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hdu1024 Max Sum Plus Plus

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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18131    Accepted Submission(s): 5931


Problem Description
Now I think you have got an AC in Ignatius.L‘s "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).

But I`m lazy, I don‘t want to write a special-judge module, so you don‘t have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
 

Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
 

Output
Output the maximal summation described above in one line.
 

Sample Input
1 3 1 2 3 2 6 -1 4 -2 3 -2 3
 

Sample Output
6 8
http://www.cnblogs.com/kuangbin/archive/2011/08/04/2127085.html
自己看吧。
#include <stdio.h>
#include <string.h>
#define INF 0x7fffffff
#define MAX 1000100

int dp[MAX] , num[MAX] , preMax[MAX];

int max(int a , int b)
{
	return a>b?a:b ;
}

int main()
{
	int m , n;
	while(~scanf("%d%d",&m,&n))
	{
		for(int i = 1 ; i <= n ; ++i)
		{
			scanf("%d",&num[i]) ;
			preMax[i] = 0 ;
			dp[i] = 0 ;
		}
		dp[0] = 0 ;
		preMax[0] = 0 ;
		int mm ;
		for(int i = 1 ; i <= m ; ++i)
		{
			mm = -INF;
			for(int j = i ; j <= n ; ++j)
			{
				dp[j] = max(dp[j-1]+num[j],preMax[j-1]+num[j]) ;
				preMax[j-1] = mm ;
				mm = max(dp[j],mm) ;
			}
		}
		printf("%d\n",mm) ;
	}
	return 0 ;
}


hdu1024 Max Sum Plus Plus

原文:http://blog.csdn.net/lionel_d/article/details/43611507

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