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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010; //图的最大点数量
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/**
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共提供两组数据,样例1为不连通用例,样例2为连通用例
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样例1:不连通,5号结点为独立的
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5 4
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1 2
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2 3
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3 4
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1 4
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样例2:连通,不存在独立结点
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5 4
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1 2
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2 3
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3 4
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1 5
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检测各种算法是否能准确获取结果
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*/
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int n; //n个人
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int m; //m个亲戚
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int p; //询问p对亲戚关系
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int x, y; //输入两个人之间的关系
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int fa[N]; //并查集数组
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//要深入理解这个递归并压缩的过程
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int find(int x) {
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if (fa[x] != x)//如果x不是族长,递归找父亲,副产品就是找回的结果更新掉自己的家族信息。
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fa[x] = find(fa[x]);//非常经典的更新,路径压缩大法!
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//返回族长是谁
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return fa[x];
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}
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//加入家族集合中
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void join(int c1, int c2) {
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int f1 = find(c1), f2 = find(c2);
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if (f1 != f2)fa[f1] = f2;//各自找家长,如果家长不一样,就把C1的族长,认C2的族长为爸爸,C1的族长强烈表示不满意
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}
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int cnt;
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int main() {
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//n个人员,m个关系
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cin >> n >> m;
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//并查集初始化
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for (int i = 1; i <= n; i++)fa[i] = i; //自己是自己的老大
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//录入m种关系,使用并查集来判断图的连通性
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for (int i = 1; i <= m; i++) {
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cin >> x >> y;
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//加入并查集
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join(x, y);
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}
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//图已经搭好了,接下来看它们根节点是否相同,如只有一个相同的根节点,则说明是一个连通图
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for (int i = 1; i <= n; i++) if (fa[i] == i)cnt++;
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if (cnt == 1)printf("图是连通的\n");
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else printf("图不是连通的\n");
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return 0;
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}
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