PAT-1134. Vertex Cover (25)

1134. Vertex Cover (25)

时间限制

600 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

这道题的关键是容易超时，自己写的代码仔细想想确实会超时，再参考了别人的代码，感觉妙哉。将每条边编号，记录每个顶点相关的边，将询问的顶点相关的边都标记了，再遍历所有的边，发现没有标记的说明没有全覆盖。

也可以使用set容器，将所有询问顶点放入set中，遍历所有的边，判断set中是否有边的两个任意一个顶点，若没有则没有全覆盖。

#include <bits/stdc++.h>

using namespace std;

int N, M, K;
struct Node {
    int x;
    int y;
};
vector<Node> nodVec, nodVec1;

void deleteNode(int x, vector<Node> nodV) {
}

int main()
{
    cin>> N>> M;
    int x, y;
    for(int i = 0; i < M; i++) {//10^4
        scanf("%d%d", &x, &y);
        Node node;
        node.x = x;
        node.y = y;
        nodVec.push_back(node);
    }
    cin>> K;
    nodVec1 = nodVec;
    for(int i = 0; i < K; i++) {//10^2
        scanf("%d", &x);
        for(int j = 0; j < x; j++) {//10^4
            scanf("%d", &y);
            vector<Node>::iterator it = nodVec1.begin();
            while(it != nodVec1.end()) {//10^4每次都会遍历所有的边，虽然有erase，但是总量还是很大。
                if(y == it->x || y == it->y) {
                    it = nodVec1.erase(it);
                }
                else {
                    it++;
                }
            }
        }
        if(nodVec1.empty()) {
            printf("Yes
");
        }
        else {
            printf("No
");
        }
        nodVec1 = nodVec;
    }
    return 0;
}

#include<bits/stdc++.h>

using namespace std;

#define MAX 10004

int N, M, K;
vector<int> nodes[MAX];
int flag[MAX];

int main() {
    cin>> N>> M;
    int x, y;
    for(int i = 0; i < M; i++) {//10^4
        scanf("%d%d", &x, &y);
        nodes[x].push_back(i);
        nodes[y].push_back(i);
    }
    cin>> K;
    for(int i = 0; i < K; i++) {//10^2
        for(int j = 0; j < M; j++) {//10^4
            flag[j] = 0;
        }
        scanf("%d", &x);
        for(int j = 0; j < x; j++) {//10^4*2;这里每次遍历一个顶点的所有边，总共最多遍历两次所有的边
            scanf("%d", &y);
            vector<int>::iterator it;
            for(it = nodes[y].begin(); it != nodes[y].end(); it++) {
                flag[*it] = 1;
            }
        }
        int flag2 = 1;
        for(int j = 0; j < M; j++) {
            if(flag[j] == 0) {
                flag2 = 0;
                printf("No
");
                break;
            }
        }
        if(flag2 == 1) {
            printf("Yes
");
        }
    }
    return 0;
}

#include<bits/stdc++.h>

using namespace std;

int N, M, K;

#define MAX 10004

struct Node {
    int x, y;
}nodes[MAX];

set<int> nodeSet;

int main() {
    cin>>N>>M;
    for(int i = 0; i < M; i++) {//10^4
        scanf("%d%d", &nodes[i].x, &nodes[i].y);
    }
    cin>>K;
    int x, y;
    for(int i = 0; i < K; i++) {//10^2
        cin>> x;
        nodeSet.clear();
        for(int j = 0; j < x; j++) {//10^4
            scanf("%d", &y);
            nodeSet.insert(y);
        }
        int flag = 1;
        for(int i = 0; i < M; i++) {//10^4*log10^4
            if(nodeSet.find(nodes[i].x) == nodeSet.end() && nodeSet.count(nodes[i].y) == 0) {
                flag = 0;
                break;
            }
        }
        if(!flag) {
            printf("No
");
        }
        else {
            printf("Yes
");
        }
    }
    return 0;
}