这个题其实想了好一会才想到满足复杂度的前缀和,其实可以用bfs做,没想到真是太傻了,虽然想到了因为数组越界的问题wa了,就找不出错来,后来才想到数组越界。
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cstdio>
using namespace std;
int mp[5010][5010];
int f[5010][5010];
int main()
{
int n, m;
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
{
int u, v;
scanf("%d%d", &u, &v);
mp[u][v] = 1;
}
f[0][0] = 0;
f[1][1] = mp[0][0];
for(int i = 1; i <= 5000; i++)
f[1][i + 1] = f[1][i - 1 + 1] + mp[0][i];
for(int i = 1; i <= 5000; i++)
f[i + 1][1] = f[i - 1 + 1][1] + mp[i][0];
for(int i = 1; i <= 5000; i++)
{
for(int j = 1; j <= 5000; j++)
{
f[i + 1][j + 1] = f[i - 1 + 1][j + 1] + f[i + 1][j - 1 + 1] - f[i - 1 + 1][j - 1 + 1] + mp[i][j];
}
}
for(int i = 0; i < m; i++)
{
int u, v;
scanf("%d%d", &u, &v);
if(mp[u][v])
printf("%d\n", 0);
else{
int l = 1, r = 5000;
while(l < r){
int j = (l + r) >> 1;
int t1 = u + j + 1;
t1 = max(0, t1);
t1 = min(5001, t1);
int t2 = v + j + 1;
t2 = max(0, t2);
t2 = min(5001, t2);
int t3 = u - j - 1 + 1;
t3 = max(0, t3);
t3 = min(t3, 5001);
int t4 = v - j - 1 + 1;
t4 = min(t4, 5001);
t4 = max(t4, 0);
int tmp = f[t1][t2] - f[t1][t4] - f[t3][t2] + f[t3][t4];
if(tmp)
{
r = j;
}
else
{
l = j + 1;
}
}
printf("%d\n", l);
}
}
return 0;
}