113 lines
3.6 KiB
C++
113 lines
3.6 KiB
C++
#include <set>
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#include <cmath>
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#include <vector>
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#include <cstdio>
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#include <cstring>
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#include <algorithm>
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using namespace std ;
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typedef long long LL ;
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#define rep( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
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#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
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#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
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#define clr( a , x ) memset ( a , x , sizeof a )
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const int MAXN = 1005 ;
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const int L = 33 ;
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int G[MAXN][MAXN] ;
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int down[MAXN][MAXN] ;
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int prefix[MAXN][MAXN] ;
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int prefix_dn[MAXN][MAXN] ;
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int prefix_lt[MAXN][MAXN] ;
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int prefix_rt[MAXN][MAXN] ;
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int ans[L][MAXN][MAXN] ;
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int n , m , q , sqrtD ;
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char s[MAXN] ;
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void scanf ( int &x , char c = 0 ) {
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while ( ( c = getchar () ) < '0' || c > '9' ) ;
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x = c - '0' ;
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while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
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}
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void read () {
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For ( i , 1 , n ) {
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scanf ( "%s" , s + 1 ) ;
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For ( j , 1 , m ) {
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G[i][j] = ( s[j] == 'X' ) ;
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prefix[i][j] = prefix[i][j - 1] + G[i][j] ;
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}
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}
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}
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void solve1 () {
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int x , y , D ;
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sqrtD = ( int ) sqrt ( ( double ) n ) ;
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For ( k , 1 , sqrtD ) {
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For ( i , 1 , n ) {
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For ( j , 1 , m ) {
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down[i][j] = G[i][j] ;
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if ( i - k > 0 ) down[i][j] += down[i - k][j] ;
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prefix_dn[i][j] = prefix_dn[i][j - 1] + down[i][j] ;
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prefix_lt[i][j] = down[i][j] ;
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if ( i - k > 0 ) {
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if ( j - k <= 0 ) prefix_lt[i][j] += prefix_dn[i - k][j - 1] - prefix_dn[i - k][0] ;
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else {
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prefix_lt[i][j] += prefix_dn[i - k][j - 1] - prefix_dn[i - k][j - k] ;
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prefix_lt[i][j] += prefix_lt[i - k][j - k] ;
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}
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}
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}
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rev ( j , m , 1 ) {
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prefix_rt[i][j] = down[i][j] ;
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if ( i - k > 0 ) {
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if ( j + k > m ) prefix_rt[i][j] += prefix_dn[i - k][m] - prefix_dn[i - k][j] ;
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else {
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prefix_rt[i][j] += prefix_rt[i - k][j + k] ;
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prefix_rt[i][j] += prefix_dn[i - k][j + k - 1] - prefix_dn[i - k][j] ;
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}
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}
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}
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For ( j , 1 , m ) ans[k][i][j] = prefix_lt[i][j] + prefix_rt[i][j] - down[i][j] ;
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}
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}
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while ( q -- ) {
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scanf ( x ) ;
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scanf ( y ) ;
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scanf ( D ) ;
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if ( D <= sqrtD ) {
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//printf ( "%d %d\n" , prefix_lt[x][y][D] , prefix_rt[x][y][D] ) ;
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printf ( "%d\n" , ans[D][x][y] ) ;
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}
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else {
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int L = y , R = y , sum = 0 ;
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while ( x > 0 ) {
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sum += prefix[x][R] - prefix[x][L - 1] ;
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x -= D ;
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L = max ( 1 , L - D ) ;
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R = min ( m , R + D ) ;
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}
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printf ( "%d\n" , sum ) ;
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}
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}
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}
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void solve2 () {
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int x , y , D ;
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while ( q -- ) {
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scanf ( x ) ;
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scanf ( y ) ;
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scanf ( D ) ;
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int L = y , R = y , sum = 0 ;
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while ( x > 0 ) {
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sum += prefix[x][R] - prefix[x][L - 1] ;
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x -= D ;
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L = max ( 1 , L - D ) ;
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R = min ( m , R + D ) ;
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}
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printf ( "%d\n" , sum ) ;
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}
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}
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int main () {
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while ( ~scanf ( "%d%d%d" , &n , &m , &q ) ) {
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read () ;
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if ( q > 1000 ) solve1 () ;
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else solve2 () ;
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}
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return 0 ;
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}
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