动态规划：HDU1003-Max Sum（最大子序列和）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 81735    Accepted Submission(s): 18797

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

解题心得：

1、这个题要求的是一个在序列中的一串连续的子序列，要求子序列的和最大，并且要求记录子序列的起始位置和终止位置。

2、可以将前面的数加在一起看作一个数，当前面的数为负数的时候则舍去，将当前这个数作为新的起点。用一个变量Max记录一下最大的值，当最大值被替换的时候记录一下终点和起点。理解了还是很容易的。

#include<bits/stdc++.h>using namespace std;const int maxn = 1e5+100;int num[maxn],dp[maxn];int Start,End,now_start,Max;int main(){    int t;    int T;    scanf("%d",&t);    T = t;    while(t--)    {        memset(dp,0,sizeof(dp));        num[0] = -1;        now_start = 1;        Max = -1000000;        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&num[i]);        for(int i=1;i<=n;i++)        {            if(num[i] + dp[i-1] >= num[i])//前面的数不是负数                dp[i] = num[i] + dp[i-1];            else            {                dp[i] = num[i];//前面一个数是负数，将现在这个数作为新的起点                now_start = i;//当前起点，当出现最大值的用得到            }            if(dp[i] >= Max)//出现最大值，记录最大值，终点和起点            {                Start = now_start;                End = i;                Max = dp[i];            }        }        printf("Case %d:\n",T-t);        if(t != 0)            printf("%d %d %d\n\n",Max,Start,End);        else            printf("%d %d %d\n",Max,Start,End);    }    return 0;}