杭电ACM1024

最新推荐文章于 2019-08-31 13:23:03 发布

java_wliang

最新推荐文章于 2019-08-31 13:23:03 发布

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分类专栏：算法文章标签：杭电ACM1024

本文链接：https://blog.csdn.net/java_wliang/article/details/14214303

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动态规划还是不怎么会灵活使用，对于决策方程的推导需要不断练习啊，参考网上的关于本题的部分推导写出来的

Max Sum Plus Plus

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 S ₁ , S ₂ , S ₃ , S ₄ ... S _x , ... S _n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S _x ≤ 32767). We define a function sum(i, j) = S _i + ... + S _j (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(i ₁ , j ₁ ) + sum(i ₂ , j ₂ ) + sum(i ₃ , j ₃ ) + ... + sum(i _m , j _m ) maximal (i _x ≤ i _y ≤ j _x or i _x ≤ j _y ≤ j _x 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(i _x , j _x )(1 ≤ x ≤ m) instead. ^_^

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
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

#include<stdio.h>
#include<limits.h>

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

int main() {
	int n , m , MaxSum;
	int *dp , *max , *num;
	int i , j;
	while(scanf("%d%d", &m, &n) != EOF) {
		dp = new int[n+10];
		max = new int[n+10];
		num = new int[n+10];
		for(i=1; i<=n; i++) {
			scanf("%d",&num[i]);
			dp[i] = 0;
			max[i] = 0;
		}
		dp[0] = 0;
		max[0] = 0;
		for(i=1; i<=m; i++) {
			MaxSum = INT_MIN;
			for(j=i; j<=n; j++) {
				dp[j] = Max(dp[j-1]+num[j] , max[j-1]+num[j]);
				max[j-1]= MaxSum;
				MaxSum = Max(dp[j] , MaxSum);
			}
		}
		printf("%d\n" , MaxSum);
		delete[] dp;
		delete[] max;
		delete[] num;
	}
	return 0;
}