DLUTOJ #1394 Magic Questions

传送门

Time Limit: 3 Sec  Memory Limit: 128 MB

Description

Alice likes playing games. So she will take part in the movements of M within N days, and each game is represented in an integer between 1 and M. Roommates have Q magic questions: How many different kinds of games does Alice participate between Lth day and Rth day(including Lth day and Rth day)?

Input

You will be given a number of cases; each case contains blocks of several lines. The first line contains 2 numbers of N and M. The second line contains N numbers implying the game numbers that Alice take part in within N days. The third line contains a number of Q. Then Q lines is entered. Each line contain two numbers of L and R.

1≤N,M,Q≤100000

Output

There should be Q output lines per test case containing Q answers required.

Sample Input

Sample Output

HINT

这是今年校赛的K题，一道经典题目，但现场没A。

在线可以用主席树，目前还不会。有一个巧妙的利用数状数组的离线解法，比较好写。

要点是：

1.将查询按右端点从小到大排序。

2.将每个数上一次出现的位置记录下来。当这个数再次出现时，将它上次出现位置上的计数消除。

Implementation:

主体是个双指针。

#include <bits/stdc++.h>
using namespace std;

const int N(1e5+5);
int n, m, q, a
, pos
, bit
, ans
;

void add(int x, int v){
for(; x<=n; bit[x]+=v, x+=x&-x);
}

int sum(int x){
int res=0;
for(; x; res+=bit[x], x-=x&-x);
return res;
}

struct P{
int l, r, id;
P(int l, int r, int id):l(l),r(r),id(id){}
P(){};
bool operator<(const P&b)const{return r<b.r;}
}p
;

int main(){
// ios::sync_with_stdio(false);
for(; ~scanf("%d%d", &n, &m); ){
for(int i=1; i<=n; i++) scanf("%d", a+i);
scanf("%d", &q);
for(int l, r, i=0; i<q; i++) scanf("%d%d", &l, &r), p[i]={l, r, i};

sort(p, p+q);   //error-prone
memset(bit, 0, sizeof(bit));
memset(pos, 0, sizeof(pos));
for(int i=1, j=0, k; j<q&&i<=n; ){  //error-prone
for(; i<=p[j].r; i++){
if(pos[a[i]]) add(pos[a[i]], -1);
pos[a[i]]=i;
add(i, 1);
}
for(k=j; k<q&&p[k].r==p[j].r; k++)  //error-prone
ans[p[k].id]=sum(p[k].r)-sum(p[k].l-1);
j=k;
}
for(int i=0; i<q; i++) printf("%d\n", ans[i]); //error-prone
}
return 0;
}