hdu1003 MAX SUM——最大子序列解题报告

原题：

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 282943    Accepted Submission(s): 67221

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

 

Author

Ignatius.L

直接暴力写的话，时间复杂度为O(n^2),会超时。

用DP的思想：若我们已经找到一个最大的子序列a,其中m为其中最后一个元素，那么在m之前的元素和一定大于零，（如果小于零，那么最大子序列应该更新为m），所以我们可以从0-n-1遍历元素位置，然后用flag_start和flag_end维护最大子序列的初始元素与末尾元素，用_max维护子序列和的最大值。

下面上代码：（不懂可以看注释）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
int a[100010];
int sum=0;
int _max=0x80000000,flag_start,flag_end,here_start,here_end;//0x80000000就是16进制表示的int最小值
int main()
{
	int t,_case;
	cin>>t;
	int caseNum=0;
	while(t--)
	{
		caseNum++; 
		cin>>_case;
		memset(a,0,100010);//清零数组
		sum=0;
		_max=0x80000000;
		flag_start=flag_end=here_start=0;
		here_end=-1;//因为在刚开始sum会加一次a[0]导致here_end++一次，所以这里要设为-1。
		for(int i=0;i<_case;i++)
		{
			cin>>a[i];
		}
		for(int i=0;i<_case;i++)
		{
			//如果i位置之前sum<0则，更新sum值为a[i],同时here_start与here_end都要设置为i这个点
			if(sum<0)
			{
				sum=a[i];
				here_start=i;
				here_end=i;
			}
			else{
			sum+=a[i];
			here_end++;//因为这里,here_end要预设为-1
		}
			//维护max的值和起点与终点 
			if(sum>_max)
			{
				_max=sum;
				flag_end=here_end;
				flag_start=here_start;
			}
		}
		printf("Case %d:\n",caseNum);
		printf("%d %d %d\n", _max,flag_start+1,flag_end+1);
		if(t)	cout<<endl; //最后一个case不需要再打印blank line
	 }
	 return 0; 
}