이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define pii pair<ll,ll>
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define freeopen freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
#define mod 1000000000
#define pb push_back
#define mk make_pair
#define ll long long
#define foor(x,vec) for(auto x:vec ){cout<<x<<" ";}
#define fi first
#define se second
#define MAXN 200069
#define lld long double
#define cha ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ffl fflush(stdout)
#define sst string
ll mvx[]={0,0,-1,1};
ll mvy[]={1,-1,0,0};
map<sst,ll> cnt;
ll n,m,s,d;
void solve(){
cin>>n>>m>>s>>d;
char c[n+10][m+10];
ll dp[n+10][m+10];
REP(i,0,n+3){
REP(j,0,m+3)dp[i][j]=0;
}
REP(i,1,n)REP(j,1,m)cin>>c[i][j];
REP(i,1,n){
REP(j,1,m){
dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
if(c[i][j]=='S'){
dp[min(i+d,n)][min(j+d,m)]++;
dp[max(i-d-1,(ll)0)][max(j-d-1,(ll)0)]++;
dp[max(i-d-1,(ll)0)][min(j+d,m)]--;
dp[min(i+d,n)][max(j-d-1,(ll)0)]--;
}
}
}
ll ans=0;
REP(i,1,n){
REP(j,1,m){
dp[i][j]+=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
if(dp[i][j]>=s && c[i][j]=='M')ans++;
}
}
cout<<ans<<endl;
}
int main(){
ll tc;
tc=1;
// cin>>tc;
while(tc--){
solve();
}
}
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