C. Vertex Cover (25)

時間限制 600 ms
內存限制 65536 kB
代碼長度限制 8000 B
判題程序 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

題目大意：給n個結點m條邊，再給k個集合。對這k個集合逐個進行判斷。每個集合S里面的數字都是結點編號，

求問整個圖所有的m條邊兩端的結點，是否至少一個結點出自集合S中。如果是，輸出Yes否則輸出No

分析：用vector v[n]保存某結點屬于的某條邊的編號，比如a b兩個結點構成的這條邊的編號為0，

則v[a].push_back(0)，v[b].push_back(0)——表示a屬于0號邊，b也屬于0號邊。對于每一個集合做判斷，

遍歷集合中的每一個元素，將當前元素能夠屬于的邊的編號i對應的hash[i]標記為1，表示這條邊是滿足有一個結點出自集合S中的。

最后判斷hash數組中的每一個值是否都是1，如果有不是1的，說明這條邊的兩端結點沒有一個出自集合S中，則輸出No。否則輸出Yes～

代碼：

#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; }

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