poj-1050-To the Max
来源:互联网 发布:淘宝试用报名技巧 编辑:程序博客网 时间:2024/05/18 02:22
To the Max
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 48613 Accepted: 25704
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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 -18 0 -2
Sample Output
15
题意:给出一个矩阵,让求出最大子矩阵和
思路:先单独对每行求最大字段和
然后,把第i行后的各行对应列的元素加到第i行的对应列元素,每加一行,就求一次最大字段和,这样就把子矩阵的多行压缩为一行了,一行了就是最大字段和了
也可以这样理解:
如矩阵:
a11 a12 a13
a21 a22 a23
a31 a32 a33
如图,先求第一行最大子段和,再求第一行跟第二行合起来的最大子段和,如a21+a11, a22+a12, a23+a13 的最大子段和,再求第一到第三合起来的最大子段和,如a11+a21+a31, a12+a22+a32, a13+a23+a33的最大子段和,如此以a11为首第一行到最后一行之间的子矩阵可以求出一个最大子矩阵值,接下来第二行最大子段和,第二行和第三行的最大子段和,第三行最大字段和………..以此类推,直到求出整个矩阵的合起来的最大子段和,求出他们之中最大的那个和就是解。
代码:
#include <Stdio.h>int n,max;int map[105][105]; int find(int k)//寻找最大值{ int i,j,t=0; for(i=1;i<=n;i++) { if(t>0){ t+=map[k][i]; }//t>0时加 else t=map[k][i];//小于等于0,t可以等于0也可以等于map[k][i] if(t>max) max=t;//找出最大值 } return max;} int main(){ int i,j,k; max=-0x3f3f3f3f; scanf("%d",&n); for(i=1;i<=n;i++){ for(j=1;j<=n;j++){ scanf("%d",&map[i][j]); } } for(i=1;i<=n;i++)//以i行为首向下 { find(i);//先求出第i行的最大字段和 for(j=i+1;j<=n;j++)//行 { for(k=1;k<=n;k++)//列 { map[i][k]+=map[j][k]; } find(i); } } printf("%d\n",max); return 0;}
- POJ 1050 To the Max
- poj 1050 To the Max
- POJ 1050 To the Max
- poj 1050 To the Max
- Poj 1050 To the Max
- POJ 1050 To the Max
- POJ 1050 To the Max
- POJ 1050 To the Max
- poj 1050 To the Max
- poj 1050 To the Max
- Poj 1050 To the Max
- POJ 1050 To the Max
- poj 1050 to the max
- POJ 1050 To the Max
- poj 1050 to the max
- poj-1050- To the Max
- POJ-1050-To the Max
- POJ 1050 To the Max
- POJ 2486 Apple Tree (树形 dp)
- Python——包管理工具Pip
- 2.2.2微信小程序内容组件 text(文本)
- hdu2059 龟兔赛跑
- 高考与机器学习训练测试
- poj-1050-To the Max
- c学习记录
- 2.2.3微信小程序内容组件 进度条:progress
- 多线程+deque实现生产者消费者模型以及最终实现
- Win7系统64位环境下使用Apache——Apache2.4整合Tomcat与mod_jk
- hdu1159 Common Subsequence
- 2.3.1微信小程序按钮组件:button
- Ordering Tasks (拓扑排序)
- 全排列