1076 Forwards on Weibo

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

The main idea of the topic ： Forward wechat , Fans forward , Fans' fans can also forward , Fans who reach a certain delivery level will not forward , Ask how many people can pass at most .

Ideas ： The relationship between fans can be seen as the distance between cities , Find the minimum distance between each city , Then, if the minimum distance is less than the corresponding number of layers, the corresponding number of people can be counted , The box floyd Algorithm , The only thing to note is that the format of the input is the person he cares about , Not the people who care about him .

#include <bits/stdc++.h>
using namespace std;
#define inf 9999999
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    int i;
    int a[n+1][n+1];
    fill(a[0],a[0]+(n+1)*(n+1),inf);// Initialization operation 
    for(i=1;i<=n;i++)a[i][i]=0;
    int num,j;
    for(i=1;i<=n;i++){
        scanf("%d",&num);
        for(j=0;j<num;j++){// Regulations a[i][j] by i Yes j Contribution of nodes .
            int val;
            scanf("%d",&val);
            a[i][val]=1;
        }
    }
    int k;
    for(i=1;i<=n;i++){// Find the nearest relationship of each intermediary 
        for(j=1;j<=n;j++){
            for(k=1;k<=n;k++){
                if(a[j][k]>a[j][i]+a[i][k]){
                    a[j][k]=a[j][i]+a[i][k];
                }
            }
        }
    }
    int cha;
    scanf("%d",&cha);
    int nums[n+1];
    memset(nums,0,sizeof(nums));
    for(i=1;i<=n;i++){// Enumerate the contribution of each node to this point , That is, the number of objects he follows and forwards 
        for(j=1;j<=n;j++){
            if(j!=i&&a[i][j]<=m)nums[j]++;
        }   
    }
    for(i=0;i<cha;i++){
        scanf("%d",&num);
        printf("%d\n",nums[num]);
    }
    system("pause");
}