求有向图G中各顶点的入度与出度。建议分别采用邻接矩阵和邻接表这两种不同的存储结构完成。 输入格式: 首先输入一个正整数T，表示测试数据的组数，然后是T组测试数据。每组测试第一行输入2个整数n、m（2≤n≤26，1≤m≤n(n-1)/2），分别表示顶点数、边数；然后输入m行，每行包含两个顶点Ai、Bi（大写字母表示），表示Ai到Bi有一条有向边。 输出格式: 对于每组测试，输出n行，依顶点的字典序在每行上输出各顶点的入度和出度（数据之间留一个空格）。用c语言写

邻接矩阵存储：

#include <stdio.h>
#include <string.h>

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int n, m;
        scanf("%d%d", &n, &m);
        int indegrees[n], outdegrees[n];
        memset(indegrees, 0, sizeof(indegrees));
        memset(outdegrees, 0, sizeof(outdegrees));
        int matrix[n][n];
        memset(matrix, 0, sizeof(matrix));
        for (int i = 0; i < m; i++)
        {
            char u, v;
            scanf(" %c %c", &u, &v);
            int x = u - 'A', y = v - 'A';
            matrix[x][y] = 1;
        }
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (matrix[i][j])
                {
                    outdegrees[i]++;
                    indegrees[j]++;
                }
            }
        }
        for (int i = 0; i < n; i++)
        {
            printf("%c %d %d\n", 'A' + i, indegrees[i], outdegrees[i]);
        }
    }
    return 0;
}

邻接表存储：

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

typedef struct Node
{
    int val;
    struct Node *next;
} Node;

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int n, m;
        scanf("%d%d", &n, &m);
        Node *indegrees[n], *outdegrees[n];
        memset(indegrees, 0, sizeof(indegrees));
        memset(outdegrees, 0, sizeof(outdegrees));
        Node *nodes[n];
        memset(nodes, 0, sizeof(nodes));
        for (int i = 0; i < n; i++)
        {
            nodes[i] = (Node *)malloc(sizeof(Node));
            nodes[i]->val = i;
            nodes[i]->next = NULL;
        }
        for (int i = 0; i < m; i++)
        {
            char u, v;
            scanf(" %c %c", &u, &v);
            int x = u - 'A', y = v - 'A';
            Node *node = (Node *)malloc(sizeof(Node));
            node->val = y;
            node->next = nodes[x]->next;
            nodes[x]->next = node;
        }
        for (int i = 0; i < n; i++)
        {
            Node *node = nodes[i]->next;
            while (node != NULL)
            {
                int j = node->val;
                outdegrees[i]++;
                indegrees[j]++;
                node = node->next;
            }
        }
        for (int i = 0; i < n; i++)
        {
            printf("%c %d %d\n", 'A' + i, indegrees[i], outdegrees[i]);
        }
    }
    return 0;
}