51nod 1084 矩阵取数问题 V2

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递归式：
if x1 != x2 | dp[step + 1][x1][x2] = max{dp[step][x1’][x2’]} + a[x1][y1] + a[x2][y2]
if x1 == x2 | dp[step + 1][x1][x2] = max{dp[step][x1’][x2’]} + a[x1][y1]。
使用step减少空间使用
如图：
这里写图片描述
初始值：
dp[0][x][y] = 0;

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <map>using namespace std;#define LL long long#define INF 0x3f3f3f3f#define PI acos(-1.0)#define E 2.71828#define MOD 1000000007#define N 210#define M 5010int n,m;int p[N][N];int dp[N*2][N][N];int main(){    scanf("%d%d",&m,&n);    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)            scanf("%d",&p[i][j]);    memset(dp,0,sizeof(dp));    for(int k = 1; k < n+m; k++)    {        for(int i = 1; i<=n && i<=k; i++)        {            for(int j = 1; j<=n && j<=k; j++)            {                dp[k][i][j]=max(dp[k][i][j],dp[k-1][i-1][j-1]+p[i][k-i+1]+(i==j?0:p[j][k-j+1]));                dp[k][i][j]=max(dp[k][i][j],dp[k-1][i-1][j]+p[i][k-i+1]+(i==j?0:p[j][k-j+1]));                dp[k][i][j]=max(dp[k][i][j],dp[k-1][i][j-1]+p[i][k-i+1]+(i==j?0:p[j][k-j+1]));                dp[k][i][j]=max(dp[k][i][j],dp[k-1][i][j]+p[i][k-i+1]+(i==j?0:p[j][k-j+1]));                //printf("dp[%d][%d][%d] = %d\n",k,i,j,dp[k][i][j]);            }        }    }    printf("%d\n",dp[n+m-1][n][n]);    return 0;}

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