HDU-1003 Max Sum-動態規劃-難度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

代碼:

/*HDU 1003 Max Sum */
/*
 *其實就是求最大子數組之和,不過需要保存起始點和結束點
 */
#include <cstdio>
#include <iostream>
#include <climits>
using namespace std;

const int  MAXN = 100002;
int dp[MAXN], first[MAXN];//first[i]保存dp[i]的起始點的位置,結束點的位置是他自己i

int main()
{
#ifdef _LOCAL
	freopen("F://input.txt", "r", stdin);
#endif
	int T, n;
	cin >> T;
    for(int j = 1; j <= T; j++)
	{
		cin >> n;
		int temp, position = 0;
		dp[0] = INT_MIN; //初始化邊界條件,有負數的最小值不能是0或-1
		for(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;
	}
}