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-coloringSolution:
No
6-coloring
No
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。
#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;
int main()
{
int n, m, k;
cin >> n >> m;
vector <vector<int>>path(n);
while (m--)
{
int a, b;
cin >> a >> b;
path[a].push_back(b);
path[b].push_back(a);
}
cin >> k;
while (k--)
{
vector<int>color(n, );
unordered_set<int>nums;
for (int i = ; i < n; ++i)
{
cin >> color[i];
nums.insert(color[i]);
}
bool flag = true;
for (int i = ; i < n && flag; ++i)
{
for (auto j : path[i])
{
if (color[i] == color[j])
{
flag = false;
break;
}
}
}
if (flag)
printf("%d-coloring\n", nums.size());
else
printf("No\n");
}
return ;
}
