PAT Advanced 1134 Vertex Cover (25) [hash散列]
题目
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. Afer 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
题目分析
给出图的每条边顶点信息,给出几组顶点集合,判断顶点集合是否是vertex cover(vertex cover指:一个顶点集合,图每条边的顶点至少有一个在这个顶点集合中)
解题思路
算法 1
- 定义顶点结构体edge,两个顶点left,right
- 定义vector v,存放图每条边的信息
- 定义int ves[N],存放每个查询顶点集合中顶点出现次数
- 每个顶点集合的判断,需要遍历所有边信息
- 每条边的两个顶点至少有一个在顶点集合中,满足条件,打印Yes
- 只要有一条边的两个顶点都不在顶点集合中,不满足条件,打印No
算法2
- 定义一个vector数组,数组下标表示顶点,vector中存放的是该顶点所在边的编号
- 每次校验一个顶点集合,定义一个int hash[M]数组,下标为图的边,值为边的顶点是否至少有一个在顶点集合中,若边满足条件置为1
- 遍历hash[M]数组,是否所有元素都为1,表示每条边至少有一个顶点在顶点集合中,该顶点集合是vertex cover
Code
Code 01(算法1 最优)
#include <iostream>
#include <vector>
using namespace std;
struct edge {
int left,right;
};
int main(int argc,char * argv[]) {
int N,M,K,L,V;
scanf("%d %d", &N,&M);
edge es[M];
for(int i=0; i<M; i++) {
scanf("%d %d",&es[i].left,&es[i].right);
}
scanf("%d",&K);
for(int i=0; i<K; i++) {
scanf("%d", &L);
int ves[N]= {0};
for(int j=0; j<L; j++) {
scanf("%d",&V);
ves[V]++;
}
int j;
for(j=0; j<M; j++) {
if(ves[es[j].left]==0&&ves[es[j].right]==0)break;
}
if(j!=M)printf("No\n");
else printf("Yes\n");
}
return 0;
}
Code 02(算法2)
#include <iostream>
#include <vector>
using namespace std;
int main(int argc,char * argv[]) {
int N,M,K,L,V;
scanf("%d %d", &N,&M);
vector<int> es[N];
int f,r;
for(int i=0; i<M; i++) {
scanf("%d %d",&f,&r);
es[f].push_back(i);
es[r].push_back(i);
}
scanf("%d",&K);
for(int i=0; i<K; i++) {
scanf("%d", &L);
int hash[M]= {0};
for(int j=0; j<L; j++) {
scanf("%d",&V);
for(int t=0; t<es[V].size(); t++) {
hash[es[V][t]]=1;
}
}
int j;
for(j=0; j<M; j++) {
if(hash[j]==0) {
break;
}
}
if(j!=M)printf("No\n");
else printf("Yes\n");
}
return 0;
}
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