1007_Maximum Subsequence Sum (25分)[动态规划/最大和子序列]

题意: 输出子序列的最大和以及子序列的首尾元素.如果存在多个最大和子序列取位置最靠左的那个.如果序列元素全<0,和为0,输出给定序列的首尾元素.

#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxLen = 10000 + 10;
int val[maxLen], dp[maxLen], front[maxLen], N, maxSum, maxR;

int main()
{
    cin >> N;
    memset(dp, 0, sizeof(dp));
    memset(val, 0, sizeof(val));
    memset(front, 0, sizeof(front));
    for (int i = 0; i < N; i++) {
        cin >> val[i];
        dp[i] = val[i];
        front[i] = i;
    }
    maxSum = dp[0]; maxR = 0; //初始化
    for (int i = 1; i < N; i++) {
        if (dp[i - 1] >= 0) { //动态规划
            dp[i] = dp[i] + dp[i - 1];
            front[i] = front[i - 1];
        }
        if (dp[i] > maxSum) { //记录:最靠左的最大子序列
            maxSum = dp[i];
            maxR = i;
        }
    }
    if (maxSum < 0) {
        cout << 0 << " " << val[0] << " " << val[N - 1] << endl;
    }
    else {
        cout << maxSum << " " << val[front[maxR]] << " " << val[maxR] << endl;
    }
    return 0;
}