2019 kickstart A轮 B 题
题目链接
下面证明曼哈顿距离具有以下性质。其他细节参考
这篇博客
绝对值不等式
[|x_1-x_2|+|y_1-y_2|>=|(x_1-x_2)-(y_1-y_2)|
]
[|x_1-x_2|+|y_1-y_2| = |x_1-x_2|+|y_2-y_1| >= |(x_1-x_2) - (y2-y1)| = |(x_1+y_1)-(x_2+y_2)|
]
分四种情况讨论可得，
[|x_1-x_2|+|y_1-y_2|=|(x_1-x_2)-(y_1-y_2)|或|(x_1+y_1)-(x_2+y_2)|
]
由上式可得:
[|x_1-x_2|+|y_1-y_2|=max(|(x_1-x_2)-(y_1-y_2)|,|(x_1+y_1)-(x_2+y_2)|)
]
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 255;
#define P pair<int,int>
#define mk make_pair
const int mod = 1e4;
int n,m;
char s[N][N];
int dist[N][N];
int dx[4]={-1,0,0,1};
int dy[4]={0,-1,1,0};
void bfs()
{
queue<P>q;
memset(dist,-1,sizeof(dist));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(s[i][j]=='1')
{
q.push(mk(i,j));
dist[i][j] = 0;
}
}
}
int numk = 1;
while(q.size())
{
int siz = q.size();
for(int i=1;i<=siz;i++)
{
int x = q.front().first;
int y = q.front().second;
q.pop();
for(int j=0;j<4;j++)
{
int X = x + dx[j];
int Y = y + dy[j];
if(X >= 1 && X <= n && Y >= 1 && Y <= m && dist[X][Y] == -1)
{
dist[X][Y] = numk;
q.push(mk(X,Y));
}
}
}
numk++;
}
}
int check(int k)
{
int addMin = 1e9,addMax = -1e9,subMin = 1e9,subMax = -1e9;
int flag = 0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(dist[i][j] > k)
{
int add = i + j;
int sub = i - j;
addMin = min(addMin,add);
subMin = min(subMin,sub);
addMax = max(addMax,add);
subMax = max(subMax,sub);
flag = 1;
}
}
}
if(flag==0)
return 1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(abs(i+j-addMin)<=k &&
abs(i+j-addMax)<=k &&
abs(i-j-subMin)<=k &&
abs(i-j-subMax)<=k)
{
return 1;
}
}
}
return 0;
}
int main() {
int T;
scanf("%d",&T);
int kase = 0;
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%s",s[i]+1);
}
bfs();
int l = 0,r = n + m;
while(l<r)
{
int mid = (l+r)/2;
if(check(mid))
{
r = mid;
}
else
{
l = mid + 1;
}
}
printf("Case #%d: %d
",++kase,l);
}
return 0;
}