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.
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 1), 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 Klines 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.
Output Specification:
For each coloring, print in a line k-coloring if it is a proper k-coloring for some positive k, 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
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
Sample Output:
4-coloring No 6-coloring No
Solution:
这道题很水，就是一个简单的判断图的边的两个顶点是不是同一个颜色，把图的精髓都没考出来，唯一有点考点的就是需要使用一个set来统计有几种颜色。
1 #include <iostream> 2 #include <vector> 3 #include <unordered_set> 4 using namespace std; 5 int main() 6 { 7 int n, m, k; 8 cin >> n >> m; 9 vector <vector<int>>path(n); 10 while (m--) 11 { 12 int a, b; 13 cin >> a >> b; 14 path[a].push_back(b); 15 path[b].push_back(a); 16 } 17 cin >> k; 18 while (k--) 19 { 20 vector<int>color(n, 0); 21 unordered_set<int>nums; 22 for (int i = 0; i < n; ++i) 23 { 24 cin >> color[i]; 25 nums.insert(color[i]); 26 } 27 bool flag = true; 28 for (int i = 0; i < n && flag; ++i) 29 { 30 for (auto j : path[i]) 31 { 32 if (color[i] == color[j]) 33 { 34 flag = false; 35 break; 36 } 37 } 38 } 39 if (flag) 40 printf("%d-coloring ", nums.size()); 41 else 42 printf("No "); 43 } 44 return 0; 45 }