2018年冬季PAT甲级考试 7-3 Vertex Coloring （25 分）

最新推荐文章于 2021-02-04 17:44:29 发布

chenxli23

最新推荐文章于 2021-02-04 17:44:29 发布

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本文链接：https://blog.youkuaiyun.com/Hickey_Chen/article/details/84933842

本文探讨了图论中的k-着色问题，即如何使用最多k种颜色为图的顶点上色，使得任意两个相邻顶点颜色不同。通过输入图的顶点数、边数以及具体的边连接信息，程序检查给定的着色方案是否为有效k-着色。

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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 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.

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

思路：这道题也不难，一开始用矩阵存边，只得到15分...后来就干脆采用pair存储一条边，这样也方便后面的判断，减少了复杂度。

程序：

#include <cstdio>
#include <vector>
#include <set>
#include <algorithm>
#include <utility>
using namespace std;
int a[10001];
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	fill(a,a+n,-1);
	vector<pair<int,int> > v;
	for(int i = 0; i < m; i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);
		v.push_back(make_pair(a,b));
	}
	int k;
	scanf("%d",&k);
	for(int i = 0; i < k; i++)
	{
		set<int> st;
		for(int j = 0; j < n; j++)
		{
			int color;
			scanf("%d",&color);
			st.insert(color);
			a[j] = color;
		}
		bool flag = true;
		for(int j = 0; j < m; j++)
		{
			if(a[v[j].first] == a[v[j].second])
			{
				flag = false;
				printf("No\n");
				break;
			}
		}
		if(flag)
		  printf("%d-coloring\n",st.size());
	}
	return 0;
}