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HDU 1026 Ignatius and the Princess I（bfs +記錄路徑）

編輯：關於C++

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14184 Accepted Submission(s): 4474
Special Judge

Problem Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166's room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:

1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.

Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.

Input The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.

Output For each test case, you should output God please help our poor hero. if Ignatius can't reach the target position, or you should output It takes n seconds to reach the target position, let me show you the way.(n is the minimum seconds), and tell our hero the whole path. Output a line contains FINISH after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.

Sample Input

5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX1
5 6
.XX...
..XX1.
2...X.
...XX.
XXXXX.

Sample Output

It takes 13 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
FINISH
It takes 14 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
14s:FIGHT AT (4,5)
FINISH
God please help our poor hero.
FINISH

/*

起點(0,0)  終點(n-1,m-1),
  'x'代表不能走，
  '.'可以走，走過來需耗費一秒，
  數字代表可以走，不過走過來一秒還需要加上這個數字

優先隊列，對於記錄路徑開一個path數組記錄這一個點是從哪一個方向走來的就可以了，然後遞歸輸出路徑




*/


#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i = a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf(%s, n)
#define sf(n)       scanf(%d, &n)
#define sff(a,b)    scanf(%d %d, &a, &b)
#define sfff(a,b,c) scanf(%d %d %d, &a, &b, &c)
#define pf          printf
#define bug         pf(Hi
)

using namespace std;

#define N 105



struct stud{
  stud(int xx,int yy,int s):x(xx),y(yy),step(s){};
  stud(){};
  int x,y,step;
  bool operator<(const stud b) const
  {
  	return step>b.step;
  }
};

int n,m;

priority_queueq;

int path[N][N];
int vis[N][N];
int step[4][2]={1,0,-1,0,0,1,0,-1};
int all;
char c[N][N];

bool judge(int x,int y)
{
	if(x>=0&&x=0&&y(%d,%d)
,++all,xx,yy,x,y);

	if(c[x][y]>='0'&&c[x][y]<='9')
	{
		int i=c[x][y]-'0';
		while(i--)
			printf(%ds:FIGHT AT (%d,%d)
,++all,x,y);
	}

}

bool bfs()
{
	int i,j;
	stud cur,next;
	while(!q.empty()) q.pop();
	mem(vis,0);
	vis[0][0]=1;
	cur.x=0;
	cur.y=0;
	cur.step=0;
	q.push(cur);
	while(!q.empty())
	{
		cur=q.top();
		q.pop();
		if(cur.x==n-1&&cur.y==m-1)
		{
		    printf(It takes %d seconds to reach the target position, let me show you the way.
,cur.step);
		    all=0;
		    show(n-1,m-1);
		    return true;
		}

		for(i=0;i<4;i++)
		{
           int xx=cur.x+step[i][0];
           int yy=cur.y+step[i][1];
           if(!judge(xx,yy)) continue;
           if(vis[xx][yy]) continue;
           if(c[xx][yy]=='X') continue;

           next.x=xx;
           next.y=yy;
           next.step=cur.step+1;
           if(c[xx][yy]>='0'&&c[xx][yy]<='9')
		   next.step+=c[xx][yy]-'0';
           q.push(next);
           vis[xx][yy]=cur.step+1;
           path[xx][yy]=i;
		}
	}
  return false;
}

int main()
{
   int i,j;
   while(~scanf(%d%d,&n,&m))
   {
   	  for(i=0;i