n(1 <= N <= 50,000) 个牛,重wi (1 <= W_i <= 10,000),力气si (1 <= S_i <= 1,000,000,000),堆成一个竖线,risk值为每只牛上面的w之和-它的si,使它的最大值最小,输出最小值。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define ll long long
using namespace std;
struct cow{
ll w,s;
bool operator < (const cow&b)const{
return w+s<b.w+b.s;
}
}a[50005];
int main() {
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld%lld",&a[i].w,&a[i].s);
sort(a+1,1+a+n);
ll ans=-a[1].s,tot=0;
for(int i=2;i<=n;i++){
tot+=a[i-1].w;
ans=max(ans,tot-a[i].s);
}
printf("%lld",ans);
return 0;
}