zoukankan      html  css  js  c++  java
  • codeforces 242C

    C. King's Path
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the i-th row and j-th column as (i, j).

    You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as n segments. Each segment is described by three integers ri, ai, bi (ai ≤ bi), denoting that cells in columns from number ai to number bi inclusive in the ri-th row are allowed.

    Your task is to find the minimum number of moves the king needs to get from square (x0, y0) to square (x1, y1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way.

    Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point.

    Input

    The first line contains four space-separated integers x0, y0, x1, y(1 ≤ x0, y0, x1, y1 ≤ 109), denoting the initial and the final positions of the king.

    The second line contains a single integer n (1 ≤ n ≤ 105), denoting the number of segments of allowed cells. Next n lines contain the descriptions of these segments. The i-th line contains three space-separated integers ri, ai, bi (1 ≤ ri, ai, bi ≤ 109, ai ≤ bi), denoting that cells in columns from number ai to number bi inclusive in the ri-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily.

    It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105.

    Output

    If there is no path between the initial and final position along allowed cells, print -1.

    Otherwise print a single integer — the minimum number of moves the king needs to get from the initial position to the final one.

    Sample test(s)
    input
    5 7 6 11
    3
    5 3 8
    6 7 11
    5 2 5
    output
    4
    input
    3 4 3 10
    3
    3 1 4
    4 5 9
    3 10 10
    output
    6
    input
    1 1 2 10
    2
    1 1 3
    2 6 10
    output
    -1
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    #include<cstdlib>
    #include<string>
    #include<cstring>
    #include<map>
    #include<queue>
    #include<vector>
    using namespace std;
    int x1,x2,y11,y2,n;
    map<pair<int,int>,int> mp;
    int dx[8]={0,0,1,-1,1,1,-1,-1};
    int dy[8]={1,-1,0,0,1,-1,1,-1};
    void BFS()
    {
        queue< pair<int,int> > q;
        mp[make_pair(x1,y11)]=0;
        q.push(make_pair(x1,y11));
        while(!q.empty())
        {
            int xx=q.front().first;
            int yy=q.front().second;
            q.pop();
            for(int i=0;i<8;i++)
            if(mp[make_pair(xx+dx[i],yy+dy[i])]==-1)
            {
                mp[make_pair(xx+dx[i],yy+dy[i])]=mp[make_pair(xx,yy)]+1;
                q.push(make_pair(xx+dx[i],yy+dy[i]));
            }
        }
    }
    int main()
    {
        scanf("%d%d%d%d%d",&x1,&y11,&x2,&y2,&n);
        for(int i=1;i<=n;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&z,&x,&y);
            for(int j=x;j<=y;j++)
                mp[make_pair(z,j)]=-1;
        }
        BFS();
        printf("%d
    ",mp[make_pair(x2,y2)]);
        return 0;
    }
    

      

  • 相关阅读:
    Postfix邮件
    RAID和LVM磁盘阵列
    CF1400G
    CF1400F
    2020 AC Saber夏季赛 游记
    APIO2018 题解
    2020北京中考游记
    初中数学几何推理大梳理
    CF1373F Network Coverage
    部编人教版初中历史书事件影响/意义汇总
  • 原文地址:https://www.cnblogs.com/water-full/p/4532628.html
Copyright © 2011-2022 走看看