hdu 4707 Pet【BFS求树的深度】

Pet

                                                         Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                         Total Submission(s): 1909    Accepted Submission(s): 924
                                                                                                                             链接： Click Me !

Problem Description

One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in the room but didn’t find the hamster. He tried to use some cheese to trap the hamster. He put the cheese trap in his room and waited for three days. Nothing but cockroaches was caught. He got the map of the school and foundthat there is no cyclic path and every location in the school can be reached from his room. The trap’s manual mention that the pet will always come back if it still in somewhere nearer than distance D. Your task is to help Lin Ji to find out how many possible locations the hamster may found given the map of the school. Assume that the hamster is still hiding in somewhere in the school and distance between each adjacent locations is always one distance unit.

Input

The input contains multiple test cases. Thefirst line is a positive integer T (0<T<=10), the number of test cases. For each test cases, the first line has two positive integer N (0<N<=100000) and D(0<D<N), separated by a single space. N is the number of locations in the school and D is the affective distance of the trap. The following N-1lines descripts the map, each has two integer x and y(0<=x,y<N), separated by a single space, meaning that x and y is adjacent in the map. Lin Ji’s room is always at location 0.

Output

For each test case, outputin a single line the number of possible locations in the school the hamster may be found.

Sample Input

   
   
   
   

    
    
    
    1
10 2
0 1
0 2
0 3
1 4
1 5
2 6
3 7
4 8
6 9
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    2
   
   
   
   

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

题意：

给定N个点，标号为0~N-1，还有N-1条边，数据保证N-1条边不成环，也就是说，输入的节点为N的一棵树。根节点为0，要你求深度大于d的节点的数目。

分析：

从根节点0开始，BFS其所有的子节点，统计深度小于等于d的节点的数目cnt，那么答案就是N-cnt。水题~

#include <queue>
#include <cmath>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
typedef long long       LL;
const int MAXN = 1e6 + 50;
struct Node
{
    vector<int> son;
} nodes[MAXN];
bool vis[MAXN];
void add_edge(int a, int b)
{
    nodes[a].son.push_back(b);
}

struct Fuck
{
    int pos, step;
    Fuck() {}
    Fuck(int _p, int _s) : pos(_p), step(_s) {}
};
queue<Fuck> Que;
int Dis, N;
int BFS()
{
    memset(vis,false,sizeof(vis));
    int cnt = 0;
    Fuck Now(0, 0);
    Que.push(Now);
    vis[0] = true;
    while(!Que.empty())
    {
        Now = Que.front();
        Que.pop();
        int nowp = Now.pos, nows = Now.step;
        if(nows == Dis) continue;
        for(int i = 0; i < nodes[nowp].son.size(); i++)
        {
            int sonp = nodes[nowp].son[i];
            if(vis[sonp]) continue;
            Que.push(Fuck(sonp, nows + 1));
            vis[sonp] = true;
            cnt ++;
        }
    }
    return N - cnt - 1;
}
int main()
{
//    FIN;
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &N, &Dis);
        for(int i = 0; i < N; i++)
            nodes[i].son.clear();
        for(int i = 1; i <= N - 1; i++)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            add_edge(a, b);
            add_edge(b, a);
        }
        int ans = BFS();
        printf("%d\n", ans);
    }
    return 0;
}