HDU1231最大连续子序列&Uva108Maximum Sum最大子矩阵(尺取法)

HDU1231：

给定K个整数的序列{ N1, N2, …, NK }，其任意连续子序列可表示为{ Ni, Ni+1, …,

Nj }，其中 1 <= i <= j <= K。最大连续子序列是所有连续子序列中元素和最大的一个，

例如给定序列{ -2, 11, -4, 13, -5, -2 }，其最大连续子序列为{ 11, -4, 13 }，最大和

为20。

在今年的数据结构考卷中，要求编写程序得到最大和，现在增加一个要求，即还需要输出该

子序列的第一个和最后一个元素。

Input

测试输入包含若干测试用例，每个测试用例占2行，第1行给出正整数K( < 10000 )，第2行给出K个整数，中间用空格分隔。当K为0时，输入结束，该用例不被处理。

Output

对每个测试用例，在1行里输出最大和、最大连续子序列的第一个和最后一个元

素，中间用空格分隔。如果最大连续子序列不唯一，则输出序号i和j最小的那个（如输入样例的第2、3组）。若所有K个元素都是负数，则定义其最大和为0，输出整个序列的首尾元素。

Sample Input

6

-2 11 -4 13 -5 -2

10

-10 1 2 3 4 -5 -23 3 7 -21

6

5 -8 3 2 5 0

1

10

3

-1 -5 -2

3

-1 0 -2

0

Sample Output

20 11 13

10 1 4

10 3 5

10 10 10

0 -1 -2

0 0 0

Huge input, scanf is recommended.

Hint

Hint

题意：给出一行数字，找出一串和最大的连续的子序列。

解题思路：模拟心中的想法。当当前子序列的和小于0时这串子序列就不再继续往下扩展了，此时子序列的开头应从下一个元素开始了。如5 -8 3 2 5 0，当以5为序列开头，计算到5-8=-3<0时就不再继续扩招将3加入子序列了（想一想为什么）而是将3作为子序列新的开头往后判断。

#include<stdio.h>
int main()
{
int a[10010],t,sum,max,left,right,i,n;
while(~scanf("%d",&n)&&n)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
t=a[0];
left=right=a[0];
sum=0;
max=a[0];
for(i=0;i<n;i++)
{
if(sum<0)
{
t=a[i];
sum=a[i];
}
else
{
sum+=a[i];

}
if(sum>max)
{
max=sum;
left=t;
right=a[i];
}
}
if(max<0) printf("0 %d %d\n",a[0],a[n-1]);
else printf("%d %d %d\n",max,left,right);
}
return 0;
}

Uva108：

A problem that is simple to solve in one dimension is often much more difficult to solve in more than

one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each

conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT

is solved quite efficiently, however. In contrast, some problems belong to the same complexity class

regardless of the dimensionality of the problem.

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 subrectangle

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:

9 2

−4 1

−1 8

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 N on a line by itself indicating the size of the square

two dimensional array. This is followed by N2

integers separated by white-space (newlines and spaces).

These N2

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

4

0 -2 -7 0 9 2 -6 2

-4 1 -4 1 -1

8 0 -2

Sample Output

15

题意：给出一个矩阵，找出矩阵中和最大的子矩阵。

解题思路:着重思考一下这道题。受上道题启发，可不可以将矩阵也用处理序列的方法来解决这道题？ 举个例子：



（图虽然有点粗糙但是应该可以看得懂~）

将输入的数字进行以下处理：a[i][j]+=a[i-1][j];

使a数组中每个数表示到该行为止该列的和，如图中5对应0+9+（-4）

通过三个for循环使得每一块子矩阵都可以被表示。

再使用上题中的思路模拟求解。

#include<stdio.h>
int main()
{
int a[110][110],n,t,i,j,k,max,sum;
while(~scanf("%d",&n))
{
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
scanf("%d",&a[i][j]);
a[i][j]+=a[i-1][j];
}
max=a[1][1];
for(i=0;i<=n;i++)
{
for(j=i;j<=n;j++)
{
sum=0;
for(k=1;k<=n;k++)
{
if(sum<0) sum=0;
else if(i!=j) sum+=a[j][k]-a[i][k];
if(sum>max&&sum!=0) max=sum;
}
}
}
printf("%d\n",max);
}
return 0;
}