hdu 3926 Hand in Hand(并查集)
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3926
题意:给两个图,问是不是(同构)长得一样。然后限制条件是,这两个图虽然不一定是联通的,其子图一定是环或者链。
首先如果点或边数目不同,一定不同构,接下来用并查集构造图就可以了,然后对图中的所有点进行排序,按照子图的规模进行排序,如果规模相同,那么环优先或者链优先。
合并的时候,要注意同一小树往大树上插,否则会造成两个图排序后的结果可能不一致,误判为不同构。
#include <bits/stdc++.h>using namespace std;const int maxn = 10005;int father1[maxn], father2[maxn];int t, n1, m1, n2, m2;struct node { int Rank; bool isRing; bool operator < (const node& other) const { if(Rank == other.Rank) { return isRing == true; } return Rank < other.Rank; }}G1[maxn], G2[maxn];int Find(int x, int *father) { if(father[x] != x) { father[x] = Find(father[x], father); } return father[x];}void Union(int x, int y, int *father, node *g) { int a = Find(x, father); int b = Find(y, father); if(a == b) { g[a].isRing = true; } else { if(g[a].Rank > g[b].Rank) { father[b] = a; g[a].Rank += g[b].Rank; } else { father[a] = b; g[b].Rank += g[a].Rank; } }}bool Compare() { sort(G1 + 1, G1 + n1 + 1); sort(G2 + 1, G2 + n2 + 1); for (int i = 1; i <= n1; i ++) { if(G1[i].isRing != G2[i].isRing || G2[i].Rank != G1[i].Rank) { return false; } } return true;}int main() { int x, y; bool flag = true; int cas = 1; scanf("%d", &t); while (t --) { flag = true; scanf("%d%d", &n1, &m1); for (int i = 1; i <= n1; i ++) { father1[i] = father2[i] = i; G1[i].Rank = G2[i].Rank = 1; G1[i].isRing = G2[i].isRing = false; } for (int i = 0; i < m1; i ++) { scanf("%d%d", &x, &y); Union(x, y, father1, G1); } scanf("%d%d", &n2, &m2); if(n1 != n2 || m1 != m2) { flag = false; } for (int i = 0; i < m2; i ++) { scanf("%d%d", &x, &y); if(flag) { Union(x, y, father2, G2); } } if(flag) { flag = Compare(); } printf("Case #%d: ", cas ++); if(flag) { printf("YES\n"); } else { printf("NO\n"); } } return 0;}
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