1134. Vertex Cover (25)
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 2Sample Output:
No Yes Yes No No
思路:給每條邊 編號,用鄰接表存儲 邊的編號,遍歷Nv個節點的邊,給遇到的邊上標記。最後如果有沒標記的就是No,反之Yes。
代碼:
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdio.h>
using namespace std;
vector <int> num[100000];
int main()
{
int n, m, k;
cin >> n >> m;
int a, b;
for (int i = 0; i < m; i++)
{
cin >> a >> b;
num[a].push_back(i);
num[b].push_back(i);
}
cin >> k;
while (k--)
{
int judge = 1;
cin >> a;
vector <int> edge (m, 0);
for (int i = 0; i < a; i++)
{
cin >> b;
for (int j = 0; j < num[b].size(); j++)
edge[num[b][j]] = 1;
}
for (int i = 0; i < m; i++)
if (!edge[i]) judge = 0;
if (judge) cout << "Yes" << endl;
else cout << "No" << endl;
}
}
因爲題目是 Vertex Cover ,所以做的時候總是從點出發,然後思緒就很混亂。看了別人的代碼以後才發現只要數邊就行了 _(:з)∠)_
思路歷程:
先想到的是用鄰接矩陣存儲,遍歷Nv個節點所在行的值,把連通的地方(1)改成0,最後遍歷全圖,如果還存在邊(1) 則輸出No,反之Yes 。接着發現是10^4,想想鐵定超時。
進而改用鄰接表存儲,遍歷Nv個節點的邊的同時刪除邊,最後如果還存在邊,則爲No,反之Yes。寫的時候發現:
1:有k組數據要測試,直接刪會影響後面處理
2:刪除對應頂點的邊時,得要搜索,時間複雜度提高
之後就走了邪路,想着用點亮頂點的方法 結果第一組數據就是頂點全亮但少一條邊的情況
看題解才知道點亮邊就行了……
PS:搜的時候看到另外一種思路,保存所有邊的信息,對於每一條邊的兩個頂點,在Vn頂點的集合中查找。如果兩個頂點都不在集合中,說明這條邊沒有被覆蓋
博客網址:
http://blog.csdn.net/akibayashi/article/details/78014749
代碼:
#include<iostream>
#include<set>
using namespace std;
int main()
{
int N, M, K, edges[10002][2];
cin >> N >> M;
for (int i = 0; i<M; i++)
{
int a, b;
cin >> a >> b;
edges[i][0] = a;
edges[i][1] = b;
}
cin >> K;
for (int i = 0; i<K; i++)
{
int n;
bool flag = true;
set<int> vset;
cin >> n;
for (int j = 0; j<n; j++)
{
int input;
cin >> input;
vset.insert(input);
}
for (int j = 0; j<M; j++)
{
if (vset.find(edges[j][0]) == vset.end() && vset.find(edges[j][1]) == vset.end())
{
flag = false;
break;
}
}
if (flag)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
return 0;
}