xtu資料結構 D. Necklace

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D. NecklaceTime Limit: 5000msMemory Limit: 32768KB64-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. InputThe 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. OutputFor each query, output a line contains an integer number, representing the result of the query. Sample Input

261 2 3 4 3 531 23 52 661 1 1 2 3 531 12 43 5

Sample Output

3714136

 解題：離線樹狀數組，為什麼要把y座標從小到大進行排序呢？因為樹狀數組的特性所致，右邊的節點可能包含左邊的節點的值，所以從左往右，不斷去掉重複的數值更簡便。

離線處理即先一次性把所有詢問儲存起來。最後一次性回複。用一個數組pre[i]記錄元素d[i]距離i最近的一次出現的下標。pre[i] == -1即表示該元素d[i]目前是唯一的，不存在重複元素的。

 

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #define LL long long13 #define INF 0x3f3f3f14 using namespace std;15 const int maxn = 210000;16 struct query {17     int x,y,id;18 } q[maxn];19 LL tree[51000];20 int pre[51000],loc[1100000],n,m,d[51000];21 LL ans[maxn];22 bool cmp(const query &a,const query &b) {23     return a.y < b.y;24 }25 int lowbit(int x) {26     return x&(-x);27 }28 void update(int x,int val) {29     for(; x <= n; x += lowbit(x)) {30         tree[x] += val;31     }32 }33 LL sum(int x) {34     LL ans = 0;35     for(; x; x -= lowbit(x))36         ans += tree[x];37     return ans;38 }39 int main() {40     int t,i,j;41     scanf("%d",&t);42     while(t--) {43         memset(loc,-1,sizeof(loc));44         memset(tree,0,sizeof(tree));45         scanf("%d",&n);46         for(i = 1; i <= n; i++) {47             scanf("%d",d+i);48             pre[i] = loc[d[i]];49             loc[d[i]] = i;50             update(i,d[i]);51         }52         scanf("%d",&m);53         for(i = 1; i <= m; i++) {54             scanf("%d %d",&q[i].x,&q[i].y);55             q[i].id = i;56         }57         sort(q+1,q+m+1,cmp);58         int y = 0;59         for(i = 1; i <= m; i++) {60             for(j = y+1; j <= q[i].y; j++) {61                 if(pre[j] != -1) update(pre[j],-d[j]);62             }63             y = q[i].y;64             ans[q[i].id] = sum(q[i].y) - sum(q[i].x-1);65         }66         for(i = 1; i <= m; i++) {67             printf("%I64d\n",ans[i]);68         }69     }70     return 0;71 }

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