Codeforces-242C-King's Path（bfs）

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 r_i, a_i, b_i (a_i ≤ b_i), denoting that cells in columns from number a_i to number b_i inclusive in the r_i-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, y₁ (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 r_i, a_i, b_i (1 ≤ r_i, a_i, b_i ≤ 10⁹, a_i ≤ b_i), denoting that cells in columns from number a_i to number b_i inclusive in the r_i-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.

Examples

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

题意：给出起点和终点，m个范围，r a b表示第r行从第a到b列是可以走的，每一步可以向8个方向走，问从起点走到终点的最小步数，不能到达输出-1；

思路：用map存下地图然后直接bfs就行了啊

AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
using namespace std;
map<pair<int,int>,int> mp;
int sx,sy,ex,ey,n;
struct node{int x;int y;int step;}p,tmp;
queue<node> que;
int dir[8][2]={-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};
bool check(int x,int y)
{
    if(mp[make_pair(x,y)]==1)
        return true;
    return false;
}
int bfs()
{
    while(que.size())
    {
        tmp=que.front();
        if(tmp.x==ex&&tmp.y==ey)
            return tmp.step;
        que.pop();
        for(int i=0;i<8;i++)
        {
            int tx=tmp.x+dir[i][0];
            int ty=tmp.y+dir[i][1];
            if(check(tx,ty))
            {
                p.x=tx;p.y=ty;p.step=tmp.step+1;
                mp[make_pair(tx,ty)]=0;
                que.push(p);
            }
        }
    }
    return -1;
}
int main()
{
    while(~scanf("%d%d%d%d",&sx,&sy,&ex,&ey))
    {
        mp.clear();
        while(que.size()) que.pop();
        scanf("%d",&n);
        int r,a,b;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&r,&a,&b);
            if(a>b) swap(a,b);
            for(int j=a;j<=b;j++)
                mp[make_pair(r,j)]=1;
        }
        p.x=sx;p.y=sy;p.step=0;
        que.push(p);
        int ans=bfs();
        printf("%d\n",ans);
    }
    return 0;
}