HDU4296

Buildings

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3721    Accepted Submission(s): 1393

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
n个板，每个板有重量和强度，还有PDV值（上面的总重量-该板的强度）
对于某种叠放方式，PDV的最大值为其代表值
求该值的最小值~
贪心，越上面的板w+s越小
#include<cstdio>#include<iostream>#include<cmath>#include<algorithm>using namespace std;struct Node{    long long w,s;}fl[100005];bool cmp(const Node &a,const Node &b){    return a.w+a.s<b.w+b.s;}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=0;i<n;i++)          scanf("%lld%lld",&fl[i].w,&fl[i].s);        sort(fl,fl+n,cmp);        long long sum=0,pdv=0;        for(int i=0;i<n;i++)        {            pdv=max(pdv,sum-fl[i].s);            sum+=fl[i].w;        }        printf("%lld\n",pdv);    }    return 0;}