PAT (Advanced Level) Practice 1154 Vertex Coloring (25分) (直接存边遍历)

1.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 10​4​​), 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

2.题目分析

学死了…………一看到图就想到邻接矩阵，甚至想到DFS………………（无一例外的超时）

其实直接用vector存边遍历就行……

3.代码

1.正确代码

 

#include<iostream>
#include<cstdio>
#include<vector>
#include<unordered_set>
using namespace std;
#define MAX 10001
int n, m, k;
int color[MAX];
struct node
{
	int s;
	int e;
};
int main()
{
	int a, b;
	scanf("%d %d", &n, &m);
	vector<node>list;
	for (int i = 0; i < m; i++)
	{
		scanf("%d %d", &a, &b);
		list.push_back({ a,b });
	}
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		bool s = true;
		unordered_set<int>com;
		for (int j = 0; j < n; j++)
		{
			scanf("%d", &color[j]);
			com.insert(color[j]);
		}
		for (int j = 0; j < list.size(); j++)
		{
			if (color[list[j].e] == color[list[j].s]) { s = false; break; }
		}
		
		if (s)printf("%d-coloring\n", com.size());
		else  printf("No\n");
	}

}

2.超时代码

#include<iostream>
#include<queue>
#include<vector>
#include<string>
#include<cstring>
#include<cstdio>
#include<unordered_set>
using namespace std;
#define INF 1000001
#define MAX 10001
int edges[MAX][MAX];
int n, m,k;
int visited[MAX];
int color[MAX];

bool BFS(int v)
{
	bool success = true;
	memset(visited, 0, sizeof(visited));
	int w = 0;
	queue<int>qu;
	visited[v] = 1;
	qu.push(v);
	vector<int> list;
	while (!qu.empty())
	{
		w = qu.front();
		qu.pop();
		list.push_back(w);
	}
	for (int j = 0; j < list.size(); j++)
	{
		w = list[j];
		for (int i = 0; i <n; i++)
		{
			if (edges[w][i] != 0&&visited[i] == 0)
			{
				if (color[v] == color[i])success = false;
				visited[i] = 1;
				qu.push(i);
			}
		}
	}
	return success;
}
int main()
{
	int a,b;
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; i++)
	{
		scanf("%d %d", &a, &b);
		edges[a][b] = 1;
		edges[b][a] = 1;
	}
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		bool s = true;
		unordered_set<int>com;
		for (int j = 0; j < n; j++)
		{
			scanf("%d", &color[j]);

			if (com.find(color[j]) == com.end())com.insert(color[j]);
		}
		for (int j = 0; j < n; j++)
		{
			if (!BFS(j)) { printf("No\n"); s = false; break; }
		}
		if (s)printf("%d-coloring\n",com.size());
	}

}

3.超时代码2

#include<iostream>
#include<cstdio>
#include<unordered_set>
using namespace std;
#define MAX 10001
int edges[MAX][MAX];
int n, m, k;
int color[MAX];
int main()
{
	int a, b;
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; i++)
	{
		scanf("%d %d", &a, &b);
		edges[a][b] = 1;
		edges[b][a] = 1;
	}
	cin >> k;
	for (int i = 0; i < k; i++)
	{
		bool s = true;
		unordered_set<int>com;
		for (int j = 0; j < n; j++)
		{
			scanf("%d", &color[j]);
			if (com.find(color[j]) == com.end())com.insert(color[j]);
		}
		for (int j = 0; j < n; j++)
		{
			for (int k = 0; k < n; k++)
			{
				if (edges[j][k] == 1)
				{
					if (color[j] == color[k]) { s = false; break; }
				}
			}
			if (!s)break;
		}
		if (s)printf("%d-coloring\n", com.size());
		else  printf("No\n");
	}

}