sdut2610---Boring Counting(离线+树状数组+离散化)

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

/*************************************************************************
    > File Name: H.cpp
    > Author: ALex
    > Mail: [email protected]
    > 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[N];
int ans[N];
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[N];

int xis[N];
int tree[N];
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;
}