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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。
   
 1 #include<iostream> 2 using namespace std; 3 int main() 4 { 5     int t,len,m=0,h,n,w,l,r,p,e,i; 6     cin>>t;h=t; 7     while(t--) 8     { 9         m++;10         l=r=p=e=0;11         cin>>len;12         int a[len];13         for(i=0;i<len;i++)14             cin>>a[i];15          n=a[0],w=a[0];l=0;p=0;16          for(i=1;i<len;i++)17          {18             if(a[i]>n+a[i]){19                 n=a[i];20                 l=i;21                 r=i;22             }23             else {24                 n=n+a[i];25                 r=i;26             }27             if(n>w){28                 w=n;29                 e=r;30                 p=l;31             }32         }33         cout<<"Case "<<m<<":"<<endl<<w<<" "<<p+1<<" "<<e+1<<endl;34         if(m!=h)cout<<endl;35     }36     return 0;37 }