sample ive 需要 tails each circus rac last fail

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. 題解： 　　和國王遊戲很像對不對，有是一道貪心題，順序是按照si+wi從小到大排序！為什麽呢？我們來證明一下。 　　首先兩個相鄰的牛i,i+1,根據排序有wi+si<w(i+1)+s(i+1).我們考慮每個牛的危險度，對於不是i和i+1的其他牛，因為他們的si不會變，所需要承受的重量不會變，所以危險度就不變，i這頭牛的危險度是si-sum(Wi-1),i+1的危險度為s(i+1)-wi-sum(wi-1)。換位置，i+1這牛就是s(i+1)-sum(wi-1),i為si-(wi+1)-sum(wi-1).將做差就可以發現恒大於了。 代碼：

#include <cstdio>
#include 
 
<iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#define MAXN 50100
#define ll long long
using namespace std;
struct node{
    int w,s;
}a[MAXN*2];
int n;

bool cmp(node x,node y){
    return x.w+x.s<y.w+y.s;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d%d",&a[i].w,&a[i].s);
    sort(a+1,a+n+1,cmp);
    int ans=-(1<<30),sum=0;
    for(int i=1;i<=n;i++){
        ans=max(ans,sum-a[i].s);
        sum+=a[i].w;
    }
    printf("%d\n",ans);
    return 0;
}

Cow Acrobats