1154Vertex Coloring(25分)

Aproper vertex coloringis 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 mostkcolors is called a (proper)k-coloring.

Now you are supposed to tell if a given coloring is a properk-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 1), 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 linek-coloring

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

題意：給定一個圖，圖中每個頂點給一個顏色，要求一條邊兩端的頂點顏色不同。如果滿足要求，輸出使用的"顏色的數量-coloring"，否則輸出No

思路：因為圖中頂點最多可以有10000個，用鄰接矩陣可能會超時（筆者沒試過），所以我採用的是鄰接表。讀入整個圖的資料之後，依次遍歷各條邊，看兩端的頂點顏色是否相同。相同則直接break輸出No,否則遍歷完所有頂點之後計算顏色數量輸出。

程式碼如下：

#include<cstdio>
#include<vector>
#include<set>
using namespace std; 
vector<int> v[10005];
int n,m;
int colors[10005];
void fun(){
    int mark=0;
    for(int i=0;i<n;i++){
        for(vector<int>::iterator it=v[i].begin();it!=v[i].end();it++){
            if(colors[i]==colors[(*it)]){
                mark=1;
                break;
            }        
        }
    }
    if(mark==0){
        set<int> s;
        for(int i=0;i<n;i++){
            s.insert(colors[i]);
        }
        printf("%d-coloring\n",s.size());
    }
    else
        printf("No\n");
}
int main(){
    int a,b;
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++){
        scanf("%d%d",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    int k;
    scanf("%d",&k);
    for(int i=0;i<k;i++){
        for(int j=0;j<n;j++){
            scanf("%d",&colors[j]);
        }
        fun();
        
    }
    return 0;
}