BZOJ_4378_[POI2015]Logistyka_tree array

BZOJ_4378_[POI2015]Logistyka_tree array

Description

Maintain a sequence of length n, which is 0 at the beginning, and supports the following two operations:
1.U ka Modify the kth number in the sequence to a.
2.Z cs On this sequence, select c positive numbers each time, subtract 1 from them, and ask if you can perform s operations.
Each query is independent, i.e. the sequence is not modified per query.

Input

The first line contains two positive integers n, m (1<=n, m<=1000000), which represent the sequence length and the number of operations, respectively.
The next m lines are m operations, where 1<=k, c<=n, 0<=a<=10^9, 1<=s<=10^9.

Output

Contains several lines, for each Z query, output TAK if feasible, NIE otherwise.

Sample Input

3 8
U 1 5
U 2 7
Z 2 6
U 3 1
Z 2 6
U 2 2
Z 2 6
Z 2 1

Sample Output

NO
YES
NO
YES

---

Note that the query is on the entire sequence and not on a given interval.

Assuming that there are k numbers greater than s, the sum of the remaining weights must be greater than or equal to (ck)*s.

So we discretize the weights and maintain them separately in two tree-like arrays.

 

Code:

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 1000050
#define RR register
typedef long long ll;
int n,m,t[N],maxn=1000000000,h[N],p[N];
ll c[N][2];
char opt[10];
inline int rd() {
    RR int x=0,f=1; RR char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){x=(x<<3)+(x<<1)+s-'0';s=getchar();}
    return x*f;
}
struct A {
    int num,v,id,opt,pos;
}a[N];
inline bool cmp1(const A &x,const A &y){return x.num<y.num;}
inline bool cmp2(const A &x,const A &y){return x.id<y.id;}
void fix(int x,int v,int flg) {
    for(;x<=m;x+=x&(-x)) c[x][flg]+=v;
}
ll inq (int x, int flg) {
    ll re=0;
    for(;x;x-=x&(-x)) re+=c[x][flg];
    return re;
}
int main() {
    n=rd(); m=rd();
    int i,j;
    for(i=1;i<=m;i++) {
        scanf("%s",opt);
        if(opt[0]=='U') {
            a[i].opt=1; a[i].id=i; a[i].pos=rd(); a[i].num=rd();
        }else {
            a[i].opt=2; a[i].id=i; a[i].pos=rd(); a[i].num=rd();
        }
    }
    sort(a+1,a+m+1,cmp1); a[0].num=134234;
    for(j=0,i=1;i<=m;i++) {
        if(a[i].num!=a[i-1].num) j++;
        a[i].v=j;
        h[j]=a[i].num;
    }
    sort(a+1,a+m+1,cmp2);
    for(i=1;i<=m;i++) {
        if(a[i].opt==1) {
            int t=a[i].pos;
            if(p[t]) {
                fix(p[t],-1,1);
                fix(p[t],-h[p[t]],2);
            }
            p[t]=a[i].v;
            fix(p[t],1,1);
            fix(p[t],h[p[t]],2);
        }else {
            int k=inq(m,1)-inq(a[i].v-1,1);
            if(k>=a[i].pos) {
                puts("TAK"); continue;
            }
            ll sum=inq(a[i].v-1,2);
            puts(sum>=1ll*a[i].num*(a[i].pos-k)?"TAK":"NIE");
        }
    }
}