1154 Vertex Coloring (25point(s)) - C语言 PAT 甲级

1154 Vertex Coloring (25point(s))

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

题目大意：

输入 N 个点，M 条边的图，K 次检查，每次给出所有点的颜色，判断是否图的所有边的两端节点颜色不同

若所有边两端点颜色全不同，输出颜色的种数，否则输出 No

设计思路：

• 把图的所有边存起来
• 每次查询，遍历每条边，检查颜色
• 颜色的个数排序后去重

编译器：C (gcc)

#include <stdio.h>
#include <stdlib.h>

int cmp(const void *a, const void *b)
{
        return *((int *)a) - *((int *)b);
}

int get_cnt(int *a, int n)
{
        int i, j;
        qsort(a, n, sizeof(a[0]), cmp);
        for (j = 1; j < n && a[j] != a[j - 1]; j++)
                ;
        for (i = j - 1, j = j + 1; j < n; j++)
                if (a[i] != a[j])
                        a[++i] = a[j];
        return i + 1;
}

int main(void)
{
        int n, m, k, e[10010][2], c[10010];
        int i, j, f;

        scanf("%d%d", &n, &m);
        for (i = 0; i < m; i++)
                scanf("%d %d", &e[i][0], &e[i][1]);
        scanf("%d", &k);
        while (k--) {
                f = 1;
                for (i = 0; i < n; i++)
                        scanf("%d", &c[i]);
                for (i = 0; i < m; i++) {
                        if (c[e[i][0]] == c[e[i][1]]) {
                                f = 0;
                                break;
                        }
                }
                if (f)
                        printf("%d-coloring\n", get_cnt(c, n));
                else
                        printf("No\n");
        }
        return 0;
}