zoukankan      html  css  js  c++  java
  • HDU 4707:Pet

    Pet

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2248    Accepted Submission(s): 1098


    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
     

    迷失在幽谷中的鸟儿,独自飞翔在这偌大的天地间,却不知自己该飞往何方……

    #include <iostream>
    #include <cmath>
    #include <stdio.h>
    #include <string>
    #include <cstring>
    #include <map>
    #include <set>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <iomanip>
    #include <algorithm>
    #define maxn 100010
    using namespace std;
    struct P
    {
        int num;
        int dis;
    };
    int vis[maxn];
    vector<int> v[maxn];
    int N,D;
    void init()
    {
        for(int i=0; i<=N; i++)
            v[i].clear();
    }
    int bfs()
    {
        int ans=0;
        P tmp,n;
        queue<P> q;
        while(!q.empty())
            q.pop();
        tmp.dis=0;
        tmp.num=0;
        q.push(tmp);
        memset(vis,0,sizeof(vis));
        while(!q.empty())
        {
            n=q.front();
            q.pop();
            if(n.dis>D)ans++;
            for(int i=0; i<(int)v[n.num].size(); i++)
            {
                if(vis[v[n.num][i]]==0)
                {
                    vis[v[n.num][i]]=1;
                    tmp.dis=n.dis+1;
                    tmp.num=v[n.num][i];
                    q.push(tmp);
                }
            }
        }
        return ans;
    }
    int main()
    {
        int T;
        int x,y;
        cin>>T;
        while(T--)
        {
            scanf("%d%d",&N,&D);
            init();
            for(int i=0; i<N-1; i++)
            {
                scanf("%d%d",&x,&y);
                v[x].push_back(y);
            }
            cout<<bfs()<<endl;
        }
        return 0;
    }
    


  • 相关阅读:
    1269 匈牙利游戏 2012年CCC加拿大高中生信息学奥赛
    2577 医院设置
    2488 绿豆蛙的归宿
    3315 时空跳跃者的魔法
    1079 回家
    1365 浴火银河星际跳跃
    1074 食物链 2001年NOI全国竞赛
    2596 售货员的难题
    wetask.cn领度任务全新试用体验
    多线程--生产者消费者--简单例子
  • 原文地址:https://www.cnblogs.com/im0qianqian/p/5989350.html
Copyright © 2011-2022 走看看