题目链接:​​1099​

每个  r  -  c  为可能多加的--

以它排序--然后就行啦--


代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node{
int r,o,c;
}dian[100100];
bool cmp(node xx,node yy)
{
return xx.c>yy.c;
}
int main()
{
int n;scanf("%d",&n);
for (int i=0;i<n;i++)
{
scanf("%d%d",&dian[i].r,&dian[i].o);
dian[i].c=dian[i].r-dian[i].o;
}
sort(dian,dian+n,cmp);
long long s=0,ans=0;
for (int i=0;i<n;i++)
{
s+=dian[i].r;
ans=max(s,ans);
s-=dian[i].c;
ans=max(s,ans);
}
printf("%lld\n",ans);
return 0;
}