HDU 3874 离线处理，树状数组

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题意：查询区间不相同数的集合。

思路：对于每一个查询区间，按照右端点排序，这样我们每次查询的时候，如果这个数在之前出现过，那么的话将前面的删除，重新在新的位置加入。这样每一种数只会被统计一次，查询直接用树状数组进行就行了，相当于前缀和。

#include<cstdio>#include<iostream>#include<algorithm>#include<vector>#include<cmath>#include<map>#include<cstring>using namespace std;typedef long long ll;typedef pair<int,int> P;#define fi first#define se second#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x)#define PI acos(-1.0)#define ITER set<int>::iteratorconst int Mod = 1e9 + 7;const int maxn = 1e6 + 10;struct Node{    bool operator < (const Node& t)const{return y < t.y;}    int x,y,id;}b[maxn];ll a[maxn],ans[maxn];ll tree[maxn];int lowbit(int x){return x & (-x);}ll get(int x){ll ret = 0;for(int i = x;i > 0; i -= lowbit(i))ret += tree[i];return ret;}void update(int x,ll y){for(int i = x; i < maxn; i += lowbit(i))tree[i] += y;}int vis[maxn];int main(){    int Tcase;scanf("%d",&Tcase);    while(Tcase --)    {        int n,m;scanf("%d",&n);for(int i = 1;i <= n;i ++)scanf("%I64d",&a[i]);clr(tree,0);        scanf("%d",&m);for(int i = 1;i <= m; i ++)scanf("%d%d",&b[i].x,&b[i].y),b[i].id = i;sort(b + 1,b + m + 1);        clr(vis,0);int pos = 1;        for(int i = 1; i <= m; i ++)        {            while(pos <= b[i].y)            {                ll x = a[pos];                if(vis[x])                    update(vis[x],-x),vis[x] = pos,update(pos,x);                else vis[x] = pos,update(pos,x);                pos ++;            }            ans[b[i].id] = get(b[i].y) - get(b[i].x - 1);        }        for(int i = 1;i <= m;i ++)printf("%I64d\n",ans[i]);    }    return 0;}

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