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Time Limit: 3 Sec Memory Limit: 128 MB
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)?
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
There should be Q output lines per test case containing Q answers required.
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[N], pos[N], bit[N], ans[N]; 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[N]; 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; }
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原文地址：http://www.cnblogs.com/Patt/p/5410711.html
