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  • cdoj 15 Kastenlauf dfs

    Kastenlauf

    Time Limit: 20 Sec  Memory Limit: 256 MB

    题目连接

    http://acm.uestc.edu.cn/#/problem/show/3

    Description

    Once every year, Jo and his friends want to visit the local fair in Erlangen, called Bergkirchweih. This year, they want to make a Kastenlauf (box run). They start at Jo's home, and have one box (Kasten) of beer (with twenty bottles). As they are very thirsty, they drink one bottle of beer every 50 metres.

    As the way from Jo's home to the Bergkirchweih is pretty long, they need more beer than they have initially. Fortunately, there are stores selling beer on the way. When they visit a store, they can drop their empty bottles and buy new bottles, but their total number of full bottles will not be more than twenty (because they are too lazy to carry more than one full box).

    You are given the coordinates of the stores, of Jo's home and of the location of the Bergkirchweih. Write a program to determine whether Jo and his friends can happily reach the Bergkirchweih, or whether they will run out of beer on the way.

    Input

    Input starts with one line containing the number of test cases t (t≤50).

    Each test case starts with one line, containing the number n of stores selling beer (with 0≤n≤100). The next n+2 lines cointain (in this order) the location of Jo's home, of the stores, and of the Bergkirchweih. The location is given with two integer coordinates x and y, (both in meters, −32768≤x,y≤32767).

    As Erlangen is a rectangularly laid out city, the distance between two locations is the difference of the first coordinate plus the difference of the second coordinate (also called Manhattan-Metric).

    Output

    For each test case print one line, containing either happy (if Jo and his friends can happily reach the Bergkirchweih), or sad (if they will run out of beer on the way).

    Sample Input

    2
    2
    0 0
    1000 0
    1000 1000
    2000 1000
    2
    0 0
    1000 0
    2000 1000
    2000 2000

    Sample Output

    happy
    sad

    HINT

    题意

    给你很多个点,要求每个点之间的距离小于等于1000就可以走过去,然后问你能否从起点走到终点

    题解:

    数据范围很小,直接爆搜

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define test freopen("test.txt","r",stdin)  
    #define maxn 200000
    #define mod 10007
    #define eps 1e-9
    int Num;
    char CH[20];
    const int inf=0x3f3f3f3f;
    const ll infll = 0x3f3f3f3f3f3f3f3fLL;
    inline ll read()
    {
        ll x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    //**************************************************************************************
    struct node
    {
        int x,y;
    };
    node a[200];
    int n;
    int flag[200];
    void dfs(int x)
    {
        if(flag[x])
            return;
        flag[x]=1;
        for(int i=1;i<=n;i++)
        {
            if(i==x)
                continue;
            if(abs(a[i].x-a[x].x)+abs(a[x].y-a[i].y)<=1000)
                dfs(i);
        }
    }
    void solve()
    {
        memset(flag,0,sizeof(flag));
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        n+=2;
        for(int i=1;i<=n;i++)
            a[i].x=read(),a[i].y=read();
        dfs(1);
        if(flag[n])
            cout<<"happy"<<endl;
        else
            cout<<"sad"<<endl;
    }
    int main()
    {
        //test;
        int t=read();
        for(int cas=1;cas<=t;cas++)
        {
            solve();
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4544843.html
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