PAT1007:Maximum Subsequence Sum

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原文链接：http://www.cnblogs.com/0kk470/p/7643173.html

本文探讨了最大子序列和问题的解决方案，采用动态规划思想，通过遍历序列并维护最大和及其对应的起始和结束索引，最终输出最大子序列的和及边界元素。

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1007. Maximum Subsequence Sum (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a sequence of K integers { N₁, N₂, ..., N_K }. A continuous subsequence is defined to be { N_i, N_i+1, ..., N_j } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

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

Sample Output:

10 1 4

思路

DP的思想。
这道题最关键点在于:确定最大值子序列的左右索引left和right。那么：
1.从左到右遍历序列时，如果遍历到第i个数时的累加和已经小于0，那么说明从这个数左边开始的所有可能的左半部序列到这个数的和都已经小于0，对于这个数右半部分的序列只减不增，还不如舍弃掉。因此最大和子序列肯定只会在这个数右边。所以暂存下新的左索引templeft = i + 1。
2.当遍历到第i个数时当前最大值tempsum已经大于输出最大值maxsum时，maxsum = tempsum,那么需要更新左右索引left = templeft 和 right = i.
3.对于遍历后maxsum < 0，说明整个序列只可能全是负数，不存在最大正数和的情况，根据题目要求令maxsum = 0,左右索引即为元序列的左右索引。

代码

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    int N;
    while(cin >> N)
    {
      vector<int> nums(N);
      int left = 0,right = N - 1,maxsum = -233,tempsum = 0,templeft = 0;
      for(int i = 0;i < N;i++)
      {
          cin >> nums[i];
          tempsum += nums[i];
          if(tempsum < 0)
          {
              tempsum = 0;
              templeft = i + 1;
          }
          else if (tempsum > maxsum)
          {
              left = templeft;
              maxsum = tempsum;
              right = i;
          }

      }
      if(maxsum < 0)
        maxsum = 0;
      cout << maxsum << " " << nums[left] << " " << nums[right] << endl;
    }
}