PAT_A1122#Hamiltonian Cycle
作者:互联网
Source:
Description:
The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".
In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers N (2), the number of vertices, and M, the number of edges in an undirected graph. Then Mlines follow, each describes an edge in the format
Vertex1 Vertex2
, where the vertices are numbered from 1 to N. The next line gives a positive integer K which is the number of queries, followed by Klines of queries, each in the format:n V1 V2 ... Vn
where n is the number of vertices in the list, and Vi's are the vertices on a path.
Output Specification:
For each query, print in a line
YES
if the path does form a Hamiltonian cycle, orNO
if not.
Sample Input:
6 10 6 2 3 4 1 5 2 5 3 1 4 1 1 6 6 3 1 2 4 5 6 7 5 1 4 3 6 2 5 6 5 1 4 3 6 2 9 6 2 1 6 3 4 5 2 6 4 1 2 5 1 7 6 1 3 4 5 2 6 7 6 1 2 5 4 3 1
Sample Output:
YES NO NO NO YES NO
Keys:
- 图的存储和遍历
Code:
1 /* 2 Data: 2019-06-03 21:33:04 3 Problem: PAT_A1122#Hamiltonian Cycle 4 AC: 32:44 5 6 题目大意: 7 H圈定义:简单圈且包含全部顶点; 8 判断所给圈是否为H圈 9 */ 10 11 #include<cstdio> 12 #include<algorithm> 13 using namespace std; 14 const int M=220; 15 int grap[M][M],vis[M],hc[M]; 16 17 int main() 18 { 19 #ifdef ONLINE_JUDGE 20 #else 21 freopen("Test.txt", "r", stdin); 22 #endif 23 24 int n,m,k,v1,v2; 25 scanf("%d%d", &n,&m); 26 fill(grap[0],grap[0]+M*M,0); 27 for(int i=0; i<m; i++) 28 { 29 scanf("%d%d", &v1,&v2); 30 grap[v1][v2]=1; 31 grap[v2][v1]=1; 32 } 33 scanf("%d", &m); 34 for(int i=0; i<m; i++) 35 { 36 scanf("%d", &k); 37 fill(vis,vis+M,0); 38 for(int j=0; j<k; j++){ 39 scanf("%d", &hc[j]); 40 if(j==0) v1=hc[j]; 41 if(j==k-1) v2=hc[j]; 42 if(j!=0) vis[hc[j]]++; 43 if(vis[hc[j]]==2) 44 vis[0]=1; 45 } 46 if(k==n+1 && v1==v2 && vis[0]==0) 47 { 48 for(int j=1; j<k; j++) 49 { 50 v2=hc[j]; 51 if(grap[v1][v2]==0){ 52 vis[0]=1; 53 break; 54 } 55 v1=v2; 56 } 57 if(vis[0]==0){ 58 printf("YES\n"); 59 continue; 60 } 61 } 62 printf("NO\n"); 63 } 64 65 return 0; 66 }
标签:A1122,PAT,Hamiltonian,int,NO,number,each,Cycle,cycle 来源: https://www.cnblogs.com/blue-lin/p/10970562.html