HDU 4296 Buildings 제3 7 회 ACM/ICPC 청 두 지역 사 이 버 전 1009 문제(욕심)

Buildings
Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 155    Accepted Submission(s): 72
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 <= 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
3 10 6 2 3 5 4 2 2 2 2 2 3 10 3 2 5 3 3
 
 
Sample Output
1 0 2
 
 
Source
2012 ACM/ICPC Asia Regional Chengdu Online
 
 
Recommend
liuyiding
 
욕심 이라는 제목.
물 많 습 니 다.정렬 방법 은 코드 를 봅 니 다.
나 는 하나하나 시험 해 보 았 지만 아직 증명 되 지 않 았 다.

//1009

#include<stdio.h>

#include<iostream>

#include<map>

#include<set>

#include<algorithm>

#include<string.h>

#include<stdlib.h>

using namespace std;

const int MAXN=100010;

struct Node

{

    int w,s;

}tt[MAXN];





bool cmp2(Node a,Node b)

{

    return a.w - b.s< b.w - a.s;

}

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        for(int i=0;i<n;i++)

          scanf("%d%d",&tt[i].w,&tt[i].s);

        sort(tt,tt+n,cmp2);

        long long ans=0;

        long long s=0;





        for(int i=0;i<n;i++)

        {

            if(s-tt[i].s>ans)ans=s-tt[i].s;

            s+=tt[i].w;

        }



        printf("%I64d
",ans);

    }

    return 0;

}