POJ 2312 Battle City Priority multi-column +BFS

Source: http://poj.org/problem?id=2312

Test instructions: The title background is the battle of the tanks when you were young, and how many steps you need at least from the beginning to the end. It is known that s and R can not go, E is empty, can go, B is brick, only after the knock can pass.

Idea: It is easy to see that this is a wide-search topic, but because of the time to go e and go B is not the same, so you can not use the normal queue to save points. Because to go to B, we have to get rid of bricks before they pass, so we can understand that to go B takes two steps, and walk e refers to the need for 1 steps. Therefore, you cannot use a normal queue to save. We can use the priority queue in the STL to save, each time from the team head is a few steps of the point on it. So the first search is the optimal solution.

Code:

#include <iostream> #include <cstdio> #include <queue> #include <string.h> #include <

Algorithm> #include <string> using namespace std;
	#define CLR (Arr,val) memset (arr,val,sizeof (arr)) struct point{int x,y,step;
	BOOL operator < (const point & P) const{return step > P.step;
} 
};
const int N = 305;
Char Map[n][n];
int sx,sy,ex,ey,flag[n][n],row,col;
int addx[4] = {0,0,1,-1};
int addy[4] = {1,-1,0,0};
	int BFs () {priority_queue<point> qq;
	Point P; p.x = SX; P.Y = sy;
	P.step = 0;
	Flag[sx][sy] = 1;
	Qq.push (P);
		while (!qq.empty ()) {Point Topp = Qq.top ();
		Qq.pop ();
		if (topp.x = = Ex && topp.y = = ey) {return topp.step;
			   } else{for (int i = 0; i < 4; ++i) {int newx = topp.x + addx[i];
			   int newy = topp.y + addy[i];
				   if (newx >= 0 && newx < row && newy >= 0 && newy < col &&!flag[newx][newy]) { if (map[newx][newy] = = ' S ' | | map[newx][newy] = = 'R ') continue;
				  Point Newp;
				  newp.x = newx;
				  NEWP.Y = Newy;
				  Flag[newx][newy] = 1;
				  if (map[newx][newy] = = ' B ') {newp.step = Topp.step + 2;
				  } else Newp.step = Topp.step + 1;
			   Qq.push (NEWP);
}}}} return-1;
	} int main () {//freopen ("1.txt", "R", stdin);
	   while (scanf ("%d%d", &row,&col)! = EOF) {if (row + col = = 0) break;
	   CLR (flag,0);
	   CLR (map, ' 0 ');
			  for (int i = 0; i < row, ++i) {for (int j = 0; J < col; ++j) {cin >> map[i][j];
				if (map[i][j] = = ' Y ') {sx = i;
			  sy = j;
				} else if (map[i][j] = = ' T ') {ex = i;
			  EY = j;
	  }}} int ans = BFS ();
	printf ("%d\n", ans);
} return 0; }