CodeForces 761F. Dasha and Photos

Dasha decided to have a rest after solving the problem D and began to look photos from previous competitions.

Let's call photos as the matrix with the size n × m, which consists of lowercase English letters.

Some k photos especially interested her, because they can be received from photo-template by painting a rectangular area in a certain color. Let's call such photos special.

More formally the i-th special photo is received from the photo-template by replacing all characters on some rectangle with upper left corner of the cell with coordinates (ai, bi) and lower right corner in the cell with coordinates (ci, di) to the symbol ei.

Dasha asks you to find the special photo so that the total distance from it to all other special photos is minimum. And calculate this distance.

Determine the distance between two photos as the sum of distances between all corresponding letters. The distance between two letters is the difference module of their positions in the alphabet. For example, the distance between letters 'h' and 'm' equals |8 - 13| = 5, because the letter 'h' is the 8-th in the alphabet, the letter 'm' is the 13-th.

Input

The first line contains three integers n, m, k (1 ≤ n, m ≤ 103, 1 ≤ k ≤ 3·105) — the number of strings in the photo-template, the number of columns and the number of special photos which are interesting for Dasha.

The next n lines contains the string with m length which consists of little Latin characters — the description of the photo-template.

Each of the next k lines contains the description of the special photo in the following format, "ai bi ci di ei"(1 ≤ ai ≤ ci ≤ n, 1 ≤ bi ≤ di ≤ m), where (ai, bi) — is the coordinate of the upper left corner of the rectangle, (ci, di) — is the description of the lower right corner, and ei — is the little Latin letter which replaces the photo-template in the described rectangle.

Output

In the only line print the minimum total distance from the found special photo to all other special photos.

Examples

input

3 3 2aaaaaaaaa1 1 2 2 b2 2 3 3 c

output

10

input

5 5 3abcdeeabcddeabccdeabbcdea1 1 3 4 f1 2 3 3 e1 3 3 4 i

output

59

Note

In the first example the photos are following:

bba    aaabba    accaaa    acc

The distance between them is 10.

题意：给一个基础n*m的小写字符串矩阵，和K个新矩阵

第i个矩阵把矩形(a[i],b[i],c[i],d[i])全部变成e[i]

定义两个字符的距离为它们之差

定义两个矩阵的距离是整个矩阵所有对应位置的字符距离之和

求一个新矩阵，让它到其他矩阵距离和最小，输出距离

题解：这题就是暴力...

我们只需要求出两个东西

p(i,j)表示所有新矩阵到原图距离前缀和，sum(x,i,j)表示全部替换成x之后到原图距离前缀和，然后乱搞。

复杂度O（26nm）

感谢http://blog.csdn.net/WorldWide_D/article/details/54807874的作者给我的题解！（怎么感觉代码基本就是变量名换了一下QAQ）

#include <bits/stdc++.h>using namespace std;const int MAXN = 1010;const int MAXM = 300010;long long ans = 1e16, s[26][MAXN][MAXN], p[MAXN][MAXN];int cnt[26], sum[26][MAXN][MAXN], a[MAXM], b[MAXM], c[MAXM], d[MAXM], n, m, K;char mp[MAXN][MAXN], e[MAXM];inline long long cal(long long f[MAXN][MAXN], int i) { return f[ c[ i ] ][ d[ i ] ] - f[ a[ i ] - 1 ][ d[ i ] ] - f[ c[ i ] ][ b[ i ] - 1 ] + f[ a[ i ] - 1 ][ b[ i ] - 1 ]; }int main(){scanf( "%d%d%d", &n, &m, &K );for( int i = 1 ; i <= n ; i++ ) scanf( "%s", mp[ i ] + 1 );for( int i = 1 ; i <= K ; i++ ){scanf( "%d%d%d%d", &a[ i ], &b[ i ], &c[ i ], &d[ i ] );while( e[ i ] < 'a' || e[ i ] > 'z' ) e[ i ] = getchar();e[ i ] -= 'a';sum[ e[ i ] ][ a[ i ] ][ b[ i ] ]++;sum[ e[ i ] ][ a[ i ] ][ d[ i ] + 1 ]--;sum[ e[ i ] ][ c[ i ] + 1 ][ b[ i ] ]--;sum[ e[ i ] ][ c[ i ] + 1 ][ d[ i ] + 1 ]++;}for( int i = 1 ; i <= n ; i++ )for( int j = 1 ; j <= m ; j++ ){int tot = 0, ret = 0;for( int x = 0 ; x < 26 ; x++ ){cnt[ x ] = sum[ x ][ i ][ j ] += sum[ x ][ i - 1 ][ j ] + sum[ x ][ i ][ j - 1 ] - sum[ x ][ i - 1 ][ j - 1 ];ret += cnt[ x ];tot += cnt[ x ] * x;}cnt[ mp[ i ][ j ] - 'a' ] += K - ret;tot += ( K - ret ) * ( mp[ i ][ j ] - 'a' );for( int x = 0 ; x < 26 ; x++ ){if( x == mp[ i ][ j ] - 'a' ) p[ i ][ j ] = p[ i ][ j - 1 ] + p[ i - 1 ][ j ] - p[ i - 1 ][ j - 1 ] + tot;s[ x ][ i ][ j ] = s[ x ][ i - 1 ][ j ] + s[ x ][ i ][ j - 1 ] - s[ x ][ i - 1 ][ j - 1 ] + tot;if( x ) cnt[ x ] += cnt[ x - 1 ];tot += cnt[ x ] - ( K - cnt[ x ] );}}for( int i = 1 ; i <= K ; i++ )ans = min( ans, p[ n ][ m ] - cal( p, i ) + cal( s[ e[ i ] ], i ) );cout << ans << endl;return 0;}