POJ 3494

单调栈，枚举每一个点，计算它上面共有多少个1，最后获取left与right值，计算max值

时间复杂度为O( n ),只是代码打得不好，时间比较慢，仍然AC

本题主要考察对搜索的理解，是应该枚举点，还是枚举边

#include <cstdio>using namespace std;const int MAX = 2010;typedef struct _Node{intdata;intn;intleft;intright;}Node;Nodedata[MAX][MAX];intstack[ MAX ];intstacki;introw,col;voidsetn(){for( int i = 0 ; i != col ; ++i )if( data[0][i].data == 1 )data[0][i].n = 1;elsedata[0][i].n = 0 ;for( int i = 1 ; i != row ; ++i )for( int j = 0 ; j != col ; ++j )if( data[i][j].data == 1 )data[i][j].n = data[i-1][j].n + 1;elsedata[i][j].n = 0;}voidsetleftright(){for( int i = 0 ; i != row ; ++i ){stacki = 0 ;for( int j = 0 ; j != col ; ++j ){while( stacki != 0 && data[ i ][ stack[ stacki - 1 ] ].n > data[i][j].n )data[ i ][ stack[ --stacki ] ].right = j - 1 ;stack[ stacki++ ] = j ;}while( stacki != 0 )data[i][ stack[ --stacki ] ].right = col - 1;stacki = 0 ;for( int j = col - 1 ; j >= 0 ; --j ){while( stacki != 0 && data[ i ][ stack[ stacki - 1 ] ].n > data[i][j].n )data[ i ][ stack[ --stacki ] ].left = j + 1 ;stack[ stacki++ ] = j ;}while( stacki != 0 )data[i][ stack[ --stacki ] ].left = 0 ;}}intgetmax(){intmax = 0;intmaxmid;for( int i = 0 ; i != row ; ++i )for( int j = 0 ; j != col ; ++j )if( data[i][j].data == 1 ){maxmid = data[i][j].right - data[i][j].left + 1 ;if( maxmid * data[i][j].n > max )max = maxmid * data[i][j].n;}return max;}intmain(){while( scanf("%d%d",&row,&col) != EOF ){for( int i = 0 ; i != row ; ++i )for( int j = 0 ; j != col ; ++j )scanf("%d",&data[i][j].data );setn();setleftright();printf("%d/n",getmax());}}