POJ 3045 Cow Acrobats

Cow Acrobats

Description

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

* Line 1: A single line with the integer N. 

* Lines 2..N+1: Line i+1 describes cow i with two space-separated integers, W_i and S_i. 

Output

* Line 1: A single integer, giving the largest risk of all the cows in any optimal ordering that minimizes the risk.

Sample Input

3

10 3

2 5

3 3

Sample Output

2

Hint

OUTPUT DETAILS: 

Put the cow with weight 10 on the bottom. She will carry the other two cows, so the risk of her collapsing is 2+3-3=2. The other cows have lower risk of collapsing.

Source

​​USACO 2005 November Silver​​

算法分析：

题意：

ｎ头牛，每一头牛有重量和力量，现在ｎ头牛垂直叠罗汉，每头牛的风险等于他头上牛的重量总和减去这只牛的力量。问怎么排列，使最大风险牛的风险最小（注意不一定是最下面的。

分析：

一开始想的枚举，但觉得会超时，而后觉得贪心应该对的，贪心策略直觉应该是力量和体重总和，

证明：两头牛a和b，a在下面，a风险=b.w-a.s; b在下面，b风险=a.w-b.s;如果要使a风险小于b风险，则移项后：a.w+a.s<b.w+b.s

代码实现：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
using namespace std;  
const double eps = 1e-8;  
typedef long long LL;  
typedef unsigned long long ULL;  
const int INF = 0x3f3f3f3f;  
const int INT_M_INF = 0x7f7f7f7f;  
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;  
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};  
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};  
const int MOD = 1e9 + 7;  
const double pi = acos(-1.0);  
const int MAXN=5010;  
const int MAXM=100010;
const int M=50010;
struct node
{
    int w,s;
}c[M];
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)
   {
      LL sum=0;
      LL maxx=-LL_M_INF;
      for(int i=0;i<n;i++)
      {
         scanf("%d%d",&c[i].w,&c[i].s);
      }
        sort(c,c+n,cmp); 
        for(int i=0;i<n;i++)   //注意，最底下的风险不一定是最大的
      {
         LL t=sum-c[i].s;
         sum+=c[i].w;
         maxx=max(maxx,t);
      }
      cout<<maxx<<endl;   
   }
   return 0;
}