HDU 4296 buildings

最新推荐文章于 2020-05-23 22:26:13 发布

原创最新推荐文章于 2020-05-23 22:26:13 发布 · 494 阅读

0 ·

CC 4.0 BY-SA版权

ACM_刷题专栏收录该内容

11 篇文章

订阅专栏

探讨了通过贪心算法解决楼板排列问题的方法，旨在寻找使得整栋楼的pdv值最小的楼板排列方式。

题目链接：点击打开链接

题目大意：一栋楼有n层，每层楼板有个w值和s值，这座建筑有个pdv值=最底层以上（不包括最底层）的w的总和-最底层的s值。现在要你排列每层楼板，使得这栋楼的pdv值最小，并且输出这个pdv值。

Input

　　There’re several test cases.
　　In each test case, in the first line is a single integer N (1 <= N <= 10 ⁵) denoting the number of building’s floors. The following N lines specify the floors. Each of them contains two integers w _i and s _i (0 <= w _i, s _i <= 100000) separated by single spaces.
　　Please process until EOF (End Of File).

Output

　　For each test case, your program should output a single integer in a single line - the minimal PDV of the whole building.
　　If no floor would be damaged in a optimal configuration (that is, minimal PDV is non-positive) you should output 0.

Sample Input

Sample Output

  
   1
0
2
  
  
   思路：试着证明一下其贪心依据。
  
  
   假设x，y是两个相邻的楼板，x在上，y在下。“W总”表示x以上的的楼板的的w值的总和。此时，pdv1 = ‘W总’+Wx - Sy。如果交换这两个楼板，那么pdv2 = 
   ‘W总’+Wy - Sx，要使得交换后pdv值比原来的要小的充要条件是pdv2< pdv1,即‘W总’+Wx - Sy < ‘W总’+Wy - Sx,化简得Wx + Sx<Wy + Sy.
  
  
   所以得到的贪心依据是w与s的和最小。
  
  
   代码如下：
   #include<iostream>
#include<algorithm>
using namespace std;
#define LL long long 
const int MAX = 100000 + 100;
struct node
{
	int w;
	int s;
}A[MAX];
int cmp( node a , node b)
{
	return a.s + a.w < b.s + b.w ;
}
int main()
{
	int i,n;
	while(cin >> n)
	{
		for(i = 0; i < n ; i++)
			cin >> A[i].w >> A[i].s;
		sort(A,A+n,cmp);
		LL sum = 0 ,pdv = 0;
		for(i = 0; i < n;i++)
		{
			pdv = max(pdv,sum-A[i].s);
			sum += A[i].w;
		}
		printf("%I64d\n",pdv);
	}
	return 0;
}