你这个思路本来就会 tle
wa 大概是因为这一行：
`m[j-1][l-1]==i+1;`
by 半只蒟蒻 @ 2023-10-22 12:17:01
@[半只蒟蒻](/user/112049) 大概口胡一下比较正确的思路：反正我们只求一个点，所以倒叙枚举地毯，然后依次判断是否覆盖这个点
by 半只蒟蒻 @ 2023-10-22 12:18:11
1e14的复杂度...
by fu__men @ 2023-10-22 12:20:31
```cpp
#include<bits/stdc++.h>
using namespace std;
long long n=0,x=0,y=0,a[10001]={},b[10001]={},g[10001]={},k[10001]={};
int main(){
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i]>>b[i]>>g[i]>>k[i];
}
cin>>x>>y;
for(int i=n;i>0;i--){
if((x>=a[i])&&(x<=(a[i]+g[i]))&&(y>=b[i])&&(y<=(b[i]+k[i]))){
cout<<i<<endl;
return 0;
}
}
cout<<-1<<endl;
return 0;
}
```
by han_jian__wu_wei_han @ 2023-10-22 12:23:38
逆天复杂度
by wangjue233 @ 2023-10-22 12:29:06
@[han_jian__wu_wei_han](/user/746444) 不建议挂代码(这不就等于讨论区tj吗
by 半只蒟蒻 @ 2023-10-22 12:43:59
@[半只蒟蒻](/user/112049) OK
by han_jian__wu_wei_han @ 2023-10-22 12:51:18
关于复杂度……
by wangchuhan @ 2023-10-28 11:25:01