判断同构图。题意得好好理解一下。
因为每个结点最大度为二，则任意连通分支必为链或者环 。可以对所有结点进行排序比较。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<map> 9 #include<iomanip> 10 #include<climits> 11 #include<string.h> 12 #include<cmath> 13 #include<stdlib.h> 14 #include<vector> 15 #include<stack> 16 #include<set> 17 #define INF 1e7 18 #define MAXN 10010 19 #define maxn 1000010 20 #define Mod 1000007 21 #define N 1010 22 using namespace std; 23 typedef long long LL; 24 25 struct graph{ 26 int son; 27 bool ring; 28 }g1[MAXN], g2[MAXN]; 29 int num1, num2; 30 int fa1[MAXN], fa2[MAXN]; 31 32 void init() 33 { 34 for (int i = 0; i < MAXN; ++i) { 35 fa1[i] = fa2[i] = i; 36 g1[i].son = g2[i].son = 1; 37 g1[i].ring = g2[i].ring = false; 38 } 39 } 40 41 int findset(int x, int fa[]) 42 { 43 return x == fa[x] ? x : findset(fa[x],fa); 44 } 45 46 void join(int x,int y,int fa[], graph g[]) 47 { 48 int root1, root2; 49 root1 = findset(x, fa); 50 root2 = findset(y, fa); 51 if (root1 == root2) 52 g[root1].ring = true; 53 else { 54 if (g[root1].son >= g[root2].son) { 55 fa[root2] = root1; 56 g[root1].son += g[root2].son; 57 } 58 else { 59 fa[root1] = root2; 60 g[root2].son += g[root1].son; 61 } 62 } 63 } 64 65 bool cmb(const graph &g1, const graph &g2)//子结点优先+先链后环排序 66 { 67 if (g1.son < g2.son) return true; 68 else if (g1.son == g2.son && g1.ring < g2.ring) 69 return true; 70 return false; 71 } 72 73 74 bool cmp(int num, graph g1[], graph g2[]) //判断图是否同构 75 { 76 sort(g1 + 1, g1 + num + 1, cmb); //排序 77 sort(g2 + 1, g2 + num + 1, cmb); 78 for (int i = 1; i <= num; ++i) 79 if (g1[i].son != g2[i].son || (g1[i].son == g2[i].son && g1[i].ring != g2[i].ring)) 80 return false; 81 return true; 82 } 83 84 int main() 85 { 86 int kase, T = 0; 87 int link1, link2; 88 int hand1, hand2; 89 int ans1, ans2; 90 bool flag; 91 scanf("%d",&kase); 92 while (kase--) { 93 flag = true; 94 init(); 95 cin >> num1 >> link1; 96 for (int i = 1; i <= link1; ++i) { 97 cin >> hand1 >> hand2; 98 join(hand1,hand2,fa1,g1); 99 } 100 cin >> num2 >> link2; 101 if (link1 != link2) flag = false; 102 for (int i = 1; i <= link2; ++i) { 103 cin >> hand1 >> hand2; 104 if (flag == false) continue; 105 else join(hand1,hand2,fa2,g2); 106 } 107 flag = cmp(num2, g1, g2); 108 if (flag == false) 109 printf("Case #%d: NO ", ++T); 110 else 111 { 112 printf("Case #%d: YES ", ++T); 113 } 114 115 } 116 return 0; 117 }