2*2+c3*3+c4*4+1]);
if(c2) dp[c1][c2][c3][c4] = max(dp[c1][c2][c3][c4], dfs(c1, c2-1, c3, c4)+a[c1*1+c2*2+c3*3+c4*4+1]);
if(c3) dp[c1][c2][c3][c4] = max(dp[c1][c2][c3][c4], dfs(c1, c2, c3-1, c4)+a[c1*1+c2*2+c3*3+c4*4+1]);
if(c4) dp[c1][c2][c3][c4] = max(dp[c1][c2][c3][c4], dfs(c1, c2, c3, c4-1)+a[c1*1+c2*2+c3*3+c4*4+1]);
return dp[c1][c2][c3][c4];
}

int main(void)
{
while(cin >> n >> m)
{
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
cnt1 = cnt2 = cnt3 = cnt4 = 0;
for(int i = 1; i <= m; i++)
{
int t;
scanf("%d", &t);
if(t == 1) cnt1++;
if(t == 2) cnt2++;
if(t == 3) cnt3++;
if(t == 4) cnt4++;
}
printf("%d\n", dfs(cnt1, cnt2, cnt3, cnt4));
}
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<cmath>
using namespace std;
typedef long long ll;
const ll maxn = 505;
int dp[55][55][55][55];
int a[maxn];

int main(void)
{
int n, m;
while(cin >> n >> m)
{
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int cnt1, cnt2, cnt3, cnt4;
cnt1 = cnt2 = cnt3 = cnt4 = 0;
for(int i = 1; i <= m; i++)
{
int t;
scanf("%d", &t);
if(t == 1) cnt1++;
if(t == 2) cnt2++;
if(t == 3) cnt3++;
if(t == 4) cnt4++;
}
dp[0][0][0][0] = a[1];
for(int i = 0; i <= cnt1; i++)
for(int j = 0; j <= cnt2; j++)
for(int k = 0; k <= cnt3; k++)
for(int l = 0; l <= cnt4; l++)
{
int val = a[1+i+j*2+k*3+l*4];
if(i) dp[i][j][k][l] = max(dp[i][j][k][l], dp[i-1][j][k][l]+val);
if(j) dp[i][j][k][l] = max(dp[i][j][k][l], dp[i][j-1][k][l]+val);
if(k) dp[i][j][k][l] = max(dp[i][j][k][l], dp[i][j][k-1][l]+val);
if(l) dp[i][j][k][l] = max(dp[i][j][k][l], dp[i][j][k][l-1]+val);
}
printf("%d\n", dp[cnt1][cnt2][cnt3][cnt4]);
}
return 0;
}