Source of the topic: http://acm.nyist.net/JudgeOnline/problem.php?pid=104
max sum
- describe
-
Given a two-dimensional matrix (r*c) composed of integers, it is now necessary to find a sub-matrix of it, so that the sum of all elements in this sub-matrix is the largest, and this sub-matrix is called the largest sub-matrix.
Example:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
Its maximum submatrix is:9 2
-4 1
-1 8
The sum of its elements is 15.- enter
-
Enter an integer n (0<n<=100) in the first line, indicating that there are n groups of test data;
each group of test data:
the first line has two integers r, c (0<r, c<=100), r, c represent the rows and columns of the matrix, respectively;
followed by r rows, each row has c integers;
- output
- Sum of elements of the largest submatrix of the output matrix.
- sample input
-
1 4 4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
- Sample output
-
15
- source
- [Miao Dongdong] Original
- Uploaded by
- Miao Dongdong
这个题一开始我只一直在想怎么把矩阵弄成子矩阵， 后来实在太麻烦， 后来搜了一下答案，发现原来这么简单。
哎，做的题太少，想法太死板，代码借鉴了http://blog.chinaunix.net/uid-22263887-id-1778903.html
记得前面做了一个最大连续子段和的题，这题就跟那个题一个解法，只不过你需要先把矩阵转化为一维的数据
//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，
然后从第二行开始 继续这样操作
//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.
# include <stdio.h>
# include <stdio.h>
# include <memory.h>
const int Max = 100;
int array[Max][Max];
int temp[Max];
int main()
{
int n, r, c, i, j, k, l, b, max_sum;
freopen("in.txt", "r", stdin);
scanf("%d", &n);
while(n--){
scanf("%d %d", &r, &c);
for(i = 0; i < r; i++){
for(j = 0; j < c; j++){
scanf("%d",&array[i][j]);
}
}
max_sum = -100000;
//从第一层开始，每一层都作为一个起点
for(i = 0; i < r; i++){
memset(temp, 0, sizeof(temp));
for(j = i; j < r; j++){
//这里就是那个求最大子段和的算法。
for(l = 0, b = 0; l < c; l++){
temp[l] += array[j][l];
if(b > 0) b += temp[l];
else b = temp[l];
if(b > max_sum) max_sum = b;
}
}
}
printf("%d\n", max_sum);
}
return 0;
}