06_最大连续和

问题来源：刘汝佳《算法竞赛入门经典--训练指南》 P61 问题8：

问题描述：给出一个长度为n的序列A1,A2,...,An，求一个连续子序列Ai,Ai+1,...,Aj，使得元素总和最大。

分析：设dp[i]为以i结尾的最大连续和，则d[i] = Max{0,d[i-1]}+Ai;

例题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

例题： hdu 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

题意：给n个数，求一个连续子序列的最大和(若有多个，取最左边的一个)，并求出最大和的开始节点和终节点。

思路：大致还是上面的思路，只不过多了开始节点和终结点，处理下就可以了

代码：

 1 #include "stdio.h"

 2 #include "string.h"

 3 #define N 100005

 4 #define INF 0x3fffffff

 5 

 6 int a[N];

 7 struct node  //用结构体存，比较方便

 8 {

 9     int l,r;

10     int k;

11 }dp[N];

12 

13 int main()

14 {

15     int T,Case;

16     int n;

17     int i,id;

18     scanf("%d",&T);

19     for(Case=1; Case<=T; Case++)

20     {

21         scanf("%d",&n);

22         for(i=1; i<=n; i++)

23             scanf("%d",&a[i]);

24         for(i=0; i<=n; i++)

25             dp[i].l = dp[i].r = dp[i].k = 0;

26         id = 1;

27         dp[1].l = dp[1].r = 1;

28         dp[1].k = a[1];

29         for(i=2; i<=n; i++)

30         {

31             if(dp[i-1].k >= 0)

32             {

33                 dp[i].k = dp[i-1].k+a[i];

34                 dp[i].l = dp[i-1].l;

35                 dp[i].r = i;

36             }

37             else

38             {

39                 dp[i].k = a[i];

40                 dp[i].l = i;

41                 dp[i].r = i;

42             }

43             if(dp[i].k > dp[id].k)

44                 id = i;

45         }

46         printf("Case %d:\n%d %d %d\n",Case,dp[id].k,dp[id].l,dp[id].r);

47         if(Case!=T) printf("\n");  //格式处理

48     }

49     return 0;

50 }