HDU 4296 Buildings

Buildings

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2935    Accepted Submission(s): 1114


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 wi and strength si. When floors are stacked up, each floor has PDV(Potential Damage Value) equal to (Σwj)-si, where (Σwj) 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 wi and si (0 <= wi,
si <= 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

 

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ACcode：

/*
题意：有 n 个地板，每个地板 i 有两个权值 Wi, Si，且 PDV(i) =  (ΣWj) - Si ( j 表示在 i 上面的地板)。问如何调整顺序，使得【max(PDV)】最小。

思路：假设i,j相邻，并且i上面的重量为sum,若i在上面，则有pi=sum-si,pj=sum+wi-sj;若j在上面，则有pi'=sum+wj-si,pj'=sum-sj;

显然有pi<pi',pj>pj',于是令pj<pi',就有sum+wi-sj<sum+wj-si，即wi+si<wj+sj;
*/
#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 100005
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
using namespace std;
struct Node{
int w;
int s;
void in(){
rd2(w,s);
}
}my[maxn];
bool cmp(Node a,Node b){
return a.w+a.s<b.w+b.s;
}
int sum;
int main(){
int n;
while(rd(n)!=EOF){
sum=0;
FOR(i,0,n-1)
my[i].in();
sort(my,my+n,cmp);
ll sum=0,maxx=0;
FOR(i,0,n-1){
maxx=max(maxx,sum-my[i].s);
sum+=my[i].w;
}
printf("%I64d\n",maxx);
}
return 0;
}
/*
3
10 6
2 3
5 4
2
2 2
2 2
3
10 3
2 5
3 3
*/