题解 | #染色#

#include <iostream>
#include <vector>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, m;
    cin >> n >> m;

    // 为黄(0), 蓝(1), 红(2) 三种颜色创建差分数组
    vector<vector<int>> diff(3, vector<int>(n + 2, 0));

    for (int i = 0; i < m; ++i) {
        int l, r, c;
        cin >> l >> r >> c;
        // 颜色编号 c (1,2,3) 对应数组下标 (0,1,2)
        diff[c - 1][l]++;
        diff[c - 1][r + 1]--;
    }

    // 记录每罐油漆含有的颜色种类数
    vector<int> yellow_count(n + 1, 0);
    vector<int> blue_count(n + 1, 0);
    vector<int> red_count(n + 1, 0);

    // 计算黄色的覆盖情况
    for (int i = 1; i <= n; ++i) {
        diff[0][i] += diff[0][i - 1];
        if (diff[0][i] > 0) yellow_count[i] = 1;
    }
    // 计算蓝色的覆盖情况
    for (int i = 1; i <= n; ++i) {
        diff[1][i] += diff[1][i - 1];
        if (diff[1][i] > 0) blue_count[i] = 1;
    }
    // 计算红色的覆盖情况
    for (int i = 1; i <= n; ++i) {
        diff[2][i] += diff[2][i - 1];
        if (diff[2][i] > 0) red_count[i] = 1;
    }

    int green_count = 0;
    // 统计绿色油漆罐的数量
    for (int i = 1; i <= n; ++i) {
        if (yellow_count[i] > 0 && blue_count[i] > 0 && red_count[i] == 0) {
            green_count++;
        }
    }

    cout << green_count << "\n";

    return 0;
}

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.io.IOException;

public class Main {
    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

        in.nextToken();
        int n = (int) in.nval;
        in.nextToken();
        int m = (int) in.nval;

        // 为黄(0), 蓝(1), 红(2) 三种颜色创建差分数组
        int[][] diff = new int[3][n + 2];

        for (int i = 0; i < m; i++) {
            in.nextToken();
            int l = (int) in.nval;
            in.nextToken();
            int r = (int) in.nval;
            in.nextToken();
            int c = (int) in.nval;
            // 颜色编号 c (1,2,3) 对应数组下标 (0,1,2)
            diff[c - 1][l]++;
            diff[c - 1][r + 1]--;
        }

        boolean[] hasYellow = new boolean[n + 1];
        boolean[] hasBlue = new boolean[n + 1];
        boolean[] hasRed = new boolean[n + 1];

        // 计算黄色的覆盖情况
        int currentYellow = 0;
        for (int i = 1; i <= n; i++) {
            currentYellow += diff[0][i];
            if (currentYellow > 0) hasYellow[i] = true;
        }

        // 计算蓝色的覆盖情况
        int currentBlue = 0;
        for (int i = 1; i <= n; i++) {
            currentBlue += diff[1][i];
            if (currentBlue > 0) hasBlue[i] = true;
        }
        
        // 计算红色的覆盖情况
        int currentRed = 0;
        for (int i = 1; i <= n; i++) {
            currentRed += diff[2][i];
            if (currentRed > 0) hasRed[i] = true;
        }

        int greenCount = 0;
        // 统计绿色油漆罐的数量
        for (int i = 1; i <= n; i++) {
            if (hasYellow[i] && hasBlue[i] && !hasRed[i]) {
                greenCount++;
            }
        }
        
        System.out.println(greenCount);
    }
}

import sys

# 快速输入
input = sys.stdin.readline

def solve():
    n, m = map(int, input().split())

    # 为黄(0), 蓝(1), 红(2) 三种颜色创建差分数组
    diff = [[0] * (n + 2) for _ in range(3)]

    for _ in range(m):
        l, r, c = map(int, input().split())
        # 颜色编号 c (1,2,3) 对应数组下标 (0,1,2)
        diff[c - 1][l] += 1
        diff[c - 1][r + 1] -= 1

    has_yellow = [False] * (n + 1)
    has_blue = [False] * (n + 1)
    has_red = [False] * (n + 1)

    # 计算黄色的覆盖情况
    current_yellow = 0
    for i in range(1, n + 1):
        current_yellow += diff[0][i]
        if current_yellow > 0:
            has_yellow[i] = True
            
    # 计算蓝色的覆盖情况
    current_blue = 0
    for i in range(1, n + 1):
        current_blue += diff[1][i]
        if current_blue > 0:
            has_blue[i] = True

    # 计算红色的覆盖情况
    current_red = 0
    for i in range(1, n + 1):
        current_red += diff[2][i]
        if current_red > 0:
            has_red[i] = True

    green_count = 0
    # 统计绿色油漆罐的数量
    for i in range(1, n + 1):
        if has_yellow[i] and has_blue[i] and not has_red[i]:
            green_count += 1
    
    print(green_count)

solve()