pat 1134 顶点覆盖问题

题目描述：

A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex of the set. Now given a graph with several vertex sets, you are supposed to tell if each of them is a vertex cover or not.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N and M (both no more than 104), being the total numbers of vertices and the edges, respectively. Then M lines follow, each describes
an edge by giving the indices (from 0 to N-1) of the two ends of the edge.

After the graph, a positive integer K (<= 100) is given, which is the number of queries. Then K lines of queries follow, each in the format:

Nv v[1] v[2] ... v[Nv]

where Nv is the number of vertices in the set, and v[i]'s are the indices of the vertices.

Output Specification:

For each query, print in a line "Yes" if the set is a vertex cover, or "No" if not.
Sample Input:

10 11
8 7
6 8
4 5
8 4
8 1
1 2
1 4
9 8
9 1
1 0
2 4
5
4 0 3 8 4
6 6 1 7 5 4 9
3 1 8 4
2 2 8
7 9 8 7 6 5 4 2

Sample Output:

No
Yes
Yes
No
No

这道题就是判断是否当前输入的顶点集合是一个顶点覆盖，既边集中每个边至少有一个顶点在该覆盖中，一开始用了比较暴力的做法，即逐个边进行比较，发现超时了：

#include<iostream>
#include<stack>
#include<math.h>
#include<stdio.h>
typedef struct{
int start;
int end;
}node;
using namespace std;
int main(){
node n[10001];
int N,M;
cin>>N>>M;
for(int i=0;i<M;i++){
cin>>n[i].start>>n[i].end;
}
int K;
cin>>K;
for(int i=0;i<K;i++){
int query[10001];
int test[10001]={0};
int Nv;
bool flag=true;
cin>>Nv;
for(int j=0;j<Nv;j++)
cin>>query[j];
for(int j=0;j<Nv;j++){
for(int k=0;k<M;k++){
if(query[j]!=n[k].start&&query[j]!=n[k].end){
continue;
}
else{
test[k]=1;
}
}
}
for(int j=0;j<M;j++){
if(test[j]==0){
flag=false;
break;
}
}
if(flag) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return 0;
}很显然这种暴力的做法复杂度较高，如果是最坏的情况下会比较很多次，因此后来想了很久，终于想出了将每个边进行标号处理，然后存在每个顶点信息中，这样就简化了许多。
#include<iostream>
#include<stack>
#include<math.h>
#include<stdio.h>
typedef struct{
int length=0;
int ele[10001];
}node;
node n[10001];
using namespace std;
int main(){
int N,M;
cin>>N>>M;
for(int i=0;i<M;i++){
int start,end;
cin>>start>>end;
n[start].ele[n[start].length++]=i;
n[end].ele[n[end].length++]=i;
}
int K;
cin>>K;
for(int i=0;i<K;i++){
int Nv;
cin>>Nv;
int test[10001]={0};
for(int j=0;j<Nv;j++){
int q;
cin>>q;
for(int k=0;k<n[q].length;k++)
test[n[q].ele[k]]++;
}
bool flag=true;
for(int j=0;j<M;j++){
if(test[j]==0){
flag=false;
break;
}
}
if(flag) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return 0;
}