HDU-1003 Max Sum-动态规划-难度2

Max Sum


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 129864    Accepted Submission(s): 30100




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

代码:

/*HDU 1003 Max Sum *//* *其实就是求最大子数组之和,不过需要保存起始点和结束点 */#include <cstdio>#include <iostream>#include <climits>using namespace std;const int  MAXN = 100002;int dp[MAXN], first[MAXN];//first[i]保存dp[i]的起始点的位置,结束点的位置是他自己iint main(){#ifdef _LOCALfreopen("F://input.txt", "r", stdin);#endifint T, n;cin >> T;    for(int j = 1; j <= T; j++){cin >> n;int temp, position = 0;dp[0] = INT_MIN; //初始化边界条件,有负数的最小值不能是0或-1for(int i = 1; i <= n; i++){cin >> temp;if(dp[i - 1] >= 0){dp[i] = dp[i - 1] + temp;first[i] = first[i - 1];}else{dp[i] = temp;first[i] = i;}dp[i] = dp[i - 1] > 0 ? dp[i - 1] + temp : temp;if(dp[i] > dp[position]) position = i;}if(j > 1) cout << endl;cout << "Case " << j << ":" << endl << dp[position] <<  " "  << first[position] << " " << position << endl;}}