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 … 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
此题题意为一个简单圆包含了所有顶点，就有圆上的定点除了首尾一样，其余不能有重复，个数必须为n+1，否则必有小圆，在点与点连接时判断有无线
利用map数组存地图，
collection数组记录已访问
代码如下：
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
#define max 201
#define query_max 100001
vector<int> vec[query_max];
int map[max][max];
int collection[max];
int main(){
int n,edge;
scanf("%d%d",&n,&edge);
int v1,v2;
for(int i=0;i<edge;++i){
scanf("%d%d",&v1,&v2);
map[v1][v2]=1;
map[v2][v1]=1;
}
int query;
scanf("%d",&query);
int size,value;
for(int i=0;i<query;++i){
scanf("%d",&size);
vec[i].resize(0);
for(int j=0;j<size;++j){
scanf("%d",&value);
vec[i].push_back(value);
}
if(vec[i].size()!=(n+1)){
// printf("%d %d\n",vec[i].size(),n+1);
printf("NO\n");
// printf("ssss\n");
}else{
if(vec[i][0]!=vec[i][vec[i].size()-1]){
printf("NO\n");
}else{
memset(collection,0,sizeof(collection));
collection[vec[i][0]]=1;
int q=1;
for(;q<vec[i].size();++q){
if(collection[vec[i][q]]==1&&q!=vec[i].size()-1){
printf("NO\n");
break;
}
if(map[vec[i][q]][vec[i][q-1]]!=1){
printf("NO\n");
break;
}else{
collection[vec[i][q]]=1;
}
}
if(q==vec[i].size()){
printf("YES\n");
}
}
}
}
return 0;
}
