HDOJ 1003 经典DP--求最大字串和

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 242314 Accepted Submission(s): 57209

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

具体代码实现：

/*简单动态规划问题：*/import java.util.*;class Main{public static void main(String args[]){Scanner sc=new Scanner(System.in);int n=sc.nextInt();for(int i=1;i<=n;i++){int m=sc.nextInt();int start=1;//初始化起始位置为1int end=1;//结束位置也是1int max=-1;int sum=0;int t=1;//标记位置为第一项for(int j=1;j<=m;j++){int x=sc.nextInt();sum=sum+x;if(sum>max){max=sum;start=t;end=j;}if(sum<0){t=j+1;//当前面几项的和小于0时，令标记位置为下一项也就是j+1sum=0;}}System.out.println("Case "+i+":"+"\r\n"+max+" "+start+" "+end);if(i!=n){System.out.println();}}}}/*为什么sum<0，就舍弃，重新开始扫描呢？我的想法是：我用start表示子序列的起始下标，end?表示子序列的终止下标。原理是，当我们得到一个子序列，如果子序列的第一个数是非正数，那么可以舍去.当一个子序列的前n个元素和为非正数时，是否也可以舍去呢？答案是可以的。假设k是i到j中任意一个下标。Sum( a, b )表示子序列第a个元素到第b个元素之和。由于加到第j个元素，子序列才开始为负数，所以Sum( i, k ) > 0，Sum( i, k ) + Sum( k, j ) = Sum( i, j ) ,所以Sum( k, j ) < Sum( i, j ) < 0所以如果把?k到j的序列附加到j之后的序列上，只会使序列越来越小。所以i到j的序列都可以舍去。*/

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