cdoj 15 Kastenlauf dfs

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原文链接：http://www.cnblogs.com/qscqesze/p/4544843.html

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#数据结构与算法

本文解析了一道名为Kastenlauf的编程竞赛题目，该题要求通过搜索算法判断角色能否在限定条件下从起点到达终点，途中需考虑补给站的位置及携带物品的限制。

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();
    }
}