hdu 3874 Necklace

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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3874

题目大意:求区间和,但相同的数只加一次.

题目思路:
这题直接用线段树是做不出来,会退化成O(nlogn),除非线段树中不同的数只存一个,这样就变成最简单的区间求和,用树状数组也能做.但是我们需要边建树边询问,遇到树中有相同的数时,先把树中的数删档,再在当前位置添加该数,之后在判断是否有询问的右端点是当前位置,有的话就询问,这样就能保证每个值只加一次,当然之前要对询问离线处理,把询问按照右端点排序.

代码:

#include <stdlib.h>#include <string.h>#include <stdio.h>#include <ctype.h>#include <math.h>#include <time.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <string>#include <iostream>#include <algorithm>using namespace std;#define ull unsigned __int64#define ll __int64//#define ull unsigned long long//#define ll long long#define son1 New(p.xl,xm,p.yl,ym),(rt<<2)-2#define son2 New(p.xl,xm,min(ym+1,p.yr),p.yr),(rt<<2)-1#define son3 New(min(xm+1,p.xr),p.xr,p.yl,ym),rt<<2#define son4 New(min(xm+1,p.xr),p.xr,min(ym+1,p.yr),p.yr),rt<<2|1#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define middle (l+r)>>1#define MOD 1000000007#define esp (1e-8)const int INF=0x3F3F3F3F;const double DINF=10000.00;//const double pi=acos(-1.0);const int N=50010,M=200010;int n,m,tot;ll a[N],cnt[M],sum[N];int vis[N<<5];struct node{int x,y,num;void write(){scanf("%d%d",&x,&y);}bool operator < (const node& p) const {return y < p.y;}}q[M];int lowbit(int x){return x&(-x);}void Add(int x,ll c){while(x<=n) sum[x]+=c,x+=lowbit(x);}ll Sum(int x){ll r=0;while(x>0) r+=sum[x],x-=lowbit(x);return r;}int main(){//freopen("1.in","r",stdin);//freopen("1.out","w",stdout);int i,j,k;int T,cas;scanf("%d",&T);for(cas=1;cas<=T;cas++){scanf("%d",&n);for(i=1;i<=n;i++) scanf("%d",&a[i]);scanf("%d",&m);for(i=1;i<=m;i++) q[i].write(),q[i].num=i;sort(q+1,q+m+1);memset(vis,0,sizeof(vis));memset(sum,0,sizeof(sum));memset(cnt,0,sizeof(cnt));for(i=k=1;i<=n;i++){if(vis[a[i]]) Add(vis[a[i]],-a[i]);Add(vis[a[i]]=i,a[i]);while(q[k].y==i){cnt[q[k].num]+=Sum(i)-Sum(q[k].x-1);k++;}}for(i=1;i<=m;i++) printf("%I64d\n",cnt[i]);}return 0;}