A proper vertex coloring is a labeling of the graph's vertices with colors such that no two vertices sharing the same edge have the same color. A coloring using at most k colors is called a (proper) k-coloring.
Now you are supposed to tell if a given coloring is a proper k-coloring.
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 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 colorings you are supposed to check. Then K lines follow, each contains N colors which are represented by non-negative integers in the range of int. The i-th color is the color of the i-th vertex.
For each coloring, print in a line k-coloring if it is a proper k-coloring for some positive k, or No if not.
- 10 11
- 8 7
- 6 8
- 4 5
- 8 4
- 8 1
- 1 2
- 1 4
- 9 8
- 9 1
- 1 0
- 2 4
- 4
- 0 1 0 1 4 1 0 1 3 0
- 0 1 0 1 4 1 0 1 0 0
- 8 1 0 1 4 1 0 5 3 0
- 1 2 3 4 5 6 7 8 8 9
- 4-coloring
- No
- 6-coloring
- No
解题思路:给定的序列,任意两个相连的点不能涂相同的颜色, 满足条件的序列中有几个不同颜色的点就是几色问题。
- #include
- #include
- #include
- #include
-
- using namespace std;
-
- const int N = 1e4 + 10;
-
- int n , m;
- vector< vector<int> >v(N);
- int a[N];
-
- bool check()
- {
- for(int i = 0;i < n;i ++)
- {
- for(int j = 0;j < v[i].size();j ++)
- {
- if(a[i] == a[v[i][j]]) return false;
- }
- }
- return true;
- }
-
- int main()
- {
- cin >> n >> m;
- for(int i = 0;i < m;i ++)
- {
- int x , b;
- cin >> x >> b;
- v[x].push_back(b);
- v[b].push_back(x);
- }
-
- int k;
- cin >> k;
- while(k --)
- {
- unordered_map<int , int>mp;
- for(int i = 0;i < n;i ++)
- {
- cin >> a[i];
- mp[a[i]] ++;
- }
-
- int cnt = mp.size();//可能方案数
-
- bool flag = check();
- if(flag) printf("%d-coloring\n" , cnt);
- else puts("No");
- }
- return 0;
- }