hdu 2870 - Largest Submatrix
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题目:统计一个字母矩阵中最大的相同字母的面积,有些字母可以换成其他字母。
分析:dp,单调队列。计算分三种分别换成a,b,c求出最大的子矩阵,求出最大即可。
然后就是单调队列优化的查询算法了,确定每个点右(左)边第一个比他小的点;
T(N)=O(N^2)。
说明:(2011-09-19 08:15)。
#include <stdio.h>#include <stdlib.h>#include <string.h>char maps[ 1005 ][ 1005 ];int smap[ 1005 ][ 1005 ];int S[ 1005 ];int L[ 1005 ];int R[ 1005 ];int SMQ[ 1005 ];int calc( char c, int k ){ switch( k ) { case 1: if ( c == 'a' || c == 'w' || c == 'y' || c == 'z' ) return 1;else return 0; case 2: if ( c == 'b' || c == 'w' || c == 'x' || c == 'z' ) return 1;else return 0; case 3: if ( c == 'a' || c == 'x' || c == 'y' || c == 'z' ) return 1;else return 0; } return 0;}int main(){ int n,m; while ( ~scanf("%d%d",&n,&m) ) { for ( int i = 1 ; i <= n ; ++ i ) scanf("%s",&maps[ i ][ 1 ]); int Max = 0; for ( int k = 1 ; k <= 3 ; ++ k ) { for ( int i = 1 ; i <= n ; ++ i ) for ( int j = 1 ; j <= m ; ++ j ) smap[ i ][ j ] = calc( maps[ i ][ j ], k ); memset( S, 0, sizeof( S ) ); for ( int i = 1 ; i <= n ; ++ i ) { for ( int j = 1 ; j <= m ; ++ j ) if ( smap[ i ][ j ] ) ++ S[ j ]; else S[ j ] = 0; S[ 0 ] = S[ m+1 ] = -1; SMQ[ 0 ] = 0; int tail1 = 0; for ( int i = 1 ; i <= m+1 ; ++ i ) { while ( S[ SMQ[ tail1 ] ] > S[ i ] ) R[ SMQ[ tail1 -- ] ] = i; SMQ[ ++ tail1 ] = i; } SMQ[ 0 ] = m+1; int tail2 = 0; for ( int i = m ; i >= 0 ; -- i ) { while ( S[ SMQ[ tail2 ] ] > S[ i ] ) L[ SMQ[ tail2 -- ] ] = i; SMQ[ ++ tail2 ] = i; } for ( int i = 1 ; i <= m ; ++ i ) if ( Max < S[ i ]*(R[ i ]-L[ i ]-1) ) Max = S[ i ]*(R[ i ]-L[ i ]-1); } } printf("%d\n",Max); } return 0;}
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