Max Sum(dp)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178388    Accepted Submission(s): 41628

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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题意 ： 给出一个序列，输出最大子序列和，简单的dp

dp数组储存的是以这个值结尾的最长的子序列和

dp[i] = max(dp[i-1]+num[i] , num[i]);

但是因为要保存起始和终止点的位置，所以可以用结构体来储存dp；

下面是代码

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 #define N 100005
 #define ll long long
 struct DP{
     ll sum;
     int l;
     int r;
     bool operator < (const DP d) const
     {
         if(sum!=d.sum) return d.sum<sum;
         else if(l!=d.l) return l<d.l;
         else return r<d.r;
     }
 }dp[N];
 ll num[N];
 int main()
 {
     int T;
     scanf("%d",&T);
     for(int cnt =  ; cnt < T ; cnt++)
     {
         int n;
         scanf("%d",&n);
         for(int i =  ;i < n ;i++)
             dp[i].sum = , dp[i].l = i,dp[i].r = i;
         for(int i =  ;i < n ;i++)
         {
             scanf("%lld",&num[i]);
             if(i==) dp[i].sum = num[],dp[i].l = ,dp[i].r = ;

             else
             {
                 if(dp[i-].sum+num[i]>=num[i])
                 {
                     dp[i].sum = dp[i-].sum+num[i];
                     dp[i].l = dp[i-].l;
                     dp[i].r = i;
                 }
                 else
                 {
                     dp[i].sum = num[i];
                     dp[i].l = i;
                     dp[i].r = i;
                 }
             }
         }
             sort(dp,dp+n);
             if(cnt!=) puts("");
             printf("Case %d:\n",cnt+);
             printf("%lld %d %d\n",dp[].sum,dp[].l+,dp[].r+);
     }
     return ;
 }