1154 Vertex Coloring (25 分)(set,hash)

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

题目大意:

给出一个图(先给出所有边,后给出每个点的颜色),问是否满足:所有的边的两个点的颜色不相同

分析:

把所有边存起来,把所有点的颜色存起来(存的过程中放入set统计颜色个数),枚举所有边,检查是否每条边的两点个颜色是否相同,若全不相同,则输出颜色个数,否则输出No~

原文链接:https://blog.csdn.net/liuchuo/article/details/84973030

题解

#include <bits/stdc++.h>

using namespace std;
struct node
{
    int t1,t2;
};
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,m,k;
    cin>>n>>m;
    vector<node> v(m);
    for(int i=0;i<m;i++)
        cin>>v[i].t1>>v[i].t2;
    cin>>k;
    while(k--){
        int a[10010]={0};
        bool flag=true;
        set<int> st;
        for(int i=0;i<n;i++){
            cin>>a[i];
            st.insert(a[i]);
        }
        for(int i=0;i<m;i++){
            if(a[v[i].t1]==a[v[i].t2]){
                flag=false;
                break;
            }
        }
        if(flag)
            printf("%d-coloring\n",st.size());
        else
            printf("No\n");
    }
    return 0;
}
posted @ 2022-01-25 23:54  勇往直前的力量  阅读(41)  评论(0)    收藏  举报