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
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 #define N 100005
 6 #define ll long long
 7 struct DP{
 8     ll sum;
 9     int l;
10     int r;
11     bool operator < (const DP d) const
12     {
13         if(sum!=d.sum) return d.sum<sum;
14         else if(l!=d.l) return l<d.l;
15         else return r<d.r;
16     }
17 }dp[N];
18 ll num[N];
19 int main()
20 {
21     int T;
22     scanf("%d",&T);
23     for(int cnt = 0 ; cnt < T ; cnt++)
24     {
25         int n;
26         scanf("%d",&n);
27         for(int i = 0 ;i < n ;i++)
28             dp[i].sum = 0, dp[i].l = i,dp[i].r = i;
29         for(int i = 0 ;i < n ;i++)
30         {
31             scanf("%lld",&num[i]);
32             if(i==0) dp[i].sum = num[0],dp[i].l = 0,dp[i].r = 0;
33             
34             else 
35             {
36                 if(dp[i-1].sum+num[i]>=num[i])
37                 {
38                     dp[i].sum = dp[i-1].sum+num[i];
39                     dp[i].l = dp[i-1].l;
40                     dp[i].r = i;
41                 }
42                 else 
43                 {
44                     dp[i].sum = num[i];
45                     dp[i].l = i;
46                     dp[i].r = i;
47                 }
48             }
49         }
50             sort(dp,dp+n);
51             if(cnt!=0) puts("");
52             printf("Case %d:\n",cnt+1);
53             printf("%lld %d %d\n",dp[0].sum,dp[0].l+1,dp[0].r+1);
54     }            
55     return 0;
56 }

 

posted on 2015-08-10 10:54  若流芳千古  阅读(391)  评论(0)    收藏  举报

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