Cow Acrobats POJ - 3045

 Farmer John‘s N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join the circus. Their hoofed feet prevent them from tightrope walking and swinging from the trapeze (and their last attempt at firing a cow out of a cannon met with a dismal failure). Thus, they have decided to practice performing acrobatic stunts.

The cows aren‘t terribly creative and have only come up with one acrobatic stunt: standing on top of each other to form a vertical stack of some height. The cows are trying to figure out the order in which they should arrange themselves ithin this stack.

Each of the N cows has an associated weight (1 <= W_i <= 10,000) and strength (1 <= S_i <= 1,000,000,000). The risk of a cow collapsing is equal to the combined weight of all cows on top of her (not including her own weight, of course) minus her strength (so that a stronger cow has a lower risk). Your task is to determine an ordering of the cows that minimizes the greatest risk of collapse for any of the cows.Input

Output

Sample Input

3
10 3
2 5
3 3

Sample Output

2

Hint

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<cstdio>
 5 #include<cstring>
 6 using namespace std;
 7 typedef long long ll;
 8 
 9 struct node{
10     int we,st;
11     bool operator<(const node& i)const{
12         return (we+st)<(i.we+i.st);
13     }
14 }a[50005];
15 
16 int n;
17 
18 int main()
19 {   while(~scanf("%d",&n)){
20        for(int i=0;i<n;i++) scanf("%d%d",&a[i].we,&a[i].st);
21        sort(a,a+n);
22        ll pre=0,ma=-1000000000;
23        for(int i=0;i<n;i++){
24              ma=max(pre-a[i].st,ma);
25              pre+=a[i].we;
26        }
27        printf("%lld\n",ma);
28     }
29     return 0;
30 }

Cow Acrobats POJ - 3045

原文：http://www.cnblogs.com/zgglj-com/p/6838856.html