hdoj3926-Hand in Hand(同构图的判断)
题目链接
思路
1.这个图就是同构图的判断,对于题目中的孩子的牵手方式,因为每个人只有两只手,所以可以看成图中每个节点的读书不大于2,因此连接方式只能是环或者链;
2.然后我们只用并查集来判断每个图含有多少环,多少链,然后对于环和链,用其中的一个节点来记录环或者链所含有节点个数;
3.然后将所有节点按是否成环和节点的数目来排序,最后再遍历判断一次结果。
code
#include <iostream> #include <algorithm> #include <cstring> #include <fstream> using namespace std; const int MAX = 10005; class Node{ public: int isCircle, count;//是否成环,以及节点的个数 }; Node node1[MAX], node2[MAX]; int f[MAX]; int cmp(Node a, Node b) { if(a.isCircle == b.isCircle) { return a.count > b.count; } return a.isCircle > b.isCircle; } void init() { for(int i = 0; i < MAX; ++ i) { f[i] = i; } } int find(int x) { if(x != f[x]) { f[x] = find(f[x]); } return f[x]; } int main() { //ifstream cin("data.in"); int t, cnt = 1; cin >> t; while(t --) { int n1, n2, m1, m2; bool ok = true; for(int i = 0; i < MAX; ++ i) { node1[i].isCircle = 0;//初始化每个孩子为单独个体,且不成环,看做节点数为1的链 node2[i].isCircle = 0; node1[i].count = 1; node2[i].count = 1; } cin >> n1 >> m1; init(); for(int i = 0; i < m1; i ++) { int x, y; cin >> x >> y; x = find(x); y = find(y); if(x != y) { //当两条链不成环时,连接两条链 f[y] = x; node1[x].count += node1[y].count; node1[y].count = 0; } else { node1[y].isCircle = 1;//成环时进行标记 } } cin >> n2 >> m2; init(); for(int i = 0; i < m2; i ++) { int x, y; cin >> x >> y; x = find(x); y = find(y); if(x != y) { f[y] = x; node2[x].count += node2[y].count; node2[y].count = 0; } else { node2[y].isCircle = 1; } } if(n1 != n2 || m1 != m2) { ok = false; } else { sort(node1+1, node1+n1+1, cmp); sort(node2+1, node2+n2+1, cmp); for(int i = 1; i < n1; i ++) { if(node1[i].isCircle != node2[i].isCircle || node1[i].count != node2[i].count){ ok = false; break; } } } cout << "Case #" << cnt ++ << ": "; if(ok) { cout << "YES" << endl; } else { cout << "NO" << endl; } } return 0; }