BFS：森林着火

Description

input

input.txt

output

output.txt

After a terrifying forest fire in Berland a forest rebirth program was carried out. Due to it N rows with M trees each were planted and the rows were so neat that one could map it on a system of coordinates so that the j-th tree in the i-th row would have the coordinates of (i, j). However a terrible thing happened and the young forest caught fire. Now we must find the coordinates of the tree that will catch fire last to plan evacuation.

The burning began in K points simultaneously, which means that initially K trees started to burn. Every minute the fire gets from the burning trees to the ones that aren’t burning and that the distance from them to the nearest burning tree equals to 1.

Find the tree that will be the last to start burning. If there are several such trees, output any.

Input

The first input line contains two integers N, M (1 ≤ N, M ≤ 2000) — the size of the forest. The trees were planted in all points of the (x, y) (1 ≤ x ≤ N, 1 ≤ y ≤ M) type, x and y are integers.

The second line contains an integer K (1 ≤ K ≤ 10) — amount of trees, burning in the beginning.

The third line contains K pairs of integers: x1, y1, x2, y2, …, xk, yk (1 ≤ xi ≤ N, 1 ≤ yi ≤ M) — coordinates of the points from which the fire started. It is guaranteed that no two points coincide.

Output

Output a line with two space-separated integers x and y — coordinates of the tree that will be the last one to start burning. If there are several such trees, output any.

Sample Input

Input

3 3

1

2 2

Output

1 1

Input

3 3

1

1 1

Output

3 3

Input

3 3

2

1 1 3 3

Output

2 2

wa了很多遍~

应当用文件读写！

原因每次搜索到这里就应当比较原有的值与现在的值的大小关系，如果当前值小于原有值就应当更新！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define inf 0x3f3f3f3f
int c[4][2]= {{0,1},{1,0},{-1,0},{0,-1}};
using namespace std;
int mmap[2009][2009];
int m,n;
int k;
int tx,ty;
struct node
{
int x,y;
node(int a=0,int b=0):x(a),y(b) {}
};
node my,cur,net;
int main()
{
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
scanf("%d%d%d",&n,&m,&k);
int x,y;
queue<node>que;
memset(mmap,inf,sizeof mmap);
for(int i=0; i<k; i++)
{
scanf("%d%d",&x,&y);
que.push(node(x,y));
mmap[x][y]=0;
}
while(!que.empty())
{
cur=que.front();
que.pop();
tx=cur.x;
ty=cur.y;
for(int i=0; i<4; i++)
{
int xx=cur.x+c[i][0];
int yy=cur.y+c[i][1];
if(xx<1||xx>n||yy<1||yy>m)
continue;
if(mmap[xx][yy]>mmap[tx][ty]+1)
{
mmap[xx][yy]=mmap[tx][ty]+1;
que.push(node(xx,yy));
}
}
}
int Max=-1,x1,y1;
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
if(mmap[i][j]>Max)
{
Max=mmap[i][j];
x1=i,y1=j;
}
// cout<<"--------"<<Max<<endl;
printf("%d %d",x1,y1);
return 0;
}