HDU

题目：

HDU - 4707 Pet

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

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题意：

家在0这个位置，然后给出数字对，前一个代表父节点，后一个代表子节点，求深度大于D的节点个数。

分析：

只要求出深度小于2的结点个数，用N减去就OK了。

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代码如下：

#include <cstdio>#include <iostream>#include <queue>#include <cstring>#include <vector>using namespace std;int dep[100005];        //深度vector<int> e[100005];int N, D, cnt;void bfs(){    queue<int> q;    q.push(0);        //家的坐标入队    dep[0] = 0;        //深度为0    while(!q.empty())    {        int v = q.front();        q.pop();        if(dep[v] >= D)        //如果深度大于等于给出的就结束（因为后面加入的时候就算上D深度的节点了，这里出现D的就可以结束）            break;        int l = e[v].size();        //父节点的子节点个数        //cout << l << endl;        for(int i = 0; i < l; i++)        {            int u = e[v][i];            if(dep[u] == -1)        //如果没有访问过就进行访问            {                q.push(u);        //入队                dep[u] = dep[v] + 1;        //深度为上一个的+1                cnt++;        //在控制的D深度的范围内计算数量            }        }    }}int main(){    int T;    scanf("%d", &T);        //测试数据组数    while(T--)    {        cnt = 0;        //归零        scanf("%d %d", &N, &D);        for(int i = 0; i < 100005; i++)    //初始化        {            e[i].clear();            dep[i] = -1;        }        for(int i = 1; i < N; i++)        {            int a, b;            scanf("%d %d", &a, &b);            e[a].push_back(b);            e[b].push_back(a);        }        bfs();        cout << N - cnt - 1 << endl;    }    return 0;}

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