poj1050--最大子序列和
http://acm.pku.edu.cn/JudgeOnline/problem?id=1050
To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 21973 | Accepted: 11400 |
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.
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
40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2
Sample Output
15
与hdu1003不同的题，但却是与1003相同的思想，具体代码如下：
代码
1
#include
<
stdio.h
>
2
#include
<
stdlib.h
>
3
#include
<
string
.h
>
4
int
n,a[
101
][
101
];
5
int
s[
101
];
6
int
ma(
int
*
p){ ————最大子序列的和，并返回最大值
7
int
i,max
=-
(
1
<<
30
),b;
8
b
=
p[
1
];
9
for
(i
=
2
;i
<=
n;i
++
)
10
{
11
if
(b
<
0
)b
=
p[i];
12
else
b
=
b
+
p[i];
13
if
(max
<
b)max
=
b;
14
}
15
return
max;
16
}
17
18
int
main(){
19
int
i,j,l,k,max
=-
(
1
<<
30
),t;
20
scanf(
"
%d
"
,
&
n);
21
for
(i
=
1
;i
<=
n;i
++
)
22
for
(j
=
1
;j
<=
n;j
++
)
23
scanf(
"
%d
"
,
&
a[i][j]);
24
memset(s,
0
,
sizeof
(s));
25
for
(l
=
1
;l
<=
n;l
++
) —————代表结束行
26
for
(i
=
1
;i
<=
n;i
++
)——————代表起始行
27
{
for
(j
=
i;j
<=
l;j
++
)
28
for
(k
=
1
;k
<=
n;k
++
)
29
s[k]
+=
a[j][k];——————s[k]中存第k列从i到l的所有数的和
30
t
=
ma(s);
31
if
(max
<
t)max
=
t;
32
memset(s,
0
,
sizeof
(s));
33
}
34
printf(
"
%d\n
"
,max);
35
return
0
;
36
}
经验总结：编程过程中，注意运用化未知为已知的数学思维。
本题虽然为矩阵，但思想就是将它转化为一列数，然后再求最大子序列和。但如何转化？
7 -8 9
-4 5 6
1 2 -3
主要是将同一列中的若干数合并。比如，从第一行开始，到第2行结束，每一列的和组成的序列为：
3 -3 15
然后求此序列的最大子序列和。求出后与max比较，最后输出的一定是最大矩阵和。
除了按照程序中按初始位置和结束位置枚举外，还可以枚举每一列中的元素个数和起始位置写循环。