HDU 1003

最新推荐文章于 2022-11-20 18:24:37 发布

原创最新推荐文章于 2022-11-20 18:24:37 发布 · 478 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#动态规划

ACM-动态规划专栏收录该内容

19 篇文章

订阅专栏

本文针对HDU1003 MaxSum问题，介绍了一种有效的动态规划求解方法，通过示例详细解释如何找出给定序列中最大子序列的和及其位置。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

HDU 1003 动态规划问题，题目的链接如下：
http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 214953 Accepted Submission(s): 50572

Problem Description
Given a sequence a[1],a[2],a[3]……a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

#include <iostream>//写了三次dp函数，结果败了。。百度了一种方法，感觉和自己的差不多、、
//没时间了，不注释了，以后看的时候再来注释。
#include <cstring>
using namespace std;
int arr[100001],dp[100001];
int main() {
    int t, n, i = 1;
    cin >> t;
    while(t--) {
        cin >> n;
        for(int i = 0 ; i < n ; i++)
            cin >> arr[i];
        dp[0] = arr[0];
        int start = 0,end1 = 0,mx = -1001;
        int first = 0,second = 0;
        for(int i=0;i<n;i++){
            if(dp[i-1]>=0){
                dp[i] = dp[i-1]+arr[i];
                end1 = i;
            }
            else{
                dp[i] = arr[i];
                start = end1 = i;
            }
            if(mx<=dp[i]){
                mx = dp[i];
                first = start;
                second = end1;
            }
        }
        cout << "Case "<< i++ << ":" << endl << mx << " " << first + 1 << " " << second + 1 <<endl;
        if(t!=0) cout<<endl;
    }
    return 0;
}