hdu 3874 Necklace 求数组任意区间和(相同元素只算一次) 树状数组+离线算法

[align=left]Problem Description[/align]Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define the beautiful value of some interval [x,y] as F(x,y). F(x,y) is calculated as the sum of the beautiful value from the xth ball to the yth ball and the same value is ONLY COUNTED ONCE. For example, if the necklace is 1 1 1 2 3 1, we have F(1,3)=1, F(2,4)=3, F(2,6)=6.

Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you must tell her F(L,R) of them.
[align=left]Input[/align]The first line is T(T<=10), representing the number of test cases.
For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query.
[align=left]Output[/align]For each query, output a line contains an integer number, representing the result of the query.
[align=left]Sample Input[/align]

2
6
1 2 3 4 3 5
3
1 2
3 5
2 6
6
1 1 1 2 3 5
3
1 1
2 4
3 5

[align=left]Sample Output[/align]

3
7
14
1
3
6

//

先对y排序，记录每一个值前一个出现的位置，然后每次一个值v时，若前面出现过v则将前面一个v灭掉，然后将当前位置加上v。。。其实就是保留最后一个出现v的位置的v，前面的地方都归0.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
const int maxn=50010;
//离线算法
int n;
__int64 c[maxn];
int lowbit(int x){return x&(-x);}
void update(int x,int val)
{
for(int i=x;i<=n;i+=lowbit(i))
{
c[i]+=val;
}
}
__int64 getsum(int x)
{
__int64 cnt=0;
for(int i=x;i>=1;i-=lowbit(i))
{
cnt+=c[i];
}
return cnt;
}

int a[maxn];
map<int,int> hash;
struct Node
{
int l,r;
int idx;
};
Node qu[maxn*4];
__int64 ans[maxn*4];
bool cmp(Node h,Node k)
{
return h.r<k.r;
}
int main()
{
int ci;scanf("%d",&ci);
while(ci--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int q;scanf("%d",&q);
for(int i=0;i<q;i++)
{
int l,r;scanf("%d%d",&l,&r);
if(l>r) swap(l,r);
qu[i].l=l,qu[i].r=r;
qu[i].idx=i;
}
sort(qu,qu+q,cmp);
for(int i=1;i<=n;i++) c[i]=0;
hash.clear();
int r=1;
for(int i=0;i<q;i++)
{
while(r<=qu[i].r)
{
if(hash[a[r]]!=0)
{
update(hash[a[r]],-a[r]);
}
hash[a[r]]=r;
update(r,a[r]);
r++;
}
ans[qu[i].idx]=getsum(qu[i].r)-getsum(qu[i].l-1);
}
for(int i=0;i<q;i++) printf("%I64d\n",ans[i]);
}
return 0;
}