Hdu 2571 命运（简单DP）

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=2571
题意：主角要从左上角走到右下角，问能够取得的最大幸运值。行走规则：向下走一格，向右走一格或者走到一列的列数是当前列的整数倍。

思路：dp，由行走规则 写转移方程即可。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100+5
#define LL __int64
#define INF 9999999
int map[25][1005],n,m;
int dp[25][1005];
void init(){
for (int i = 0; i <= m; i++)
dp[0][i] = -INF;
for (int i = 0; i <= n; i++)
dp[i][0] = -INF;
dp[0][1] = 0;
dp[1][0] = 0;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
scanf("%d",&map[i][j]);
init();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++){
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
for (int k = 2; k <= m; k++)
{
if(j % k == 0)
dp[i][j] = max(dp[i][j],dp[i][j/k]);
}
dp[i][j] += map[i][j];
}
printf("%d\n",dp
[m]);
}
return 0;
}