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< N <= 200), the number of vertices, and M, the number of edges in an undirected graph. Then M lines 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 K lines 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, or “NO” 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
题目大意:给出一个图,判断给定的路径是不是哈密尔顿路径
分析:1.设置falg1 判断节点是否多走、少走、或走成环
2.设置flag2 判断这条路能不能走通
3.当falg1、flag2都为1时是哈密尔顿路径,否则不是
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#include <iostream> #include <set> #include <vector> using namespace std; int main() { int n, m, cnt, k, a[210][210] = {0}; cin >> n >> m; for(int i = 0; i < m; i++) { int t1, t2; scanf("%d%d", &t1, &t2); a[t1][t2] = a[t2][t1] = 1; } cin >> cnt; while(cnt--) { cin >> k; vector<int> v(k); set<int> s; int flag1 = 1, flag2 = 1; for(int i = 0; i < k; i++) { scanf("%d", &v[i]); s.insert(v[i]); } if(s.size() != n || k - 1 != n || v[0] != v[k-1]) flag1 = 0; for(int i = 0; i < k - 1; i++) if(a[v[i]][v[i+1]] == 0) flag2 = 0; printf("%s",flag1 && flag2 ? "YES\n" : "NO\n"); } return 0; } |
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