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: [e-mail]     > 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;}