YT05-動態歸劃求解課後題目-1004—Max Sum -（6.21日-煙臺大學ACM預備隊解題報告）

kewlgrl發表於2015-07-15
ACM

Max Sum

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 10   Accepted Submission(s) : 9

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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

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 

一開始容易想到的都是暴力方法，一個個向後面走，只要和大於max的就記錄，但是會超時


for (i=1;i<= m;i++)  
        {  
            int sum=0;  
            for (int j= i;j<m;j++)  
            {  
                sum+=data [j];  
                if (sum>max)  
                {  
                    max=sum;  
                    point1=i;  
                    point2=j;  
                }  
            }  
        } 


可以用動態規劃，對演算法進行改進，起始還是從頭開始，一直往後搜尋，當遇到和小於０的時候，就對sum重新賦值，並把記錄起始的點變為和為0的後一位，和不小於0的，就繼續搜尋下去.


#include <iostream>
using namespace std;
int main()
{
     int T,n,i,max,num,sum;
     int jilu1,jilu2,qian,hou;
//定義jilu1、jilu2為運算中記錄起始整數位置，qian、hou為最後結果
     cin>>T;
     for(int k=1;k<=T;k++)
     {
         cin>>n;
         sum=max=-1000;
         qian=hou=1;
         jilu1=jilu2=1;
         for(i=1;i<=n;i++)
         {
             cin>>num;
             if(sum>=0)		{jilu2++;sum+=num;}
else if(sum<0)
	 {sum=num;jilu1=i;jilu2=i;}
//當前面集合為負數時，不用將前面數字加入計算和，直接令sum = 新輸入元素

	if(max<sum)         	
	{
	max=sum;
	qian=jilu1;
	hou=jilu2;
	}
 }
cout<<max<<‘ '<<qian<<‘ '<<hou<<endl;
	if(k!=T)
	cout<<endl;	
     }
     return 0;
}


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