zoukankan      html  css  js  c++  java
  • TOJ 2017: N-Credible Mazes

    2017: N-Credible Mazes 分享至QQ空间

    Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByte
    Total Submit: 6            Accepted:0

    Description

    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 ?C1. 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
    数据的锅,理解题意写的代码大概都是可以AC的

    爆搜的代码

    #include<stdio.h>
    #include<algorithm>
    using namespace std;
    int a[1005][10],b[1005][10],c[1005][3];
    int main()
    {
        int n,T=1;
        while(~scanf("%d",&n),n)
        {
            int i,j,k;
            for(i=0;;i++)
            {
                scanf("%d",&a[i][0]);
                if(a[i][0]==-1)break;
                for(j=1; j<n; j++)
                    scanf("%d",&a[i][j]);
                for(j=0; j<n; j++)
                    scanf("%d",&b[i][j]);
            }
            for(k=0; k<i; k++)
                for(j=0; j<n; j++)
                    c[k][j]=0;
            c[0][0]=1;
            for(i=0; i<k; i++)
            {
                int f=0;
                for(j=0; j<n; j++)
                    if(a[i+1][j]!=a[i][j])
                    {
                        f++;
                        if(abs(a[i+1][j]-a[i][j])!=1)break;
                    }
                if(f<=1&&j==n&&c[i][0]==1)c[i+1][0]=1;
                f=0;
                for(j=0; j<n; j++)
                    if(b[i][j]!=a[i][j])
                    {
                        f++;
                        if(abs(b[i][j]-a[i][j])!=1)break;
                    }
                if(f<=1&&j==n&&c[i][0]==1)c[i][1]=1;
                f=0;
                for(j=0; j<n; j++)
                    if(b[i+1][j]!=b[i][j])
                    {
                        f++;
                        if(abs(b[i+1][j]-b[i][j])!=1)break;
                    }
                if(f<=1&&j==n&&c[i][1]==1)c[i+1][1]=1;
                f=0;
                for(j=0; j<n; j++)
                    if(b[i+1][j]!=a[i+1][j])
                    {
                        f++;
                        if(abs(b[i+1][j]-a[i+1][j])!=1)break;
                    }
                if(f<=1&&j==n&&c[i+1][1]==1)c[i+1][0]=1;
            }
            if(c[k-1][1]>=1) printf("Maze #%d can be travelled
    ",T++);
            else printf("Maze #%d cannot be travelled
    ",T++);
        }
        return 0;
    }

    map+set维护

    #include<stdio.h>
    #include<vector>
    #include<set>
    #include<map>
    using namespace std;
    map<int,vector<int> >M;
    set<int>S;
    int la(int &num,int n)
    {
        num=0;
        for(int i=1; i<=n; i++)
        {
            int x;
            scanf("%d",&x);
            if(x==-1)return 0;
            num=num*10+x;
        }
        return 1;
    }
    int dfs(int st,int ed)
    {
        if(st==ed)return 1;
        int l=M[st].size();
        for(int i=0; i<l; i++)
        {
            int v=M[st][i];
            if(!S.count(v))
            {
                S.insert(v);
                if(dfs(v,ed))return 1;
            }
        }
        return 0;
    }
    int main()
    {
        int t=1,n;
        while(~scanf("%d",&n))
        {
            if(n==0)
            {
                if(scanf("%d",&n)==EOF)
                    break;
            }
            M.clear(),S.clear();
            int st,ed;
            la(st,n);
            la(ed,n);
            int s,e;
            while(la(s,n))
            {
                la(e,n);
                M[s].push_back(e);
                M[e].push_back(s);
            }
            if(dfs(st,ed))
                printf("Maze #%d can be travelled
    ",t++);
            else
                printf("Maze #%d cannot be travelled
    ",t++);
        }
        return 0;
    }

    网上的HDUac代码

    #include<stdio.h>
    #include<string.h>
    int f,sum,e,s;
    int a[10010],b[10010];
    int mark[10010];
    void dfs(int w)
    {
        int i;
        if (w==e)
        {
            f=1;
            return ;
        }
        for (i=1; i<=sum; i++)
            if (mark[i]==0 && (a[i]==w || b[i]==w))
            {
                mark[i]=1;
                if (a[i]==w) dfs(b[i]);
                else dfs(a[i]);
            }
    }
    
    
    int main()
    {
        int Case,i,j,n,x;
        Case=0;
        while (scanf("%d",&n)!=EOF)
        {
            if (n==0) break;
    
            Case++;
            s=e=0;
            for (i=1; i<=n; i++)
            {
                scanf("%d",&x);
                s=s*10+x;
            }
            for (i=1; i<=n; i++)
            {
                scanf("%d",&x);
                e=e*10+x;
            }
    
            sum=0;
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
    
            while (scanf("%d",&x)!=EOF)
            {
                if (x==-1) break;
    
                sum++;
                a[sum]=x;
                for (i=1; i<n; i++)
                {
                    scanf("%d",&x);
                    a[sum]=a[sum]*10+x;
                }
                for (i=1; i<=n; i++)
                {
                    scanf("%d",&x);
                    b[sum]=b[sum]*10+x;
                }
    
            }
            f=0;
            memset(mark,0,sizeof(mark));
            dfs(s);
            printf("Maze #%d ",Case);
            if (f==0)
                printf("cannot be travelled
    ");
            else
                printf("can be travelled
    ");
        }
        return 0;
    }
  • 相关阅读:
    libev & libevent简介
    MyEclipse10+Flash Builder4+BlazeDS+Tomcat7配置J2EE Web项目报错(一)
    增加表空间大小的四种方法
    JavaScript获取某年某月的最后一天
    Not in 改写左连接不需要关注连接列是否重复数据
    自连接
    左链接,右连接
    In,内链接和空值
    HighCharts基本折线图
    NetBeans运行项目报错
  • 原文地址:https://www.cnblogs.com/BobHuang/p/7697718.html
Copyright © 2011-2022 走看看