2017-9-17pat甲级 C

C. Vertex Cover (25)

时间限制

600 ms

内存限制

65536 kB

代码长度限制

8000 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:

10 118 76 84 58 48 11 21 49 89 11 02 454 0 3 8 46 6 1 7 5 4 93 1 8 42 2 87 9 8 7 6 5 4 2

Sample Output:

NoYesYesNoNo

题目大意：给n个结点m条边，再给k个集合。对这k个集合逐个进行判断。每个集合S里面的数字都是结点编号，
求问整个图所有的m条边两端的结点，是否至少一个结点出自集合S中。如果是，输出Yes否则输出No
分析：用vector v[n]保存某结点属于的某条边的编号，比如a b两个结点构成的这条边的编号为0，
则v[a].push_back(0)，v[b].push_back(0)——表示a属于0号边，b也属于0号边。对于每一个集合做判断，
遍历集合中的每一个元素，将当前元素能够属于的边的编号i对应的hash[i]标记为1，表示这条边是满足有一个结点出自集合S中的。
最后判断hash数组中的每一个值是否都是1，如果有不是1的，说明这条边的两端结点没有一个出自集合S中，则输出No。否则输出Yes～
代码：
#include <iostream>#include <vector>using namespace std;int main() {    int n, m, k, nv, a, b, num;    scanf("%d%d", &n, &m);    vector<int> v[n];    for (int i = 0;i < m; i++) {        scanf("%d%d", &a, &b);        v[a].push_back(i);        v[b].push_back(i);    }    scanf("%d", &k);    for (int i = 0; i < k; i++) {        scanf("%d", &nv);        int flag = 0;        vector<int> hash(m, 0);        for (int j = 0; j < nv; j++) {            scanf("%d", &num);            for (int t = 0; t < v[num].size(); t++)                hash[v[num][t]] = 1;        }        for (int j = 0; j < m; j++) {            if (hash[j] == 0) {                printf("No\n");                flag = 1;                break;            }        }        if (flag == 0) printf("Yes\n");    }    return 0;}