1076 Forwards on Weibo (30point(s)) - C语言 PAT 甲级

1076 Forwards on Weibo (30point(s))

Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may follow many other users as well. Hence a social network is formed with followers relations. When a user makes a post on Weibo, all his/her followers can view and forward his/her post, which can then be forwarded again by their followers. Now given a social network, you are supposed to calculate the maximum potential amount of forwards for any specific user, assuming that only L levels of indirect followers are counted.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (≤1000), the number of users; and L (≤6), the number of levels of indirect followers that are counted. Hence it is assumed that all the users are numbered from 1 to N. Then N lines follow, each in the format:

M[i] user_list[i]

where M[i] (≤100) is the total number of people that user[i] follows; and user_list[i] is a list of the M[i] users that followed by user[i]. It is guaranteed that no one can follow oneself. All the numbers are separated by a space.

Then finally a positive K is given, followed by K UserID’s for query.

Output Specification:

For each UserID, you are supposed to print in one line the maximum potential amount of forwards this user can trigger, assuming that everyone who can view the initial post will forward it once, and that only L levels of indirect followers are counted.

Sample Input:

7 3
3 2 3 4
0
2 5 6
2 3 1
2 3 4
1 4
1 5
2 2 6

Sample Output:

4
5

题目大意：

输入有 N 个用户，L 最大转发层数，并按照 1 ~ N 的序号给出每个用户关注的总数和关注 ID

最后一行询问，某 ID 用户发出一条微博后最多有多少人转发

设计思路：

使用队列实现广度优先遍历

• 每人仅转发一次，即不可重复入队
• 广度遍历时，判断最大层数不超过 L 层

编译器：C (gcc)

#include <stdio.h>

int v[1010][1010];
int n, l;
int bfs(int x)
{
        int sum = 0, level[1010] = {0}, visit[1010] = {0};
        int q[1010], front, rear;
        int i;

        q[0] = x;
        front = 0;
        rear = 1;
        visit[x] = 1;
        while (front != rear) {
                x = q[front];
                front++;
                for (i = 1; i <= n; i++) {
                        if (v[x][i] && !visit[i] && level[x] < l) {
                                sum++;
                                visit[i] = 1;
                                level[i] = level[x] + 1;
                                q[rear] = i;
                                rear++;
                        }
                }
        }
        return sum;
}
int main(void)
{
        int i, j, k;

        scanf("%d%d", &n, &l);
        for (i = 1; i <= n; i++) {
               scanf("%d", &j);
               while (j--) {
                       scanf("%d", &k);
                       v[k][i] = 1;
               }
        }
        scanf("%d", &k);
        while (k--) {
                scanf("%d", &j);
                printf("%d\n", bfs(j));
        }
        return 0;
}