xtu数据结构 D. Necklace

D. Necklace

Time Limit: 5000ms

Memory Limit: 32768KB

64-bit integer IO format: %I64d      Java class name: Main

 

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.

 

Input

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.

 

Output

For each query, output a line contains an integer number, representing the result of the query.

 

Sample Input

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

Sample Output

3
7
14
1
3
6

 解题：离线树状数组，为什么要把y坐标从小到大进行排序呢？因为树状数组的特性所致，右边的节点可能包含左边的节点的值，所以从左往右，不断去掉重复的数值更简便。

离线处理即先一次性把所有询问存储起来。最后一次性回复。用一个数组pre[i]记录元素d[i]距离i最近的一次出现的下标。pre[i] == -1即表示该元素d[i]目前是唯一的，不存在重复元素的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#define LL long long
#define INF 0x3f3f3f
using namespace std;
const int maxn = 210000;
struct query {
    int x,y,id;
} q[maxn];
LL tree[51000];
int pre[51000],loc[1100000],n,m,d[51000];
LL ans[maxn];
bool cmp(const query &a,const query &b) {
    return a.y < b.y;
}
int lowbit(int x) {
    return x&(-x);
}
void update(int x,int val) {
    for(; x <= n; x += lowbit(x)) {
        tree[x] += val;
    }
}
LL sum(int x) {
    LL ans = 0;
    for(; x; x -= lowbit(x))
        ans += tree[x];
    return ans;
}
int main() {
    int t,i,j;
    scanf("%d",&t);
    while(t--) {
        memset(loc,-1,sizeof(loc));
        memset(tree,0,sizeof(tree));
        scanf("%d",&n);
        for(i = 1; i <= n; i++) {
            scanf("%d",d+i);
            pre[i] = loc[d[i]];
            loc[d[i]] = i;
            update(i,d[i]);
        }
        scanf("%d",&m);
        for(i = 1; i <= m; i++) {
            scanf("%d %d",&q[i].x,&q[i].y);
            q[i].id = i;
        }
        sort(q+1,q+m+1,cmp);
        int y = 0;
        for(i = 1; i <= m; i++) {
            for(j = y+1; j <= q[i].y; j++) {
                if(pre[j] != -1) update(pre[j],-d[j]);
            }
            y = q[i].y;
            ans[q[i].id] = sum(q[i].y) - sum(q[i].x-1);
        }
        for(i = 1; i <= m; i++) {
            printf("%I64d
",ans[i]);
        }
    }
    return 0;
}