HDU-1003 Max Sum (最大子段和问题DP)

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本文介绍了一种求解整数序列中具有最大和的连续子序列的问题，并提供了一个有效的算法实现。通过动态规划思想，文章详细解释了如何确定子序列的开始和结束位置，以及如何计算最大和。

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 276155 Accepted Submission(s): 65571

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

解答：

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm> 
#include<vector> 
#include<cmath>  
#include<sstream>
#include<cstdlib>
#include<queue>
#include<map>
const int N =100005;
int d[N],num[N]; //d[i]表示以第i个数结尾的序列的最大值 
using namespace std;
int main(){
	int T;
	cin>>T;
	for(int cases = 1;cases<=T;cases++){
		memset(num,0,sizeof(num));
		memset(d,0,sizeof(d));
		int MAX = -10000;
		int n,start,end;
		cin>>n;
		for(int i =0;i<n;i++){
			scanf("%d",&num[i]);
		}
		for(int i =0;i<n;i++){
			d[i] = max(d[i-1]+num[i],num[i]);
			if(MAX < d[i]){
				MAX = d[i];
				end = i;
			}
		}
		int sum = 0;
		for(int i = end;i>=0;i--){
			sum+=num[i];
			if(sum == MAX){
				start = i;
			}
		}
		printf("Case %d:\n",cases);
		printf("%d %d %d\n",MAX,start+1,end+1);
		if(cases != T) printf("\n"); 
	}
	return 0;
}