zoukankan      html  css  js  c++  java
  • hdu 1081 To The Max

    Problem Description
    Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
    As an example, the maximal sub-rectangle of the array:
    0 -2 -7 0
    9 2 -6 2
    -4 1 -4 1
    -1 8 0 -2
    is in the lower left corner:
    9 2
    -4 1
    -1 8
    and has a sum of 15.
    Input
    The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
    Output
    Output the sum of the maximal sub-rectangle. 
    Sample Input
    4
    0 -2 -7 0
    9 2 -6 2
    -4 1 -4 1
    -1 8 0 -2 
    Sample Output
    15
    这个题是求一个矩阵的最大子矩阵,我因为没有初始化maxx,WA了好多次,教训
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<iostream>
    using namespace std;
    int maxx;
    int num[1001];
    int map[1001][1001];
    int top;
    void dos()
    {
        int ns,i;
        ns=0;
        for(i=0;i<top;i++)
        {
            ns+=num[i];
            if(ns<0)ns=0;
            maxx=max(maxx,ns);
        }
    }
    int main()
    {
        int i,j,n,x,k;
        while(scanf("%d",&n)!=EOF)
        {
            memset(map,0,sizeof(map));
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                {
                    scanf("%d",&x);
                    map[i][j]=map[i][j-1]+x;
                }
            }
            maxx=-99999999;
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=i;j++)
                {
                    top=0;
                    for(k=1;k<=n;k++)
                    {
                        num[top++]=map[k][i]-map[k][j-1];
                    }
                    dos();
                }
            }
            cout<<maxx<<endl;
        }
        return 0;
    }
    至少做到我努力了
  • 相关阅读:
    【校招面试 之 C/C++】第1题 为什么优先使用构造函数的初始化列表
    Linux平台ORACLE INSTANT客户端安装
    ORACLE数据库链接
    ORACLE用户管理
    【转】软件开发工具介绍之 6.Web开发工具
    【转】三大UML建模工具Visio、Rational Rose、PowerDesign的区别
    ORACLE数据库查看执行计划
    数据分析方法
    ORACLE对象大小写问题
    计算机改名引发的ORA
  • 原文地址:https://www.cnblogs.com/chensunrise/p/3687211.html
Copyright © 2011-2022 走看看