pat甲级1122. Hamiltonian Cycle (25)

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1122. Hamiltonian Cycle (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 ... V_n

where n is the number of vertices in the list, and V_i'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

算法设计：

一个序列是哈密顿环必须满足下列条件，缺一不可：

1. 必须能构成一个环
2. 必须经过图中所有节点
3. 除了首尾这对相同结点可以出现2次外，其余节点在环中只能经过1次

邻接表版c++代码：

#include<bits/stdc++.h>
using namespace std;
int N,M,K;
vector<int>graph[205];//图
bool Hamiltonian(const vector<int>&v){//判断一个序列是否是哈密顿环
    if(v.size()!=N+1||v.front()!=v.back())//判断结点个数是否为图中结点总数+1、判断首尾结点是否相同
        return false;
    unordered_set<int>s;
    for(int i=0;i<v.size()-1;++i){//判断除尾节点外每个结点是否只出现一次
        if(s.find(v[i])!=s.cend())
            return false;
        s.insert(v[i]);
    }
    for(int i=1;i<v.size();++i)//判断序列中每两个相邻数之间是否有路径
        if(find(graph[v[i-1]].cbegin(),graph[v[i-1]].cend(),v[i])==graph[v[i-1]].cend())
            return false;
    return true;
}
int main(){
    scanf("%d%d",&N,&M);
    while(M--){
        int a,b;
        scanf("%d%d",&a,&b);
        graph[a].push_back(b);
        graph[b].push_back(a);
    }
    scanf("%d",&K);
    while(K--){
        scanf("%d",&M);
        vector<int>A(M);
        for(int i=0;i<M;++i)
            scanf("%d",&A[i]);
        printf("%s\n",Hamiltonian(A)?"YES":"NO");
    }
    return 0;
}

邻接矩阵版c++代码：

#include<bits/stdc++.h>
using namespace std;
int N,M,K;
bool graph[205][205];//图
bool Hamiltonian(const vector<int>&v){//判断一个序列是否是哈密顿环
    if(v.size()!=N+1||v.front()!=v.back())//判断结点个数是否为图中结点总数+1、判断首尾结点是否相同
        return false;
    unordered_set<int>s;
    for(int i=0;i<v.size()-1;++i){//判断除尾节点外每个结点是否只出现一次
        if(s.find(v[i])!=s.cend())
            return false;
        s.insert(v[i]);
    }
    for(int i=1;i<v.size();++i)//判断序列中每两个相邻数之间是否有路径
        if(!graph[v[i-1]][v[i]])
            return false;
    return true;
}
int main(){
    scanf("%d%d",&N,&M);
    while(M--){
        int a,b;
        scanf("%d%d",&a,&b);
        graph[a][b]=graph[b][a]=true;
    }
    scanf("%d",&K);
    while(K--){
        scanf("%d",&M);
        vector<int>A(M);
        for(int i=0;i<M;++i)
            scanf("%d",&A[i]);
        printf("%s\n",Hamiltonian(A)?"YES":"NO");
    }
    return 0;
}