HDU 1003 Max Sum(最大子序列和)

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

思路：这道题也是属于动态规划的题目，要求一个序列中的最大子序列的和，稍微暴力一点的dp是求出每个位置的最大子序列和，（状态转移方程为 dp[i] = max( a[i], dp[i-1]+a[i]）(其中dp[i]为第i个位置的最大子序列和),然后再去找到序列的最大子序列和与开始和结束的位置，当题目问题规模n非常大，在最坏的情况下，很容易超时(上交TLE)。因此必须进行优化。如果当前子序列的前 一 子序列和小于0，则我们可以抛弃子序列和小于0的那一部分（因为是连续的，加上一个小于的数只会使结果更小），以当前位置为当前子序列的起始位置，求出后面的子序列和。

值得注意的是题目中已经说明每个数的取值范围为-1000~1000，那么一开始初始化的最大值tmax必须比-1000小，才能将sum保存于tmax中（不然在整个序列都是负数的时候，就会出现WA）。当子序列的和大于tmax时，需要及时更新开始的位置和结束的位置以及最大值。

AC代码：

#include

int main()
{
    int T;
    int n;
    int temp;
    int tmax;//保存最大值
    int curbegin;//保存当前子序列开始的位置
    int begint;//最大子序列开始位置
    int endt;//最大子序列结束位置
    int sum;//当前子序列的和
    scanf("%d",&T);
    for(int j = 1; j <= T; j++)
    {
        scanf("%d",&n);
        //初始化
        tmax = -1001;
        curbegin = 1;
        begint = endt = 1;
        sum = 0;

        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&temp);
            if(sum < 0)
            {//从新的位置开始
                sum = temp;
                curbegin = i;
            }
            else
            {
                sum += temp;
            }
            if(tmax < sum)
            {//状态及时更新
                tmax = sum;
                begint = curbegin;
                endt = i;
            }
        }

        printf("Case %d:\n",j);
        printf("%d %d %d\n", tmax, begint, endt);
        if(j != T)
            printf("\n");
    }
    return 0;
}