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.

代码

#include <iostream>
#include<vector>
using namespace std;
int cnt,D;

void dfs(int currentNode,int currentDeep,vector<vector<int> > &matrix,bool *visited){
    if(currentDeep>D) return;
    visited[currentNode]=1;
    cnt++;
    int len=matrix[currentNode].size();
    for (int i=0;i<len ;i++ )
    {
        int nextNode=matrix[currentNode][i];
        if(!visited[nextNode]) dfs(nextNode,currentDeep+1,matrix,visited);
    }
}

int main()
{
    int caseNum,N,a,b;
    cin>>caseNum;
    while (caseNum--)
    {
        cin>>N>>D;
        vector<vector<int> > matrix;
        matrix.resize(N);
        bool visited[N]; //记得初始化
        for (int i=0;i<N ;i++ )
        {
            visited[i]=0;
            matrix[i].clear();
        }
        cnt=0;
        for (int i=0;i<N-1 ;i++ )
        {
            scanf("%d%d",&a,&b);
            matrix[a].push_back(b);
            matrix[b].push_back(a);
            visited[i]=0;
        }
        dfs(0,0,matrix,visited);
        cout<<N-cnt<<"\n";
    }
    return 0;
}

注意事项

1.cin与scanf速度相差很大，当大量数据输入时使用scanf，这是导致一直超时time limit exceed 的原因
2.对于每个测试样例，要记得对全局变量进行初始化，不然是上一次的数据
3.vector<vector<int> >二维数组的传参，传引用
4.邻接矩阵存储图，邻接矩阵使用vector<vector<int> >可以简化存储，相当于链表实现的邻接表
5.dfs使用递归，bfs使用队列