hdu_1003_Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 142281    Accepted Submission(s): 33104

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

 

问题的想法是《编程之美》上的，修改了一下提交就ok了。。

思路：

   考虑数组的第一个元素a[0],以及最大的一段数组（a[i].....a[j]）跟a[0]之间的关系，有一下几种情况：

     1. 当0=i=j时，元素a[0]本身构成和最大的一段。

     2.当0=i<j时，和最大的一段以a[0]开始。

     3.当0<i时，元素a[0]跟和最大的一段没有关系。

假设已经知道（a[1]....a[n-1]）中的最大的一段数组之和为All[1]，并且已经知道了（a[1]......a[n-1]）中包含a[1]的和最大的一段数组为start[1]。

那么有以上情况可以得出（a[0].....a[n-1]）中问题的解All[0]是三种情况的最大值max(a[0],a[0]+start[1],All[1]).

so问题符合无后效性，用动态规划方法可解决。
   

 #include<iostream>
 using namespace std;
 int main()
 {
     int t,len,m=,h,n,w,l,r,p,e,i;
     cin>>t;h=t;
     while(t--)
     {
         m++;
         l=r=p=e=;
         cin>>len;
         int a[len];
         for(i=;i<len;i++)
             cin>>a[i];
          n=a[],w=a[];l=;p=;
          for(i=;i<len;i++)
          {
             if(a[i]>n+a[i]){
                 n=a[i];
                 l=i;
                 r=i;
             }
             else {
                 n=n+a[i];
                 r=i;
             }
             if(n>w){
                 w=n;
                 e=r;
                 p=l;
             }
         }
         cout<<"Case "<<m<<":"<<endl<<w<<" "<<p+<<" "<<e+<<endl;
         if(m!=h)cout<<endl;
     }
     return ;
 }