【dp专题1】hdu1003 D

Max Sum

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

最大子段和  模板题   求起始位置可把我wrong坏了

先放复杂度为O（N*N）的代码

#include<stdio.h>
#include<string.h>
#define N 100000+10
int dp[N];
int  num[N];

int main()
{
	int n,i,j;
	int t,t2,x;
	int start,end;
	int max ;
	scanf("%d",&t);
	t2 = t;
	while(t --)
	{
		scanf("%d",&n);
		for( i = 1; i <= n; i ++)
			scanf("%d",&num[i]);
		memset(dp,0,sizeof(dp));
		max = -999999999;
		start = 1;
		end = 1;
		x = 0;
		for( i = 1; i <= n; i ++)
		{
			if(dp[i-1] >= 0)
				dp[i] = dp[i-1]+num[i];
			else
				dp[i] = num[i];
			if(dp[i] > max)
			{
				max = dp[i];
				for(j = i; j >= 1; j --)
				{
					x += num[j];
					if(x == max)
						start = j;
				}
				x = 0;
				end = i;
			}
		}
		printf("Case %d:
",t2-t);
		printf("%d %d %d
",max,start,end);
		if(t> 0)
			printf("
");
	}
	return 0;
}

复杂度为O（N）的代码（避免了超时问题）

#include<stdio.h>

int max_start,max_end,num;
int now_start,now_end;
int f;
int max;

int main()
{
	int t,t2,n,i;
	scanf("%d",&t);
	t2 = t;
	while( t --)
	{
		scanf("%d",&n);
		for(i = 1; i <= n; i ++)
		{
			scanf("%d",&num);
			if( i == 1)
			{
				f = max = num;
				now_start = max_end = i;
			}
			else
			{
				if( f >= 0)
					f += num;
				else
				{
					f = num;
					now_start = i;
				}
			}
			if( f >= max)
			{
				max = f;
				max_start = now_start;
				max_end = i;
			}
		}
		printf("Case %d:
%d %d %d
",t2-t,max,max_start,max_end);  
		if(t)
			printf("
");
	}
	return 0;
}