How to find the matrix row next to 1 in the block, and returns the number of coordinates and 1. Magic wand of PS why will find so fast, great god literacy.

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Fyi:

//item number: 0186 
//the turtle enclosure 
//the difficulty level: D; Run time limit: 1000 ms; Run space limit: 51200 KB; Code length limit: 2000000 b 
//item description 
//turtle small pretty strut walk around on the beach, and brutally announced: "my footprints everywhere are the turtles territory," 
//the turtle any action can be broken down into the following three basic action: 
//f forward to climb one 
//l place left 90 degrees 
//r back 90 degrees turn right 
//as a result of small pretty don't allow other turtles through its territory, so the closed area is surrounded by its footprint is its territory, 
//your task is to program according to the file input a series of small pretty basic action, to calculate how much small pretty were occupied territory, 
//program file called turtle, 
//to deepen the understanding of the question, we observe the following basic action sequence: 
//FLFFFLFFFRFRRFFFLF 
//# small pretty in starting point, of course, will leave footprints, the direction of the if it starts to down, can its footprint (marked by *) as the diagram below: 
//* * * * 
//* * 
//# * 
//* * * * 
//obviously, small pretty territory for a total of 15, 
//enter 
//only one string length less than 1000, and the string does not contain the 'f', 'l', 'r' characters, 
//output 
//in which there is only one positive integer that territory to the total number of lattice, 
//input sample 
//FLFFFLFFFRFRRFFFLF 
//output sample 
//15 
//other description 
//2011 Beijing haidian district secondary problem 
# include & lt; Stdio. H> 
# include & lt; Stdlib. H> 
# include & lt; String. H> 
# define MAXN 1000 
The static char n [MAXN + 2 * 2] [MAXN + 2 * 2];//record somewhere be climbed information: 0 not climb over the | 1 climb over the 
Char f;//the current direction of 'N' | 'S' | 'W' | 'E' 
Int x, y;//the current position 
Char [MAXN p + 1]; 
Int I, L; 
Int a, b, h, t; 
Int sp=0; 
Int the x1 [MAXN]; 
Int x2 (MAXN); 
Int y1 [MAXN]; 
Int yy1, xx1 xx2; 
Void push (int ay1, int ax1, int ax2) {
Y1 [sp]=ay1; 
X1=ax1 [sp]; 
X2=ax2 [sp]; 
If (spSp=sp + 1; 
} else {
Printf (" stack overflow! \n"); 
The exit (1); 
} 
} 
Void pop (void) {
Sp=sp - 1; 
Yy1=y1 [sp]; 
Xx1=x1 (sp); 
Xx2=x2 (sp); 
} 
//void show () {
//int zy, zx; 
//
//for (zy=0; Zy//for (zx=0; Zx//printf (" % d ", n [zy] [zx]); 
//} 
//printf (" \ n "); 
//} 
//printf (" \ n "); 
//} 
Void scanlinefill () {
Xx, int ax1, ax2; 
While (1) {
If (sp<=0) break;//
Pop (); 
//printf (" O yy1, xx1 xx2=% d, % d, % d \ n ", yy1, xx1, xx2); 
For (xx=xx1; Xx<=xx2; Xx++) n [yy1] [xx]=2;//fill this line 
Yy1 + +; 
If (yy1 & lt;=2 * MAXN) 
For (xx=xx1; Xx<=xx2; Xx++) {//examines from left to right next to a line each point below 
If (n [yy1] [xx]==0) {
Ax1=xx - 1; 
While (1) {//to the left to find the zero 
If (n [yy1] [ax1]!=0) break; 
Ax1 -; 
} 
Ax1 + +;//non zero point for the new line on the right the left point 
} 
If (n [yy1] [xx]==0) {
Ax2=xx + 1; 
While (1) {//to the right to find the zero 
If (n [yy1] [ax2]!=0) break; 
Ax2 + +; 
} 
Ax2 -;//not 0 points left for the new line right 
//printf (" I yy1 ax1, ax2=% d, % d, % d \ n ", yy1, ax1, ax2); 
Push (yy1 ax1, ax2);//record new scan lines to stack 
Xx=ax2 + 1;//the next cycle starting from the new point on the right side of the scan line right test 
} 
} 
} 
} 
Int main () {
The scanf (" % s ", p); 
F='S'. 
Y=MAXN; 
X=MAXN; 
N [y] [x]=1; 
L=strlen (p); 
for (i=0; iThe switch (p [I]) {
Case 'f' : 
The switch (f) {
Case 'N' : y -; N [y] [x]=1; break; 
Case 'S' : y++; N [y] [x]=1; break; 
Case 'W' : x -; N [y] [x]=1; break; 
Case 'E' : x++; N [y] [x]=1; break; 
} 
break; 
Case 'l' : 
The switch (f) {
Case 'N' : f='W'; break; 
Case 'S' : f='E'; break; 
Case 'W' : f='S'. break; 
Case 'E' : f='N'; break; 
} 
break; 
Case "r" : 
The switch (f) {
Case 'N' : f='E'; break; 
Case 'S' : f='W'; break; 
Case 'W' : f='N'; break; 
Case 'E' : f='S'. break; 
} 
break; 
} 
} 
//show (); 
For (y=0; YFor (x=0; x//show (); 
Push (1,1,2 * MAXN);//from (y1==1, the x1==1, 2 x2==* MAXN) began to 
Scanlinefill ();//to use the scan line seed filling algorithm fill 2 
//show (); 
a=0; 
For (y=0; Y<2 * MAXN + 1; Y++) {
For (x=0; x<2 * MAXN + 1; X++) {
If (n [y] [x]!=2) +; 
} 
} 
Printf (" % d \ n ", a); 
return 0; 
}