<poj-1611>The Suspects

通过并查集算法解决学生群体中SARS潜在传播者的问题。案例输入包含学生数量及群体信息,输出潜在传播者的总数。

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The Suspects
Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 34781 Accepted: 16883

Description

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others. 
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP). 
Once a member in a group is a suspect, all members in the group are suspects. 
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space. 
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

Sample Output

4
1
1

解题思路

明明说是简单并查集的使用,其实在做这个题的时候大致的思路还是对的,但是因为一点习惯上的问题
还有就是我的并查集是现学的,可能理解的不太好,一个小的细节让我调试的很久还是没看出来

#include<stdio.h>
#include<string.h>

int pre[30000];
int ch[30000];

int find(int x)
{
    int r = x;//查找上一级
    while(r != pre[r])
    {
        r = pre[r];
    }
    int i = x, t;
    while(i != r)//压缩路径
    {
        t = pre[i];
        pre[i] = r;
        i = t;
    }
    return r;
}

void join(int x, int y)//合并
{
    x = find(x);
    y = find(y);
    if(x != y)
    {
        pre[y] = x;
    }
}

int main()
{
    int n, m;
    int t;
    while(scanf("%d %d",&n, &m)!=EOF)
    {
        memset(pre, 30001, sizeof(pre));
        int count = 0;
        if(n == 0&&m == 0) break;
        for(int i = 0; i <= n-1; i++)
        {
            pre[i] = i;
        }
        for(int i = 1; i <= m; i++)
        {
            scanf("%d",&t);
            for(int j = 0; j <= t-1; j++)
            {
                scanf("%d",&ch[j]);
            }
            for(int j = 0; j < t-1; j++)
            {
                join(ch[j], ch[j+1]);
            }
        }
        for(int i = 0; i <= n-1; i++)
        {
            if(find(i) == find(0))
                count++;
        }
        printf("%d\n",count);
    }
    return 0;
}


资源下载链接为: https://pan.quark.cn/s/9e7ef05254f8 行列式是线性代数的核心概念,在求解线性方程组、分析矩阵特性以及几何计算中都极为关键。本教程将讲解如何用C++实现行列式的计算,重点在于如何输出分数形式的结果。 行列式定义如下:对于n阶方阵A=(a_ij),其行列式由主对角线元素的乘积,按行或列的奇偶性赋予正负号后求和得到,记作det(A)。例如,2×2矩阵的行列式为det(A)=a11×a22-a12×a21,而更高阶矩阵的行列式可通过Laplace展开或Sarrus规则递归计算。 在C++中实现行列式计算时,首先需定义矩阵类或结构体,用二维数组存储矩阵元素,并实现初始化、加法、乘法、转置等操作。为支持分数形式输出,需引入分数类,包含分子和分母两个整数,并提供与整数、浮点数的转换以及加、减、乘、除等运算。C++中可借助std::pair表示分数,或自定义结构体并重载运算符。 计算行列式的函数实现上,3×3及以下矩阵可直接按定义计算,更大矩阵可采用Laplace展开或高斯 - 约旦消元法。Laplace展开是沿某行或列展开,将矩阵分解为多个小矩阵的行列式乘积,再递归计算。在处理分数输出时,需注意避免无限循环和除零错误,如在分数运算前先约简,确保分子分母互质,且所有计算基于整数进行,最后再转为浮点数,以避免浮点数误差。 为提升代码可读性和可维护性,建议采用面向对象编程,将矩阵类和分数类封装,每个类有明确功能和接口,便于后续扩展如矩阵求逆、计算特征值等功能。 总结C++实现行列式计算的关键步骤:一是定义矩阵类和分数类;二是实现矩阵基本操作;三是设计行列式计算函数;四是用分数类处理精确计算;五是编写测试用例验证程序正确性。通过这些步骤,可构建一个高效准确的行列式计算程序,支持分数形式计算,为C++编程和线性代数应用奠定基础。
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