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：

 

  1 #include <iostream>

  2 #include <cstring>

  3 #include <cstdio>

  4 #include <algorithm>

  5 

  6 using namespace std;

  7 

  8 #define lson l, m, rt->left

  9 #define rson m + 1, r, rt->right

 10 

 11 const int maxn = 55555;

 12 

 13 struct Node {

 14     int sum;

 15     Node *left, *right;

 16 }*root[maxn], tree[2000000], *idx;

 17 

 18 int num[maxn], p[maxn], mp[maxn];

 19 int n, m;

 20 int tol;

 21 

 22 

 23 inline Node* nextNode() {

 24     idx->sum = 0;

 25     idx->left = idx->right = NULL;

 26     return idx++;

 27 }

 28 

 29 inline Node* copyNode(Node* temp) {

 30     idx->sum = temp->sum;

 31     idx->left = temp->left;

 32     idx->right = temp->right;

 33     return idx++;

 34 }

 35 

 36 inline bool cmp(int a, int b) {

 37     return num[a] < num[b];

 38 }

 39 

 40 void input() {

 41     scanf("%d%d", &n, &m);

 42     for(int i = 1; i <= n; i++) {

 43         scanf("%d", &num[i]);

 44         p[i] = i;

 45     }

 46     sort(p + 1, p + n + 1, cmp);

 47     int pre = -0x7f7f7f7f;

 48     tol = 0;

 49     for(int i = 1; i <= n; i++) {

 50         if(num[ p[i] ] == pre) {

 51             num[ p[i] ] = tol;

 52         }

 53         else {

 54             pre = num[ p[i] ];

 55             mp[ ++tol ] = num[ p[i] ];

 56             num[ p[i] ] = tol;

 57         }

 58     }

 59 }

 60 

 61 void build(int l, int r, Node* rt) {

 62     if(l == r) return;

 63     int m = (l + r) >> 1;

 64     rt->left = nextNode();

 65     rt->right = nextNode();

 66     build(lson);

 67     build(rson);

 68 }

 69 

 70 int query(int ll, int rr, int l, int r, Node* rtl, Node* rtr) {

 71     if(ll <= l && r <= rr) {

 72         return rtr->sum - rtl->sum;

 73     }

 74     int m = (l + r) >> 1;

 75     int ret = 0;

 76     if(ll <= m) ret += query(ll, rr, l, m, rtl->left, rtr->left);

 77     if(rr > m) ret += query(ll, rr, m + 1, r, rtl->right, rtr->right);

 78     return ret;

 79 }

 80 

 81 Node* add(int x, int l, int r, Node* rt) {

 82     Node* temp = copyNode(rt);

 83     if(l == r) {

 84         temp->sum++;

 85         return temp;

 86     }

 87     int m = (l + r) >> 1;

 88     if(x <= m) temp->left = add(x, lson);

 89     else temp->right = add(x, rson);

 90     temp->sum = temp->left->sum + temp->right->sum;

 91     return temp;

 92 }

 93 

 94 void build() {

 95     idx = tree;

 96     root[0] = nextNode();

 97     build(1, tol, root[0]);

 98     for(int i = 1; i <= n; i++) {

 99         root[i] = add(num[i], 1, tol, root[i - 1]);

100     }

101 }

102 

103 inline int bin1(int x) { //find the left-most number that is >= x

104     int left = 1, right = tol;

105     int ans = -1;

106     while(left <= right) {

107         int mid = (left + right) >> 1;

108         if(mp[mid] >= x) {

109             ans = mid;

110             right = mid - 1;

111         }

112         else left = mid + 1;

113     }

114     return ans;

115 }

116 

117 inline int bin2(int x) { // find the right-most number that is <= x

118     int left = 1, right = tol;

119     int ans = -1;

120     while(left <= right) {

121         int mid = (left + right) >> 1;

122         if(mp[mid] <= x) {

123             ans = mid;

124             left = mid + 1;

125         }

126         else right = mid - 1;

127     }

128     return ans;

129 }

130 

131 void solve() {

132     static int cas = 1;

133     printf("Case #%d:\n", cas++);

134     while(m--) {

135         int l, r, a, b;

136         scanf("%d%d%d%d", &l, &r, &a, &b);

137         a = bin1(a);

138         b = bin2(b);

139         if(a == -1 || b == -1) {

140             puts("0");

141             continue;

142         }

143         printf("%d\n", query(a, b, 1, tol, root[l - 1], root[r]));

144     }

145 }

146 

147 int main() {

148     //freopen("input.in", "r", stdin);

149     //freopen("output.out", "w", stdout);

150 

151     int __t;

152     scanf("%d", &__t);

153     while(__t--) {

154         input();

155         build();

156         solve();

157     }

158     return 0;

159 }

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