DP专题2 HDOJ 1003 Max Sum

传送门：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): 81597    Accepted Submission(s): 18767

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

Author

Ignatius.L

 

    问题描述：取一个数组中连续的m个数，使其总和最大

    算法分析：DP思路，此题不仅要求输出最大的和而且要求输出他的起始位置和结束位置。我的思路是定义状态temp[]数组，temp[i]表示以i作为尾巴的总和最大的连续数的和，这里的temp[i]可能是i一个数字，也可能是i加上i前面的连续几个数字。

    实现： 首先temp[1] = input[1] ，px[1] = 1 py[1] = 1 对于temp[i](i >= 2) ，如果temp[i-1]>=0 在i-1的基础上加上i必然是temp[i]的最大值，则temp[i] = temp[i-1] + input[i] ，px[i] = px[i-1]  py[i] = i ;如果temp[i-1] < 0 则temp[i] 只能等于input[i]否则会更小。

    所以temp[i] = input[i] + temp[i-1]>=0?temp[i-1]:0 ； 即是新的状态方程

    代码如下：

   

/*Memory: 1768 KB   Time: 31 MS  Language: GCC   Result: Accepted   This source is shared by hust_lcl*/#include <stdio.h>#include <stdlib.h>int input[100010];int temp[100010];int px[100010];int py[100010];int main(){    int n , m , i , flag , k = 1 , j;    scanf("%d",&n);    while(n--)    {        scanf("%d",&m);        for(i = 1 ; i <= m ; i ++)          scanf("%d",&input[i]);        printf("Case %d:\n",k++);        temp[1] = input[1];        px[1] = 1;        py[1] = 1;        for(i = 2 ; i <= m ; i ++ )        {            if(temp[i-1]>=0)              {                  temp[i] = input[i] + temp[i-1];                  px[i] = px[i-1];                  py[i] = i;              }            else              {                  temp[i] = input[i];                  px[i] = i ;                  py[i] = i;              }        }        flag = temp[1];        j = 1;        for(i = 1 ; i <= m ; i++)          if(temp[i] > flag)          {              flag = temp[i];              j = i;          }        printf("%d %d %d\n",flag, px[j],py[j]);        if(n>0) printf("\n");    }    return 0;}