ZOJ 1074 To the Max(DP 最大子矩阵和)
To the Max
Time Limit: 2 Seconds Memory Limit: 65536 KB
Problem
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.
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.
Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Output
15
代码如下:
# include<stdio.h>
# include<string.h>
# define N
int main(){
int a[N][N],b[N];
int n,i,j,k;
while(scanf("%d",&n)!=EOF)
{
for(i=;i<n;i++)
for(j=;j<n;j++)
scanf("%d",&a[i][j]);
int max= -;
for(i=;i<n;i++) //第i行到第j行的最大子矩阵和
{
memset(b,,sizeof(b));
for(j=i;j<n;j++)
{
int sum=;
for(k=;k<n;k++)
{
b[k] += a[j][k];
sum += b[k];
if(sum<) sum=b[k];
if(sum>max) max=sum;
}
}
}
printf("%d\n",max);
}
return ;
}
ZOJ 1074 To the Max(DP 最大子矩阵和)的更多相关文章
- ZOJ 1074 To the Max
原题链接 题目大意:这是一道好题.在<算法导论>这本书里面,有一节是介绍如何求最大子序列的.这道题有点类似,区别是从数组变成了矩阵,求最大子矩阵. 解法:完全没有算法功底的人当然不知道最大 ...
- ZOJ 3703 Happy Programming Contest(DP)
题目链接 输出路径,搞了一个DFS出来,主要是这里,浪费了好长时间. #include <cstdio> #include <string> #include <cstr ...
- ZOJ 3211 Dream City(DP)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3374 题目大意:JAVAMAN 到梦幻城市旅游见到了黄金树,黄金树上 ...
- ZOJ 2702 Unrhymable Rhymes(DP)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1702 题目大意:给定有很多数字组成的诗,譬如 “AABB”, “AB ...
- ZOJ 3471 Most Powerful(DP + 状态压缩)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4257 题目大意:有 n(2<=n<=10) 个原子,每两 ...
- ZOJ 2852 Deck of Cards DP
题意: 一一个21点游戏. 1. 有三个牌堆,分别为1X,2X,3X. 2. 纸牌A的值为1,纸牌2-9的值与牌面面相同,10(T).J.Q.K的值为10,而而joke(F)的值为 任意大大. 3. ...
- ZOJ 3623 Battle Ships 简单DP
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3623 题意:给出N种可以建造的船和对方的塔生命值L,每种船给出建造时 ...
- ZOJ 4027 Sequence Swapping(DP)题解
题意:一串括号,每个括号代表一个值,当有相邻括号组成()时,可以交换他们两个并得到他们值的乘积,问你最大能得到多少 思路:DP题,注定想得掉头发. 显然一个左括号( 的最远交换距离由他右边的左括号的最 ...
- ZOJ 3211 Dream City(线性DP)
Dream City Time Limit: 1 Second Memory Limit: 32768 KB JAVAMAN is visiting Dream City and he se ...
随机推荐
- C# WinForm窗体界面设置问题
设置方法: 一:Form对象 属性: 设计中的Name:窗体类的类名AcceptButton:窗口的确定按钮CancelButton:窗口按ESC的取消按钮 1.外观 Backcolor:背景颜色Fo ...
- DATEDIFF()(转)
SQL DATEDIFF 函数 Leave a reply SQL DATEDIFF() 函数用来返回2个时间的差.这个函数在SQL Server和MySQL中都有,但语法上有不同. SQL CASE ...
- Poj 3683-Priest John's Busiest Day 2-sat,拓扑排序
Priest John's Busiest Day Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8872 Accept ...
- 从源码编译rpi的内核
Kernel Building https://www.raspberrypi.org/documentation/linux/kernel/building.md There are two mai ...
- 日常使用 Git 的 19 个建议
如果你对git一无所知,那么我建议先去读一下Git 常用命令速查.本篇文章主要适合有一定 git 使用基础的人群. 目录: 日志输出参数 查看文件的详细变更 查看文件中指定位置的变更 查看尚未合并(m ...
- Package org.xml.sax Description
This package provides the core SAX APIs. Some SAX1 APIs are deprecated to encourage integration(集成:综 ...
- 数位DP问题整理(一)
第一题:Amount of degrees (ural 1057) 题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1057 题意:[x,y ...
- mysql加入�管理员
1.首先用超级管理员登录,然后点击权限button 2.点击加入�新用户,填写登录名和password,全局权限不用选,点击新建用户button 3.编辑新加入�的用户(编辑权限) 4.找到" ...
- Windows XP下安装WinCE6.0开发环境
Windows下怎样编译WinCE6.0及开发应用程序.以下介绍(安装之前必须保证C盘有足够的空间!20g左右!主要是由于在安装程序在安装过程中要解压): 在Visual Studio 2005之前, ...
- 构建tcpdump/wireshark pcap文件
pcap文件格式是bpf保存原始数据包的格式,很多软件都在使用,比如tcpdump.wireshark等等,了解pcap格式可以加深对原始数据包的了解,自己也可以手工构造任意的数据包进行测试. p ...