码迷,mamicode.com
首页 > 编程语言 > 详细

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

时间:2015-03-31 19:59:29      阅读:140      评论:0      收藏:0      [点我收藏+]

标签:

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; }

 

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

标签:

原文地址:http://www.cnblogs.com/yuchenlin/p/4381691.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!