1107 Social Clusters (30 分)合并有共同元素的集合

1⃣️集合个数最多是n个,即每人一个集合
2⃣️通过共同爱好找到人的编号,将相同爱好的人合并,而不是合并爱好
3⃣️并查集函数包括 查找根节点(同时压缩路径),union,init,
数组包括 isroot[], father[], 如果一个结点有多个元素,例如一
个人有多个爱好,找的是相同爱好,合并的是人,需
添加course[].通过course找到和人的编号,合并人


#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> father, isRoot;
int cmp1(int a, int b){
    
    
    return a > b;
}
int findFather(int x) {
    
    
    int a = x;
    while(x != father[x])
        x = father[x];
    while(a != father[a]) {
    
    
        int z = a;
        a = father[a];
        father[z] = x;
    }
    return x;
}
void Union(int a, int b) {
    
    
    int faA = findFather(a);
    int faB = findFather(b);
    if(faA != faB)
        father[faA] = faB;
}
void init(int n){
    
    
    for(int i = 1; i <= n; i++)
        father[i] = i;
}
int main() {
    
    
    int n, k, t, num = 0;
    int course[1001] = {
    
    0};
    scanf("%d", &n);
    father.resize(n + 1);
    isRoot.resize(n + 1);
    init(n);
    for(int i = 1; i <= n; i++) {
    
    
        scanf("%d:", &k);
        for(int j = 0; j < k; j++) {
    
    
            scanf("%d", &t);
            if(course[t] == 0)
                course[t] = i;     
            else
                Union(i, findFather(course[t]));         //和course[t]号是朋友
        }
    }
    for(int i = 1; i <= n; i++)
        isRoot[findFather(i)]++;
    for(int i = 1; i <= n; i++) {
    
    
        if(isRoot[i] != 0)
            num++;
    }
    printf("%d\n", num);
    sort(isRoot.begin(), isRoot.end(), cmp1);
    for(int i = 0; i < num; i++) {
    
    
        printf("%d", isRoot[i]);
        if(i != num - 1) printf(" ");
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/qq_42835526/article/details/113845093