DP（一）HDOJ 1003 Max Sum(java版)

题目链接：点我一下

题目：

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.

输入：

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).

输出

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.

示例输入：

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

示例输出：

Case 1:
14 1 4

Case 2:
7 1 6

思路：

比较简单的dp，很容易推出状态转移方程：sum[i] = max{sum[i-1]+a[i],a[i]}. (sum[i]记录以a[i]为子序列末端的最大连续和.)
然后用一个值记录更新sum[i]的最大值即可。
即对于a[i]这个数字，我们考虑是否将它选入之前连续的序列。
如果选，状态变为sum[i-1]+a[i] ; 如果不选，则从此开始一个新的序列，故和为a[i]。
理解方程后，代码很好写了。

代码：

import java.util.*;    public class Main{     public static void main(String[] args) {        int T = 0 , i = 0, j = 0;        Scanner cin = new Scanner(System.in);        T = cin.nextInt();        for(i=1;i<=T;i++) {            int n = 0;            int temp = -1001, ans = -1001, s = 0, e = 0, ss = 0, ee = 0;            n = cin.nextInt();            for(j=1;j<=n;j++) {                int t = cin.nextInt();                if(temp + t < t) {                    temp = t;                    s = e = j;                }                else {                    temp += t;                    e++;                }                if(ans < temp) {                    ans = temp;                    ss = s;                    ee = e;                }            }            System.out.println("Case " + i + ":");            System.out.println(ans + " " + ss + " " + ee);            if(i != T) {                System.out.println();            }        }    }}