1154. Vertex Coloring (25)-PAT甲级真题 (遍历边 哈希表)

题意

给一个图，n个节点，m条边，依次给出m条边的信息。然后询问k次，每次给出染色方案，每个节点一个颜色，要求共边的结点的颜色不同，最后输出是否k-coloring图。

思路

用edge数组记录边的端点信息。每次先数color个数，然后遍历每个边看看是否有一个边的两端颜色相同。最后按题目要求输出。

总结

• 仔细读题：each contains N colors which are represented by non-negative integers in the range of int. 这句话看漏导致初始用的bool st[10001]去数颜色数。
• 时间复杂度= k * n = 1E6

题解

//23:05
#include<bits/stdc++.h>
using namespace std;
int n, m, k, a, b, col[10001];
struct node{
	int a, b;
}edge[10001];
int main(){
	cin>>n>>m;
	for(int i = 0; i < m; i ++) scanf("%d %d", &edge[i].a, &edge[i].b);
	cin>>k;
	while(k--){
		int cnt = 0;
		unordered_map<int, bool> st;
		for(int i = 0; i < n; i ++){
			scanf("%d", &col[i]);
			if(!st[col[i]]){
				st[col[i]] = true;
				cnt++;
			}
		}
		for(int i = 0; i < m; i ++){
			if(col[edge[i].a] == col[edge[i].b]){
				puts("No");
				break;
			}
			if(i == m - 1) printf("%d-coloring\n", cnt); 
		}
	}
	return 0;
}

题目

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