HDU-1003-Max Sum(DP)

最新推荐文章于 2022-11-06 15:03:56 发布

Prim233

最新推荐文章于 2022-11-06 15:03:56 发布

阅读量332

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划文章标签： dp

本文链接：https://blog.youkuaiyun.com/qq_32680617/article/details/52760120

动态规划专栏收录该内容

70 篇文章

订阅专栏

本文介绍了一种求解整数序列中具有最大和的连续子序列的问题，并提供了一个具体的算法实现。通过动态规划方法，文章详细展示了如何计算最大子序列和及其起始和结束位置。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 221844 Accepted Submission(s): 52151

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

跟HDU-1231好像

代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
using namespace std;
const int maxn=100005;
const int INF=0x3f3f3f3f;
struct node
{
    int sum;
    int left;
    int right;
} DP[maxn];
int num[maxn];
int main()
{
    int T;
    scanf("%d",&T);
    for(int casen=1; casen<=T; casen++)
    {
        int N;
        scanf("%d",&N);
        for(int i=1; i<=N; i++)
            scanf("%d",&num[i]);
        DP[1].sum=num[1];
        DP[1].left=1;
        DP[1].right=1;
        int result=num[1];
        int left=1;
        int right=1;
        for(int i=2; i<=N; i++)
        {
            if(DP[i-1].sum+num[i]>=num[i])
            {
                DP[i].sum=DP[i-1].sum+num[i];
                DP[i].left=DP[i-1].left;
                DP[i].right=i;
            }
            else
            {
                DP[i].sum=num[i];
                DP[i].left=i;
                DP[i].right=i;
            }
            if(DP[i].sum>result)
            {
                result=DP[i].sum;
                left=DP[i].left;
                right=DP[i].right;
            }
        }
        casen==T?printf("Case %d:\n%d %d %d\n",casen,result,left,right):printf("Case %d:\n%d %d %d\n\n",casen,result,left,right);
    }
    return 0;
}