Nyoj-284-tank wars

 

Tank war time limit: 1000 MS | memory limit: 65535 kb difficulty: 3

Description

Memory of us had played the game "Battle City" in our childhood, and some people (like me) Even often play it on computer now.
What we are discussing is a simple edition of this game. given a map that consists of empty spaces, rivers, steel Wils and brick Wils only. your task is to get a bonus as soon as possible suppose that no enemies will disturb you (see the following picture ).



Your tank can't move through rivers or wall, but it can destroy brick wallby shooting. A brick wall will be turned into empty spaces when you hit, however, if your shot hit a steel wall, there will be no damage to
Wall. in each of your turns, you can choose to move to a neighboring (4 directions, not 8) empty space, or shoot in one of the four directions without a move. the shot will go ahead in that direction, until it go out of the map or hit a wall. if the shot hits
A brick wall, the wall will disappear (I. E ., in this turn ). well, given the description of a map, the positions of your tank and the target, how many turns will you take at least to arrive there?

Input

The input consists of several test cases. the first line of each test case contains two integers m and n (2 <= m, n <= 300 ). each of the following M lines contains N uppercase letters, each of which is one of 'y' (you), 'T'
(Target),'s (steel wall), 'B' (brick wall), 'R' (river) and 'E' (empty space ). both 'y' and 't'appear only once. A test case of M = n = 0 indicates the end of input, and shoshould not be processed.

Output

For each test case, please output the turns you take at least in a separate line. If you can't arrive at the target, output "-1" instead.

Sample Input

3 4YBEBEERESSTE0 0

Sample output

8

The question is very simple, that is, "input m rows and n columns". "Y" indicates the location of your tank, and "T" indicates your goal, on the way, "S" indicates the steel wall that cannot pass, and "R" indicates the river that cannot pass, "B" indicates that you are wasting a bullet (that is, taking one step is equivalent to taking two steps), and "E" can represent the open space. This is equivalent to the shortest path problem. Deep Search and wide search are all supported. However, I am doing deep search myself, and wide search is a code that sticks to a large bull.

The code for deep search is as follows:

#include<stdio.h>#include<string.h>int m,n,a[3],b[2],s[302][302];char map[302][302];void fun(int w,int e,int x,int y){     if(x<0||y<0||x==m||y==n)   return ;     if(map[x][y]=='R'||map[x][y]=='S'||map[x][y]=='Y') return ;     if(x==b[0]&&y==b[1])     {         if(s[x][y]==0)  s[x][y]=s[w][e]+1;         else         {             if(s[x][y]<s[w][e]+1) return ;             else                  s[x][y]=s[w][e]+1;         }     }     if(map[x][y]=='B')     {         if(s[x][y]==0)  s[x][y]=s[w][e]+2;         else         {             if(s[x][y]<s[w][e]+2) return ;             else                  s[x][y]=s[w][e]+2;         }     }     if(map[x][y]=='E')     {         if(s[x][y]==0)  s[x][y]=s[w][e]+1;         else         {             if(s[x][y]<s[w][e]+1) return ;             else                  s[x][y]=s[w][e]+1;         }     }     fun(x,y,x-1,y);fun(x,y,x+1,y);fun(x,y,x,y-1);fun(x,y,x,y+1);}int main(){    int k,i,j,z[5];    while(scanf("%d%d",&m,&n),m,n)    {        memset(s,0,sizeof(s));        for(i=0;i<m;i++)        {            scanf("%s",map[i]);            for(j=0;j<n;j++)            {                if(map[i][j]=='Y') {a[0]=i;a[1]=j;}                if(map[i][j]=='T') {b[0]=i;b[1]=j;}            }        }        fun(a[0],a[1],a[0]-1,a[1]);        fun(a[0],a[1],a[0],a[1]-1);        fun(a[0],a[1],a[0]+1,a[1]);        fun(a[0],a[1],a[0],a[1]+1);        if(s[b[0]][b[1]]==0)          printf("-1\n");        else                       printf("%d\n",s[b[0]][b[1]]);    }}

The specific code of guangsearch is as follows:

 #include<iostream>#include<cstdio>#include<cstdlib>#include<queue>#include<cstring>using namespace std;char map[301][301];int row,col;int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};struct node {int x;int y;int c;friend bool operator < (const node &a,const node &b){return a.c>b.c;}};node begin1,end1;priority_queue<node>data;int bfs(){int i=0;node tem;while( !data.empty() ){begin1=data.top();data.pop();if( (begin1.x==end1.x) && (begin1.y==end1.y))       return begin1.c;for(i=0;i<4;i++){tem.x=begin1.x+dir[i][0];tem.y=begin1.y+dir[i][1];if(tem.x>=0 && tem.x<row && tem.y>=0 && tem.y<col && map[tem.x][tem.y]!='1'){if(map[tem.x][tem.y]=='B'){tem.c=begin1.c+2;data.push(tem);map[tem.x][tem.y]='1';}else{if(map[tem.x][tem.y]=='E' || map[tem.x][tem.y]=='T'){tem.c=begin1.c+1;data.push(tem);map[tem.x][tem.y]='1';}}}}}return 0;}int main(){int i=0,j=0;int step;while(scanf("%d%d",&row,&col)&&(row+col)){step=0;memset(map,'\0',sizeof(map));for(i=0;i<row;i++){scanf("%s",map[i]);for(j=0;j<col;j++){if(map[i][j]=='Y'){begin1.x=i;begin1.y=j;begin1.c=0;}if(map[i][j]=='T'){end1.x=i;end1.y=j;end1.c=0;}}}data.push(begin1);map[begin1.x][begin1.y]='1';step=bfs();if(step>0)printf("%d\n",step);elseprintf("-1\n");while(!data.empty())data.pop();}return 0;}