Points in Segments (II)（区间统计 + 离散化）

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本文介绍了一种解决大规模区间统计问题的方法，通过离散化技术减少内存使用，并提出一种高效算法来处理大量区间和查询点的问题。适用于编程竞赛及实际场景。

1089 - Points in Segments (II)

Time Limit: 2 second(s)

Memory Limit: 64 MB

Given n segments (1 dimensional) and q points, for each point you have to find the number of segments which contain that point. A point p_i will lie in a segment A B if A ≤ p_i ≤ B.

For example, if the segments are (6 12), (8 8), (10 12), (8 11), (0 12) and the point is 11, then it is contained by 4 segments.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 50000) and q (1 ≤ q ≤ 50000).

Each of the next n lines contains two integers A_k B_k (0 ≤ A_k ≤ B_k ≤ 10⁸) denoting a segment.

Each of the next q lines contains an integer denoting a point. Each of them range in [0, 10⁸].

Output

For each case, print the case number in a single line. Then for each point, print the number of segments that contain that point.

Sample Input

Output for Sample Input

5 4

6 12

8 8

10 12

8 11

0 12

Case 1:

Notes

Dataset is huge, use faster I/O methods.

题意：

给出 T （<= 5）组数据，后给出 N（1 ~ 50000）和 Q（1 ~ 50000），代表有 N 个区间和 Q 个数，后给出 N 个区间和 Q 个数的询问，问 Q 这个数被覆盖了多少次，数的范围是 0 ~ 10 ^ 8。

思路：

区间统计 + 离散化。给出 50000 个数，极限情况也就有 3 X 50000 个不同的数，故数组不需要开到 10 ^ 8，只需要开到离散后的极限值就好。区间统计方法是起点++，（终点 + 1）--，最后统计后一项等于该项 + 前一项的值。最后输出结果即可。

AC：

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

using namespace std;

const int MAX = 50005 * 3;

int l[MAX], r[MAX], ask[MAX];
int num[MAX], res[MAX];


int main() {
        int t;
        scanf("%d", &t);

        for (int tt = 1; tt <= t; ++tt) {

                int n, m, ans = 0;
                scanf("%d%d", &n, &m);

                for (int i = 0; i < n; ++i) {
                        scanf("%d%d", &l[i], &r[i]);
                        num[ ans++ ] = l[i];
                        num[ ans++ ] = r[i];
                }

                for (int i = 0; i < m; ++i) {
                        scanf("%d", &ask[i]);
                        num[ ans++ ] = ask[i];
                }

                sort(num, num + ans);
                ans = unique(num, num + ans) - num;

                memset(res, 0, sizeof(res));
                for (int i = 0; i < n; ++i) {
                        int f = lower_bound(num, num + ans, l[i]) - num;
                        int t = lower_bound(num, num + ans, r[i]) - num;
                        ++res[f];
                        --res[t + 1];
                }

                for (int i = 0; i < ans; ++i)
                        res[i] += res[i - 1];

                printf("Case %d:\n", tt);
                for (int i = 0; i < m; ++i) {
                        int a = lower_bound(num, num + ans, ask[i]) - num;
                        printf("%d\n", res[a]);
                }

        }

        return 0;
}