51nod 1083 矩阵取数问题（基础dp）

先用记忆化搜索写的，然后用两重循环dp的
因为只能向右走或向下走，所以每个格子也只能从上边走过来或者从左边走过来，也即是dp[i][j]=max(dp[i-1][j],dp[i][j-1])+G[i][j];dp[i][j]表示走到第ｉ行第ｊ列的最大值

#include <cstdio>#include <cstring>#include <algorithm>using  namespace  std;const int MAXN = 510;int G[MAXN][MAXN];int dp[MAXN][MAXN];int n;int dfs(int i, int j){    if(dp[i][j] > 0)        return dp[i][j];    if(i == n && j == n)    {        return dp[i][j] = G[i][j];    }    if(i == n && j < n)        return dp[i][j] = G[i][j]+dfs(i,j+1);    else if(i < n && j == n)        return dp[i][j] = G[i][j]+dfs(i+1,j);    else        return dp[i][j] = max(G[i][j]+dfs(i+1,j),G[i][j]+dfs(i,j+1));}/*int main(){    scanf("%d",&n);    for(int i = 1; i <= n; ++i)        for(int j = 1; j <= n; ++j)            scanf("%d",&G[i][j]);    printf("%d\n",dfs(1,1));    return 0;}*/int main(){    scanf("%d",&n);    for(int i = 1; i <= n; ++i)    {        for(int j = 1; j <= n; ++j)            scanf("%d",&G[i][j]);    }    for(int i = 1; i <= n; ++i)    {        for(int j = 1; j <= n; ++j)            dp[i][j] = max(dp[i-1][j],dp[i][j-1])+G[i][j];    }    printf("%d\n",dp[n][n]);    return 0;}