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usaco堆積木(記得國王遊戲的簡化版,思路一樣)

自己證明(交換法):

usaco堆積木(記得國王遊戲的簡化版,思路一樣)
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#define ll long long
using namespace std;
struct aa
{
	ll a,b;
}s[1000009];
int n;
ll ans=-0x3f3f3f3f;
bool cmp(aa a,aa b)
{
	return a.a+a.b<b.a+b.b;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%lld%lld",&s[i].a,&s[i].b);
	sort(s+1,s+n+1,cmp);//排序
	int r=0;
	for (int i=1;i<=n;i++)
	{
		if (r-s[i].b>ans) ans=r-s[i].b;
		r+=s[i].a;
	}//程式看似沒有多麼高深,這個貪心關鍵全在前面的證明裡的,交換法證貪心!!!!!!!!!!!
	printf("%lld",ans);
	return 0;
}