【算法学习笔记】24.记忆化搜索 解题报告 SJTU_OJ 1002 二哥种花生

在吴学长的代码上做了一点简化修改，本质一样。在外面铺了一圈0，让代码更简单一点，不用考虑边界情况了。

题目: http://acm.sjtu.edu.cn/OnlineJudge/problem/1002

#include <iostream>
#include <string>
using namespace std;

int main(int argc, char const *argv[])
{
    int L,W;    cin>>L>>W;
    int** sums = new int*[L+1];
    for (int i = 0; i <= L; ++i)    sums[i] = new int[W+1];
    //memset(sums,0,sizeof(sums));
    //input and store the sum
    for (int i = 1; i <= L; ++i)
    {
        for (int j = 1; j <= W; ++j)
        {
            int t; cin>>t;
     　　　　//sum[i][j] can be considered as three components:t,up,left-sidebar
            sums[i][j] = t + sums[i-1][j] + 　　　　　　　　　　　　　　　　(sums[i][j-1] - sums[i-1][j-1]);
        }
    }
    //move the scope and calculate the maximum
    int a,b; cin>>a>>b;
    int max=0;
    for (int i = a; i <= L; ++i)
    {
        for (int j = b; j <= W; ++j)
        {
            //sum - left - up
            int tmp = 0;
            tmp = sums[i][j] - sums[i][j-b] 　　　　　　　　　　　　　　- (sums[i-a][j] - sums[i-a][j-b]);
            if(tmp > max)
                max = tmp;
        }
    }
    cout<<max<<endl;
    for (int i = 0; i < L+1; ++i)
    {
        delete[] sums[i];
    }
    delete[] sums;
    return 0;
}