ural 1146. Maximum Sum(动态规划)

时间:2024-08-07 21:33:14

1146. Maximum Sum

Time limit: 1.0 second Memory limit: 64 MB
Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.
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-hand corner and has the sum of 15.

Input

The input consists of an N × N array of integers. The input begins with a single positive integer Non a line by itself indicating the size of the square two dimensional array. This is followed by N 2integers separated by white-space (newlines and spaces). These N 2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is the sum of the maximal sub-rectangle.

Sample

input output
4
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
15

题意:输入部分第一行是一个正整数N,说明矩阵的大小。下一行后面跟着 个整数。留意这些整数的输入格式可能比较混乱。矩阵是以行优先存储的。N可能和100一样大,矩阵中每个数字的范围为 [-127, 127].

思路1:对矩阵进行穷举。找出所有的子矩阵计算矩阵和并找出最大值。

 #include<iostream>
#include<cstdio>
#include<string>
#include<cstring> using namespace std;
int s[][]= {};
int dp[][]= {}; int main()
{
// freopen("1.txt","r",stdin);
int i,j,ki,kj;
int n;
while(cin>>n)
{
for(i=; i<=n; i++)
for(j=; j<=n; j++)
{
cin>>s[i][j];
s[i][j]+=s[i-][j];
s[i-][j]+=s[i-][j-];
}//输入矩阵并且计算从(1,1)到(i-1,j)的矩阵的和值
for(i=; i<=n; i++)
s[n][i]+=s[n][i-];//计算从(1,1)到(n,i)的矩阵的和值
int max=-;//特别注意因为
for(i=; i<=n; i++)
for(j=; j<=n; j++)
{
max=-;
for(ki=; ki<i; ki++)
for(kj=; kj<j; kj++)
if(max<s[i][j]-s[i][kj]-s[ki][j]+s[ki][kj])//计算(ki,kj)到(i,j)的子矩阵的和并且和max比较;
max=s[i][j]-s[i][kj]-s[ki][j]+s[ki][kj];
dp[i][j]=max;
}
max=-;
for(i=; i<=n; i++)
for(j=; j<=n; j++)
if(max<=dp[i][j])
max=dp[i][j];
cout<<max<<endl;
}
return ;
}

思路2求最大子矩阵的和,先按列进行枚举起始列和结束列,用数据够快速求出每行起始列到结束列之间各数据的和,然后就可以降成一维的,问题就变成了求最大连续子序列了

 #include<cstdio>
#include<cstring>
#include<iostream> using namespace std; #define N 1100
#define INF 0x3f3f3f3f
int s[N][N]= {},dp[N],n; void get_sum(int x ,int y)
{
for(int i=; i<=n; i++)
{
dp[i]=s[y][i]-s[x-][i];
}
return ;
} int DP()
{
int sum=,max=-INF;
for(int i=; i<=n; i++)
{
sum+=dp[i];
max=sum>max?sum:max;
if(sum<) sum=;
}
return max;
} int main()
{
int i,j;
// freopen("1.txt","r",stdin);
while(cin>>n)
{
for(i=; i<=n; i++)
for(j=; j<=n; j++)
{
cin>>s[i][j];
s[i][j]+=s[i-][j];
}//输入矩阵并且计算从(1,j)到(i,j)的矩阵的和值
int max=-INF,ans;
for(i=; i<=n; i++)
for(j=i; j<=n; j++)
{
get_sum(i,j);
ans=DP();
max=ans>max?ans:max;
}
printf("%d\n",max);
}
return ;
}