sdutoj 2610 Boring Counting

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2610

Boring Counting

Time Limit: 3000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

    In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you task is to answer a list of queries, for each query, please tell us among [L, R], how many Pi is not less than A and not greater than B( L<= i <= R). In other words, your task is to count the number of Pi (L <= i <= R,  A <= Pi <= B).

输入

     In the first line there is an integer T (1 < T <= 50), indicates the number of test cases. 
     For each case, the first line contains two numbers N and M (1 <= N, M <= 50000), the size of sequence P, the number of queries. The second line contains N numbers Pi(1 <= Pi <= 10^9), the number sequence P. Then there are M lines, each line contains four number L, R, A, B(1 <= L, R <= n, 1 <= A, B <= 10^9)

输出

    For each case, at first output a line ‘Case #c:’, c is the case number start from 1. Then for each query output a line contains the answer.

示例输入

1
13 5
6 9 5 2 3 6 8 7 3 2 5 1 4
1 13 1 10
1 13 3 6
3 6 3 6
2 8 2 8
1 9 1 9

示例输出

Case #1:
13
7
3
6
9

提示

 

来源

 2013年山东省第四届ACM大学生程序设计竞赛

示例程序

分析：

题意是说给你一串数字，求在动态区间[L , R]内的在[A , B]范围内的数的个数。

数据很大，一般的方法会超。

官方标程：

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>

 using namespace std;

 #define lson l, m, rt->left
 #define rson m + 1, r, rt->right

 const int maxn = ;

 struct Node {
     int sum;
     Node *left, *right;
 }*root[maxn], tree[], *idx;

 int num[maxn], p[maxn], mp[maxn];
 int n, m;
 int tol;

 inline Node* nextNode() {
     idx->sum = ;
     idx->left = idx->right = NULL;
     return idx++;
 }

 inline Node* copyNode(Node* temp) {
     idx->sum = temp->sum;
     idx->left = temp->left;
     idx->right = temp->right;
     return idx++;
 }

 inline bool cmp(int a, int b) {
     return num[a] < num[b];
 }

 void input() {
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n; i++) {
         scanf("%d", &num[i]);
         p[i] = i;
     }
     sort(p + , p + n + , cmp);
     int pre = -0x7f7f7f7f;
     tol = ;
     for(int i = ; i <= n; i++) {
         if(num[ p[i] ] == pre) {
             num[ p[i] ] = tol;
         }
         else {
             pre = num[ p[i] ];
             mp[ ++tol ] = num[ p[i] ];
             num[ p[i] ] = tol;
         }
     }
 }

 void build(int l, int r, Node* rt) {
     if(l == r) return;
     int m = (l + r) >> ;
     rt->left = nextNode();
     rt->right = nextNode();
     build(lson);
     build(rson);
 }

 int query(int ll, int rr, int l, int r, Node* rtl, Node* rtr) {
     if(ll <= l && r <= rr) {
         return rtr->sum - rtl->sum;
     }
     int m = (l + r) >> ;
     int ret = ;
     if(ll <= m) ret += query(ll, rr, l, m, rtl->left, rtr->left);
     if(rr > m) ret += query(ll, rr, m + , r, rtl->right, rtr->right);
     return ret;
 }

 Node* add(int x, int l, int r, Node* rt) {
     Node* temp = copyNode(rt);
     if(l == r) {
         temp->sum++;
         return temp;
     }
     int m = (l + r) >> ;
     if(x <= m) temp->left = add(x, lson);
     else temp->right = add(x, rson);
     temp->sum = temp->left->sum + temp->right->sum;
     return temp;
 }

 void build() {
     idx = tree;
     root[] = nextNode();
     build(, tol, root[]);
     for(int i = ; i <= n; i++) {
         root[i] = add(num[i], , tol, root[i - ]);
     }
 }

 inline int bin1(int x) { //find the left-most number that is >= x
     int left = , right = tol;
     int ans = -;
     while(left <= right) {
         int mid = (left + right) >> ;
         if(mp[mid] >= x) {
             ans = mid;
             right = mid - ;
         }
         else left = mid + ;
     }
     return ans;
 }

 inline int bin2(int x) { // find the right-most number that is <= x
     int left = , right = tol;
     int ans = -;
     while(left <= right) {
         int mid = (left + right) >> ;
         if(mp[mid] <= x) {
             ans = mid;
             left = mid + ;
         }
         else right = mid - ;
     }
     return ans;
 }

 void solve() {
     static int cas = ;
     printf("Case #%d:\n", cas++);
     while(m--) {
         int l, r, a, b;
         scanf("%d%d%d%d", &l, &r, &a, &b);
         a = bin1(a);
         b = bin2(b);
         if(a == - || b == -) {
             puts("");
             continue;
         }
         printf("%d\n", query(a, b, , tol, root[l - ], root[r]));
     }
 }

 int main() {
     //freopen("input.in", "r", stdin);
     //freopen("output.out", "w", stdout);

     int __t;
     scanf("%d", &__t);
     while(__t--) {
         input();
         build();
         solve();
     }
     return ;
 }