POJ3467（預先處理）

標籤：

Cross Counting

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 1331		Accepted: 375

Description

Given a N × M grid with different colors on each cell, your task is to calculate the total amount of crosses of a specific color. We say there exists a cross of sizek centered at the cell (x,y) iff all cells lying in thex-th row or the y-th column and within a distance of k from (x,y) share the same color. Note that if two crosses have the same center but different sizes we consider they are distinct. Unfortunately, the color of each cell may varies by time and you have to respond to all the queries.

Input

There are four integers, N, M, C, Q, in the first line. (1 ≤N, M, C ≤ 100, 1 ≤ Q ≤ 10000)
The next N lines each contains M integers between 1 and C which describe the color of cells.
The following Q lines each has either the form "C i j k" indicating to change the color of cell (i,j) into k, or the form "Q c" indicating to query the total amount of crosses of colorc. (1 ≤ i ≤ N, 1 ≤ j ≤ M, 1 ≤ k,c ≤ C)

Output

Output the answer to each query.

Sample Input

5 5 3 61 3 2 3 13 3 2 3 32 2 2 2 23 3 2 3 31 3 2 3 1Q 1Q 2Q 3C 2 3 3C 3 2 3Q 3

Sample Output

0201

Source

POJ Monthly--2007.11.25, Yang Yi

題目看起來有線段樹的味道，我線段樹能力有限。

發現N,M都非常少，用O(N*M*(N+M))做預先處理。

對於每次更新。改變的是十字架。所以更新須要O(N+M)個更新。

我在外層加了一層哨兵。寫起來比較順暢。

/***********************************************************> OS     : Linux 3.13.0-24-generic (Mint-17)> Author : yaolong> Mail   : [email protected]> Time   : 2014年10月15日 星期三 07時11分26秒 **********************************************************/#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;const int N = 205;int mp[N][N];int cnt[N][N];int color[N];void update ( int i, int j ){    cnt[i][j] = 0;    for ( int k = 1;; k++ )    {        if ( mp[i + k][j] == mp[i][j] && mp[i - k][j] == mp[i][j] && mp[i][j + k] == mp[i][j] && mp[i][j - k] == mp[i][j] )        {            cnt[i][j]++;        }        else        {            return ;        }    }}int main(){    int n, m, c, q;    while ( scanf ( "%d%d%d%d", &n, &m, &c, &q ) != EOF )    {        int i, j, k;        memset ( color, 0, sizeof ( color ) );        memset ( mp, 63, sizeof ( mp ) );        memset ( cnt, 0, sizeof ( cnt ) );        for ( i = 1; i <= n; i++ )        {            for ( j = 1; j <= m; j++ )            {                scanf ( "%d", &mp[i][j] );            }        }        for ( i = 1; i <= n; i++ )        {            for ( j = 1; j <= m; j++ )            {                update ( i, j );                color[mp[i][j]] += cnt[i][j];            }        }        char tmp;        while ( q-- )        {            scanf ( " %c", &tmp );            if ( tmp == 'Q' )            {                scanf ( "%d", &i );                printf ( "%d\n", color[i] );            }            else            {                scanf ( "%d%d%d", &i, &j, &k );                if ( mp[i][j] == k )                {                    continue;                }                color[mp[i][j]] -= cnt[i][j];                mp[i][j] = k;                update ( i, j );                color[mp[i][j]] += cnt[i][j];                //cout<<cnt[i][j]<<" "<<i<<j<<endl;                for ( int ak = 1; ak <= n; ak++ )                {                    if ( i != ak )                    {                        color[mp[ak][j]] -= cnt[ak][j]; //減掉                        update ( ak, j );                        color[mp[ak][j]] += cnt[ak][j];                    }                }                for ( int ak = 1; ak <= m; ak++ )                {                    if ( j != ak )                    {                        color[mp[i][ak]] -= cnt[i][ak]; //減掉                        update ( i, ak );                        color[mp[i][ak]] += cnt[i][ak];                    }                }            }        }    }    return 0;}

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POJ3467（預先處理）