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根据题意首先找出可以当作起始点的点,并用数字作顺序标记。
之后从这些起始点开始,向右开始搜寻字母组成单词,直至到边界或到黑块
之后依旧从这些起始点开始,向下开始搜寻字母组成单词,直至到边界或到黑块
其中注意输出格式如" 1.AT",题目并不是要求在数字前加2个空格,不难发现题目样例输出中有“ 19.DEA”,可以得知,题目的意思是:数字及数字前的空格总共占3个字符的位置
/* UvaOJ 232 Emerald Sat 18 Apr, 2015 */ #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int MAXN = 10 + 2; char puzzle[ MAXN ][ MAXN ]; int isStart[ MAXN ][ MAXN ]; //bool used[ MAXN ][ MAXN ]; int rows, cols, posCount; const char BLACK_GRID = ‘*‘; int isStarted( int i, int j ) { // judge whether the puzzle[i][j] is a starting point if( puzzle[i][j] == BLACK_GRID ) { return -1; // 这个 } if( i-1<0 || j-1<0 ) { return ++posCount; } if( puzzle[i-1][j]==BLACK_GRID || puzzle[i][j-1]==BLACK_GRID ) { return ++posCount; } return -1; } void across( int x, int y ) { // whether can output a word if( ( y-1<0 || puzzle[x][y-1]==BLACK_GRID ) && puzzle[x][y]!=BLACK_GRID ) { // presention !!! printf( "%3d.", isStart[x][y] ); while( y<cols && puzzle[x][y]!=BLACK_GRID ) { printf( "%c", puzzle[x][y++] ); } printf("\n"); } } void down( int x, int y ) { // whether can output a word if( ( x-1<0 || puzzle[x-1][y]==BLACK_GRID ) && puzzle[x][y]!=BLACK_GRID ) { printf( "%3d.", isStart[x][y] ); // presention !!! while( x<rows && puzzle[x][y]!=BLACK_GRID ) { printf( "%c", puzzle[x++][y] ); } printf("\n"); } } int main() { int i, j; int counter = 0; while( scanf( "%d", &rows )!=EOF && rows ) { scanf( "%d", &cols ); posCount = 0; for( i=0; i<rows; i++ ) { getchar(); for( j=0; j<cols; j++ ) { scanf( "%c", &puzzle[i][j] ); isStart[i][j] = isStarted( i, j ); } } if( counter ) { printf( "\n" ); } printf( "puzzle #%d:\n", ++counter ); /* --------Across---------- */ printf( "Across\n" ); for( i=0; i<rows; i++ ) { for( j=0; j<cols; j++ ) { across( i, j ); } } /* --------Down---------- */ printf( "Down\n" ); for( i=0; i<rows; i++ ) { for( j=0; j<cols; j++ ) { down( i, j ); } } } return 0; }
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原文地址:http://www.cnblogs.com/Emerald/p/4438071.html