sdut2610---Boring Counting(离线+树状数组+离散化)
2015-04-06 20:21
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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
先把数据离散化,然后离线,树状数组处理
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
先把数据离散化,然后离线,树状数组处理
/************************************************************************* > File Name: H.cpp > Author: ALex > Mail: zchao1995@gmail.com > Created Time: 2015年04月06日 星期一 16时35分42秒 ************************************************************************/ #include <functional> #include <algorithm> #include <iostream> #include <fstream> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #include <queue> #include <stack> #include <map> #include <bitset> #include <set> #include <vector> using namespace std; const double pi = acos(-1.0); const int inf = 0x3f3f3f3f; const double eps = 1e-15; typedef long long LL; typedef pair <int, int> PLL; const int N = 50010; int arr ; int ans ; struct QUE { int a, b; int Id; int flag; QUE(int _a = 0, int _b = 0, int _Id = 0, int _flag = 0) { a = _a; b = _b; Id = _Id; flag = _flag; } }; vector <QUE> que ; int xis ; int tree ; int cnt, n; int search(int val) { int mid; int l = 1, r = cnt; while (l <= r) { mid = (l + r) >> 1; if (xis[mid] > val) { r = mid - 1; } else if (xis[mid] < val) { l = mid + 1; } else { break; } } return mid; } int lowbit(int x) { return x & (-x); } void add(int x) { for (int i = x; i < N; i += lowbit(i)) { ++tree[i]; } } int sum(int x) { int ans = 0; for (int i = x; i; i -= lowbit(i)) { ans += tree[i]; } return ans; } int main() { int t; scanf("%d", &t); int icase = 1; while (t--) { int m; scanf("%d%d", &n, &m); memset(tree, 0, sizeof(tree)); cnt = 0; for (int i = 1; i <= n; ++i) { que[i].clear(); scanf("%d", &arr[i]); xis[++cnt] = arr[i]; } int l, r, a, b; for (int i = 1; i <= m; ++i) { scanf("%d%d%d%d", &l, &r, &a, &b); que[r].push_back(QUE(a, b, i, 1)); que[l - 1].push_back(QUE(a, b, i, -1)); } memset(ans, 0, sizeof(ans)); sort(xis + 1, xis + 1 + cnt); cnt = unique (xis + 1, xis + 1 + cnt) - xis - 1; for (int i = 1; i <= n; ++i) { int val = search(arr[i]); add(val); int size = que[i].size(); for (int j = 0; j < size; ++j) { int a = lower_bound(xis + 1, xis + 1 + cnt, que[i][j].a) - xis; int b = lower_bound(xis + 1, xis + 1 + cnt, que[i][j].b) - xis; if (a > cnt) { continue; } --a; if (b > cnt) { b = cnt; } else if (xis[b] > que[i][j].b) { --b; } int flag = que[i][j].flag; int Id = que[i][j].Id; ans[Id] += flag * (sum(b) - sum(a)); } } printf("Case #%d:\n", icase++); for (int i = 1; i <= m; ++i) { printf("%d\n", ans[i]); } } return 0; }
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