BZOJ4378: [POI2015]Logistyka

每个数对答案的贡献=min(a[i],s)

所以用树状数组维护小于等于每个数的权值和以及个数

每次对于一个s，把小于s的所有数加起来，再把大于s的当成s，统计一下贡献，看能否实现

#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
#define N 2000010
using namespace std;
struct ppp {long long siz,sum;};
ppp c[N];
ppp operator+(ppp x,ppp y){return (ppp){x.siz+y.siz,x.sum+y.sum};}
long long n,m;
long long lowbit(long long x) {return x&(-x);}
void change(long long x,ppp v)
{
	for(;x<=m+1;x+=lowbit(x))
	c[x]=c[x]+v;
}
ppp check(long long x)
{
	ppp ret={0,0};
	for(;x;x-=lowbit(x))
	ret=ret+c[x];
	return ret;
}
long long o[N],x[N],y[N];
long long aa[N];
map<long long,long long>p;
long long a[N];
int main()
{
	scanf("%lld%lld",&n,&m);
	long long i,j;
	char s[10];
	for(i=1;i<=m;i++)
	{
		scanf("%s%lld%lld",s,&x[i],&y[i]);
		if(s[0]=='Z') o[i]=1;
		aa[i]=y[i];
	}
	sort(aa+1,aa+m+1);
	p[0]=1;
	for(i=1;i<=m;i++)
	p[aa[i]]=i+1;
	change(p[0],(ppp){n,0});
	for(i=1;i<=m;i++)
	{
		if(o[i]==0)
		{
			change(p[a[x[i]]],(ppp){-1,-a[x[i]]});
			a[x[i]]=y[i];
			change(p[a[x[i]]],(ppp){1,a[x[i]]});
		}
		else
		{
			ppp t=check(p[y[i]]);
		//	cout<<t.siz<<' '<<t.sum<<endl;
			if((n-t.siz)*y[i]+t.sum>=y[i]*x[i]) puts("TAK");
			else puts("NIE");
		}
	}
}