HDU-3874 Necklace (segment tree + offline processing)

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NecklaceTime limit:15000/5000 MS (java/others) Memory limit:65536/32768 K (java/others) Total submission (s): 4003 Accepted Submission (s): 1330 Problem Descriptionmery has a beautiful necklace. The necklace is a made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value is the same, so if the and more balls has the same beautiful value, we just Coun T 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 are only Counte D ONCE. For example, if the necklace are 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'll give you M intervals [l,r] (1<=l<=r<=n) and your 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 Bal Ls. 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 3714136for this problem, you can see my blog http://blog.csdn.net/qq_18661257/article/details/47419441 provides offline processing tutorial, basically can understand the mechanism of offline processing, Then we have to understand the topic, can not appear the number of duplicates, so the right-most value offline, the previous value of one by one can be deleted to get the correct answer, of course, we also need to note that the value range is a long long, I was in this view for nearly one hours, query () The return value of the function should also be a long long, the place where the father is here, others, we see the code basically can understand the #include <cstdio> #include <cstring> #include <algorithm> #include <vector>using namespace std; typedef long long LL; #define Lson RT << 1, L, Mid#define Rson RT << 1|1, Mid + 1, r#define root 1, 1, nconst I NT MAXN = 5e4 + 5;const int maxm = 2e5 + 5;const int maxs = 1e6 + 5;int N, M, T, PRE[MAXS]; LL ANS[MAXM],SUM[MAXN << 2];struct Node {int L, r, id; BOOL operator < (const node & object) Const {return R < OBJECT.R; }} node[maxm];void pushup (int rt) {Sum[rt] = Sum[rt << 1] + sum[rt << 1|1];} void build (int rt,int l,int R) {if (L = = r) {scanf ("%i64d", &sum[rt]); Return } int mid = (L + r) >> 1; Build (Lson); Build (Rson); Pushup (RT);} void update (int p,int RT, int l, int r) {if (L = = r) {Sum[rt] = 0; Return } int mid = (L + r) >> 1; if (P <= mid) Update (P, Lson); else Update (P, Rson); Pushup (RT);} LL query (int L, int R, int RT,int l, int r) {if (L <= l && R <= R) {return SUM[RT]; } int mid = (L + r) >> 1; LL ret = 0; if (L <= mid) ret + = query (l, R, Lson); if (R > Mid) ret + = query (L, R, Rson); return ret;} int main () {//freopen ("D://imput.txt", "R", stdin); scanf ("%d", &t); while (T--) {scanf ("%d", &n); Build (root); scanf ("%d", &m); for (int i = 1; I <= M; i + +) {scanf ("%d%d", &AMP;NODE[I].L, &AMP;NODE[I].R); Node[i].id = i; } memset (Pre,-1, sizeof (pre)); Sort (node + 1, node + M + 1); for (int i = 1,J = 1; I <= N; i++) {int tmp = query (I, I, root); if (tmp! = 0 && pre[tmp]! =-1) {//If there is a duplicate number in front of it, delete his update (pre[tmp],root); } Pre[tmp] = i; while (J <= M && NODE[J].R = = i) {//If the value on the right is equal to the current value, then you can refer to the tutorial in my blog ans[node[j].id] = query (Node [J].l, NODE[J].R, root); j + +; }} for (int i = 1; I <= M; i + +) {printf ("%i64d\n", Ans[i]); }} return 0;}

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HDU-3874 Necklace (segment tree + offline processing)