PAT甲级Maximum Subsequence Sum

PAT甲级Maximum Subsequence Sum题解

题目

原题

Given a sequence of K integers A continuous subsequence is defined to be 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.

理解：其实就是求累积和最大的子序列。

输入要求

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.

翻译：每个输入文件包含一个测试点，每个测试点占两行，第一行包含一个小于等于10000的正整数K，第二行包含K个数字，用空格分隔

输出要求

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.

翻译：对于每个测试点来说，在一行中输出最大和，以及最大子序列的头尾元素。数据必须用一个空格分隔，但行尾无多语空格。如果最大子序列不唯一，输出i和j小的（正如用例那辆），如果所有的数字都是负数，name最大的和应该是0，并且你应该输出整个数列的首位元素。

基本思路

DP问题，状态转移方程为DP[i]=max(DP[i-1],v[i]),从中找最大的DP元素即可，麻烦一些的是首位元素的确定：在找最大元素时顺便记录其对应的i（子串的尾部），然后向前探索，每进过一个元素将最大值减去该元素的值，知道为0代表找到了头部。

AC代码

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    int m = -1;
    int end;
    cin >> n;
    vector<int> v(n + 1);
    vector<int> dp(n + 1, 0);
    for (int i = 1; i <= n; i++)
    {
        cin >> v[i];
        dp[i] = max(dp[i - 1] + v[i], v[i]);
        if (dp[i] > m)
        {
            m = dp[i];
            end = i;
        }
    }
    if (m < 0)
    {
        cout << 0 << ' ' << v[1] << ' ' << v[n];
        return 0;
    }
    int begin = end;
    int temp = m;
    for (begin = end; begin > 0; begin--)
    {
        temp -= v[begin];
        if (temp == 0)
        {
            break;
        }
    }
    cout << m << ' ' << v[begin] << ' ' << v[end];
    return 0;
}