hdu 4296 Buildings

Buildings

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4259    Accepted Submission(s): 1580

Problem Description

　　Have you ever heard the story of Blue.Mary, the great civil engineer? Unlike Mr. Wolowitz, Dr. Blue.Mary has accomplished many great projects, one of which is the Guanghua Building.
　　The public opinion is that Guanghua Building is nothing more than one of hundreds of modern skyscrapers recently built in Shanghai, and sadly, they are all wrong. Blue.Mary the great civil engineer had try a completely new evolutionary building method in project of Guanghua Building. That is, to build all the floors at first, then stack them up forming a complete building.
　　Believe it or not, he did it (in secret manner). Now you are face the same problem Blue.Mary once stuck in: Place floors in a good way.
　　Each floor has its own weight w_i and strength s_i. When floors are stacked up, each floor has PDV(Potential Damage Value) equal to (Σw_j)-s_i, where (Σw_j) stands for sum of weight of all floors above.
　　Blue.Mary, the great civil engineer, would like to minimize PDV of the whole building, denoted as the largest PDV of all floors.
　　Now, it’s up to you to calculate this value.

 

Input

　　There’re several test cases.
　　In each test case, in the first line is a single integer N (1 <= N <= 10⁵) 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

310 62 35 422 22 2310 32 53 3

 

Sample Output

102

解题思路：不同的顺序都会有一个最大PDV，你的任务是把这个PDV最小化。

对于相邻放置的两块板，设两块板为i,j他们上面的重量为sum

           1) a=sum-si;b=sum+wi-sj;

           交换两个板的位置

          2)a'=sum+wj-si;b'=sum-sj;

          如果1优于2,求解得有效的条件为wj-si>wi-sj。

          所以按si+wi的和排序贪心即可。

 代码如下：

#include<iostream>#include<algorithm>#include<cstdio>using namespace std;struct Node{int w,s;}data[100005];int cmp(Node a,Node b){return a.w+a.s<b.w+b.s;}int main(){int n;while(~scanf("%d",&n)){for(int i=0;i<n;i++)scanf("%d%d",&data[i].w,&data[i].s);sort(data,data+n,cmp);long long sum=0;long long ans=0;for(int i=1;i<n;i++){sum+=data[i-1].w;if(ans<sum-data[i].s)ans=sum-data[i].s;}printf("%lld\n",ans);}return 0;}