Gym 100960G （set+树状数组）

Problem Youngling Tournament

题目大意

　　给一个序列a[i]，每次操作可以更改一个数，每次询问 将序列排序后有多少个数a[i]>=sum[i-1]。

　　n<=10^5,q<=5*10^4，a[i]<=10^12

解题分析

　　可以发现，在最优情况下，该序列长度最多为logn。

　　将询问离线后，用multiset来维护a[i],用树状数组来维护sum[i]。树状数组所存下标为离散化后的a[i]。

　　每次查询时跳跃式地在set中查找。

　　时间复杂度 O((n+m)log(n+m)+mlognlogn)

参考程序

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <string>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")
using namespace std;

#define N 100008             
#define M 50008    
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1 
#define clr(x,v) memset(x,v,sizeof(x));
#define bitcnt(x) __builtin_popcount(x)
#define rep(x,y,z) for (int x=y;x<=z;x++)
#define repd(x,y,z) for (int x=y;x>=z;x--)
const int mo  = 1000000007;
const int inf = 0x3f3f3f3f;
const int INF = 2000000000;
/**************************************************************************/ 

int n,m,tot,flag;
LL a[N],c[N],x[N*2];
int b[N];
multiset<long long> S;
struct Binary_Indexed_Tree{
    LL a[N*3];
    void clear(){
        clr(a,0);
    }
    void insert(int x,LL val){
        for (int i=x;i<=N<<1;i+=i & (-i))
            a[i]+=val;
    }
    LL sigma(int x){
        LL res=0;
        for (int i=x;i>0;i-=i & (-i))
            res+=a[i];
        return res;
    }
    LL query(int x,int y){
        return sigma(y)-sigma(x-1);
    }
}T;
int id(LL xy){
    return lower_bound(x+1,x+tot+1,xy)-x;
}
set <long long> :: iterator it,it1;
void query(){
    int ans=1; LL tmp;
    it = S.begin(); 
    it1 = it ; it1++; 
    if (*it==*it1) ans++;
    tmp=T.sigma(id(*it));
    while (1){
        it = S.lower_bound(tmp);
        if (it==S.begin()) it++;
        if (it==S.end()) break;
        it1 = it; it1--;
        tmp=T.sigma(id(*it1));
        if (*it>=tmp) ans++;
        tmp=T.sigma(id(*it));
    }
    printf("%d
",ans);

}

int main(){
    scanf("%d",&n);
    rep(i,1,n){
        scanf("%lld",&a[i]);
        x[++tot]=a[i];
    }
    scanf("%d",&m);
    rep(i,1,m){
        scanf("%d%lld",&b[i],&c[i]);
        x[++tot]=c[i];
    }
    sort(x+1,x+tot+1);
    tot=unique(x+1,x+tot+1)-x;
    T.clear();
    rep(i,1,n){
        T.insert(id(a[i]),a[i]);
        S.insert(a[i]);
    }
    query();
    rep(i,1,m){
        S.erase(S.find(a[b[i]]));
        T.insert(id(a[b[i]]),-a[b[i]]);
        S.insert(c[i]);
        T.insert(id(c[i]),c[i]);
        a[b[i]]=c[i];
        query();
    }

}