杭电oj-1003 Max Sum

最新推荐文章于 2024-08-03 14:51:53 发布

原创最新推荐文章于 2024-08-03 14:51:53 发布 · 355 阅读

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本文探讨了最大子序列和问题的两种算法实现，一种为双层循环的超时算法，另一种为优化后的单层循环算法，后者在时间效率上更优，适用于大规模数据处理。

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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

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

分析

又是最大字段和的问题。之前做过浙大PTA有个类似的（1007）。不过杭电这个对时间要求好像更高，同样的算法在杭电这个问题就会超时。

第一版（超时）

#include <iostream>
using namespace std;
int main(){
    int num;
    cin>>num;
    int c = num;
    while(num--){
        int n;
        cin>>n;
        int a[100001];
        int i,j;
        for(i = 0;i<n;i++){
            cin>>a[i];
        }
        int s = a[0],m;
        int p = 0,q = 0;
        for(i = 0;i<n;i++){
            m = a[i];
            for(j = i+1;j<n;j++){
                m += a[j];
                if(m>s&&a[j]!=0) {s = m;p = i;q = j;}
            }
        }
        cout<<"Case "<<c-num<<":"<<endl;
        cout<<s<<" "<<p+1<<" "<<q+1;
        if(num != 0) cout<<endl;
    }
    return 0;
}

第二版（通过）

双重循环改为单层。

#include <iostream>
using namespace std;
int main(){
    int num;
    cin>>num;
    int c = num;
    while(num--){
        int n;
        cin>>n;
        int a[100001];
        int i,j;
        for(i = 0;i<n;i++){
            cin>>a[i];
        }
        int s = a[0],m = 0;
        int p = 0,q = 0,t = 0;
        for(i = 0;i<n;i++){
            m += a[i];
            if(m > s) {
                s = m;
                p = t;
                q = i;
            }
            if(m < 0){
                m = 0;
                t = i+1;
            }
        }
        cout<<"Case "<<c-num<<":"<<endl;
        cout<<s<<" "<<p+1<<" "<<q+1<<endl;
        if(num != 0) cout<<endl;
    }
    return 0;
}