1154 Vertex Coloring -PAT甲级

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

一开始做的时候，用num来标记颜色的数量，用vst数组来标记此颜色是否出现过，有一个样例一直显示段错误，找了半天，发现题目中说颜色的编号在整数的范围中，并不是10000的范围，坑

后来用了集合，样例全都通过

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=10010;
vector<int>graph[N];
set<int>st;
int color[N];
int n,m,k,flag=1,num;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		graph[u].push_back(v);
		graph[v].push_back(u);
	}
	scanf("%d",&k);
	for(int i=1;i<=k;i++){
		st.clear();
		flag=1;
		for(int j=0;j<n;j++){
			scanf("%d",&color[j]);
			st.insert(color[j]);
		}
		for(int j=0;j<n;j++){
			if(graph[j].size()==0) continue;
			for(int h=0;h<graph[j].size();h++){
				if(color[j]==color[graph[j][h]]){
					flag=0;
					break;
				}
			}
			if(!flag) break;
		}
		if(flag){
			printf("%d-coloring\n",st.size());
		}else{
			printf("No\n");
		}
	}
	return 0;
}