#树状数组，离散#洛谷 3586 [POI2015]LOG

题目

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分析

考虑(geq s)的部分最多取到(s)，
设(<s)的总和为(p)，个数为(t)，
那么(p+(n-t)*sgeq c*s)就一定能取到

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代码

#include <cstdio>
#include <cctype>
#include <algorithm>
#define rr register
using namespace std;
const int N=1000011;
struct rec{int x,y,z;}q[N];
typedef long long lll;
int n,m,b[N],a[N],tot;
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
struct Tree_Array{
	lll c[N];
	inline void update(int x,int y){
		for (;x<=tot;x+=-x&x) c[x]+=y;
	}
	inline lll query(int x){
		rr lll ans=0;
		for (;x;x-=-x&x) ans+=c[x];
		return ans;
	}
}cw,ct;
signed main(){
	n=iut(),m=iut(),b[m+1]=0;
	for (rr int i=1;i<=m;++i){
		rr char c=getchar();
		while (!isalpha(c)) c=getchar();
		q[i]=(rec){iut(),iut(),c=='U'};
		b[i]=q[i].y;
	}
	for (rr int i=1;i<=n;++i) a[i]=1;
	sort(b+1,b+2+m),tot=unique(b+1,b+2+m)-b-1,ct.update(1,n);
	for (rr int i=1;i<=m;++i) q[i].y=lower_bound(b+1,b+1+tot,q[i].y)-b;
	for (rr int i=1;i<=m;++i)
	if (q[i].z){
		rr int t1=a[q[i].x],t2=q[i].y;
		cw.update(t1,-b[t1]),ct.update(t1,-1),
		cw.update(t2,b[t2]),ct.update(t2,1);
		a[q[i].x]=q[i].y;
	}else{
		rr lll W=cw.query(q[i].y-1),C=n-ct.query(q[i].y-1);
		puts((q[i].x-C)*b[q[i].y]<=W?"TAK":"NIE");
	}
	return 0;
}