1007 Maximum Subsequence Sum (25 分)【动态规划】

1007 Maximum Subsequence Sum (25 分)

Given a sequence of K integers { N​1​​, N​2​​, …, 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

本题为求最大连续子序列和的问题
可以采用进行改进后的穷举法的思路：
1.从头开始扫描数组，用sum记录当前子序列之和，用maxs记录最大连续子序列和
2.如果在扫描中遇到负数，当前子序列和sum将会减小，若sum<0，表面前面扫描的那个子序列可抛弃了（因为负数对于总和只可能拉低总和，不可能增加总和）并重新开始下一个子序列的分析，并置sum = 0；
3.若这个子序列和不断增加，那么最大子序列和maxs也不断增加

本题需要注意的问题是：如果所有数都小于0，那么认为最大的和为0，并且输出首尾元素
因此本题再进行改进，用leftindex和rightindex分别记录最大子序列的起始和结束地址，用tempindex记录当前遍历子序列的起始地址，只有当当前子序列之和大于maxs时，对leftindex，rightindex，以及maxs进行更新。

#include<iostream>
using namespace std;

int num[10010];

int main()
{
    int n;
    cin>>n;
    for(int i = 1;i<=n;i++)
    {
        cin>>num[i];
    }
    int leftindex = 1,rightindex = n,lefttemp = 1;
    int sum = 0,maxs = -1;
    for(int i = 1;i<=n;i++)
    {
        sum+=num[i];
        if(sum<0)
        {
            sum = 0;
            lefttemp = i+1;
        }
        else if(sum>maxs)
        {
            maxs = sum;
            rightindex = i;
            leftindex = lefttemp;
        }
    }
    if(maxs<0)
        maxs = 0;
    cout<<maxs<<" "<<num[leftindex]<<" "<<num[rightindex]<<endl;
    return 0;
}