Max Sum(hdoj1003)

最大子序列和问题解析

最新推荐文章于 2021-02-19 17:01:32 发布

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最新推荐文章于 2021-02-19 17:01:32 发布

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分类专栏： dp 文章标签：子串

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本文解析了一个经典的编程问题——最大子序列和问题，通过两个示例代码详细介绍了如何求解给定序列中具有最大和的子序列及其起始与结束位置。适用于初学者理解动态规划思想。

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 255008 Accepted Submission(s): 60595

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

Author

Ignatius.L

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注释

代码一中的flag我代表的意思是，当前子串的和。
代码二中的flag我代表的意思是，前一个状态的子串和。
我猜你不明白，不明白就对了。

代码一

#include<cstdio>
#include<cmath>
#include<iostream>
#include<string.h>
using namespace std;
#define MAXN 100001
#define MAXM 220
#define inf 10e22;
int a[MAXN];

int main() {
    int k;
    scanf("%d",&k);
    for (int p = 0; p < k; p++)
    {
        int n;
        scanf("%d", &n);
        for (int j = 0; j < n; j++)
        {
            scanf("%d", &a[j]);
        }
        int max = -100000,flag=0,begin=1,end=1,ansBegin=1,ansEnd=1;
        for (int i = 0; i < n; i++)
        {
            flag = flag + a[i];

            if (flag < 0) {
                begin = i+1;
                end = i+1;
            }
            else {
                end = i + 1;
            }
            if (flag > max)
            {
                max = flag;
                ansBegin = begin;
                ansEnd = end;
            }
            if (flag < 0)
            {
                flag = 0;
                begin++;
            }
        //    flag = flag < 0 ? 0 : flag;
        }

        printf("Case %d:\n",p+1);
        printf("%d %d %d\n",max,ansBegin,ansEnd);
        if (p!=k-1)
        {
            printf("\n");
        }
    }

}

代码二

#include<cstdio>
#include<cmath>
#include<iostream>
#include<string.h>
using namespace std;
#define MAXN 100001
#define MAXM 220
#define inf 10e22;
int a[MAXN];

int main() {
    int k;
    scanf("%d",&k);
    for (int p = 0; p < k; p++)
    {
        int n;
        scanf("%d", &n);
        for (int j = 0; j < n; j++)
        {
            scanf("%d", &a[j]);
        }
        int max = a[0],flag=a[0],begin=1,end=1,ansBegin=1,ansEnd=1;
        for (int i = 1; i < n; i++)
        {
            if (flag >= 0)
            {
                flag = flag + a[i];
                end++;
            }
            else {
                flag = a[i];
                begin = i + 1;
                end = i + 1;
            }
            if (flag>max)
            {
                ansBegin = begin;
                ansEnd = end;
                max = flag;
            }
        //    flag = flag < 0 ? 0 : flag;
        }

        printf("Case %d:\n",p+1);
        printf("%d %d %d\n",max,ansBegin,ansEnd);
        if (p!=k-1)
        {
            printf("\n");
        }
    }

}