hdu 1307 N-Credible Mazes

传送门

An n-tersection is defined as a location in n-dimensional space, n being a positive integer, having all non-negative integer coordinates. For example, the location (1,2,3) represents an n-tersection in three dimensional space. Two n-tersections are said to be adjacent if they have the same number of dimensions and their coordinates differ by exactly 1 in a single dimension only. For example, (1,2,3) is adjacent to (0,2,3) and (2,2,3) and (1,2,4), but not to (2,3,3) or (3,2,3) or (1,2). An n-teresting space is defined as a collection of paths between adjacent n-tersections.

Finally, an n-credible maze is defined as an n-teresting space combined with two specific n-tersections in that space, one of which is identified as the starting n-tersection and the other as the ending n-tersection.

Input

The input file will consist of the descriptions of one or more n-credible mazes. The first line of the description will specify n, the dimension of the n-teresting space. (For this problem, n will not exceed 10, and all coordinate values will be less than 10.) The next line will contain 2n non-negative integers, the first n of which describe the starting n-tersection, least dimension first, and the next n of which describe the ending n-tersection. Next will be a nonnegative number of lines containing 2n non-negative integers each, identifying paths between adjacent n-tersections in the n-teresting space. The list is terminated by a line containing only the value ?. Several such maze descriptions may be present in the file. The end of the input is signalled by space dimension of zero. No further data will follow this terminating zero.

Output

For each maze output it's position in the input; e.g. the first maze is "Maze #1", the second is "Maze #2", etc. If it is possible to travel through the n-credible maze's n-teresting space from the starting n-tersection to the ending n-tersection, also output "can be travelled" on the same line. If such travel is not possible, output "cannot be travelled" instead.

Sample Input

2 
0 0 2 2
0 0 0 1
0 1 0 2
0 2 1 2
1 2 2 2
-1
3
1 1 1 1 2 3
1 1 2 1 1 3
1 1 3 1 2 3
1 1 1 1 1 0
1 1 0 1 0 0
1 0 0 0 0 0
-1
0

Sample Output

Maze #1 can be travelled
Maze #2 cannot be travelled 

【题意】

给你一个n维空间,然后第二行给出一个起点一个终点,接下来是一些联通的点。每行两个表示这两个点相互连通。然后问你起点和终点是否连通。

【分析】

由于数据范围比较小,所以我们可以把这些n维的点离散成一维的点,然后对于这些一维的点我们接着用map优化到1e4的范围以内。然后一个并查集就搞定了。

【代码】

#include <bits/stdc++.h>
using namespace std;

int len;
int read()
{
    int ret;
    int a;
    scanf("%d",&a);
    if(a==-1) return -1;
    ret = a;
    for(int i = 1;i<len;i++)
    {
        scanf("%d",&a);
        ret = ret*10+a;
    }
    return ret;
}
map<int,int> mp;
int pre[12345];
void init()
{
    mp.clear();
    for(int i = 1;i<12345;i++)
        pre[i] = i;
}
int fin(int a)
{
    return a==pre[a]?a:(pre[a]=fin(pre[a]));
}
void uni(int a,int b)
{
    int aa = fin(a);
    int bb = fin(b);
    if(aa==bb) return;
    pre[bb]=aa;
}
int main()
{
//    freopen("in.txt","r",stdin);
    int g_case = 1;
    while(cin>>len,len)
    {
        printf("Maze #%d ",g_case++);
        init();
        int from = read(),to = read();
        mp[from] = 1;
        mp[to] = 2;
        int flag = 2;
        while((from = read())!=-1)
        {
            to = read();
            if(mp[from]==0) mp[from] = ++flag;
            if(mp[to]==0) mp[to] = ++flag;
            uni(from,to);
        }
        if(fin(1)==fin(2)) puts("can be travelled");
        else puts("cannot be travelled");
    }
    return 0;
}

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转载自blog.csdn.net/mengzhongsharen/article/details/79435439