Poj2312 Battle City BFS

Http://poj.org/problem? Id = 2312

Battle City

Time limit:1000 ms		Memory limit:65536 K
Total submissions:6903		Accepted:2336

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 the 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 Head 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

Source

Poj monthly, Lu Xiaoshi
From y to T, S and R, it takes two steps for B to ask the minimum number of steps required. Idea: BFS is relatively simple. Because it takes two steps to go through B, when I first pass through B, I will stop and add one step at a time, and then I will add another step at the next time. This ensures that BFS nodes are extended to the minimum number of steps. You can also use the priority queue to direct BFs, which is very violent .. Queue code:

/** * @author neko01 *///#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <string.h>#include <iostream>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define min3(a,b,c) min(a,min(b,c))#define max3(a,b,c) max(a,max(b,c))#define pb push_back#define mp(a,b) make_pair(a,b)#define clr(a) memset(a,0,sizeof a)#define clr1(a) memset(a,-1,sizeof a)#define dbg(a) printf("%d\n",a)typedef pair<int,int> pp;const double eps=1e-9;const double pi=acos(-1.0);const int INF=0x3f3f3f3f;const LL inf=(((LL)1)<<61)+5;const int N=305;char s[N][N];bool vis[N][N];int n,m;int sx,sy;int dir[4][2]={-1,0,0,-1,1,0,0,1};struct node{    int x,y,step;};bool inmap(int x,int y){    return x>=0&&x<n&&y>=0&&y<=m&&s[x][y]!='S'&&s[x][y]!='R'&&!vis[x][y];}int bfs(){    clr(vis);    queue<node>q;    node cur,next;    cur.x=sx;    cur.y=sy;    cur.step=0;    vis[sx][sy]=true;    q.push(cur);    while(!q.empty())    {        cur=q.front();        if(s[cur.x][cur.y]=='T') return cur.step;        q.pop();        if(s[cur.x][cur.y]=='B')        {            next=cur;            next.step++;            s[cur.x][cur.y]='E';            q.push(next);            continue;        }        for(int i=0;i<4;i++)        {            int xx=cur.x+dir[i][0];            int yy=cur.y+dir[i][1];            if(inmap(xx,yy))            {                next.x=xx,next.y=yy;                next.step=cur.step+1;                vis[xx][yy]=true;                q.push(next);            }        }    }    return -1;}int main(){    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0) break;        for(int i=0;i<n;i++)        {            scanf("%s",s[i]);            for(int j=0;j<m;j++)                if(s[i][j]=='Y')                    sx=i,sy=j;        }        printf("%d\n",bfs());    }    return 0;}

Priority queue code:

/** * @author neko01 *///#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <string.h>#include <iostream>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define min3(a,b,c) min(a,min(b,c))#define max3(a,b,c) max(a,max(b,c))#define pb push_back#define mp(a,b) make_pair(a,b)#define clr(a) memset(a,0,sizeof a)#define clr1(a) memset(a,-1,sizeof a)#define dbg(a) printf("%d\n",a)typedef pair<int,int> pp;const double eps=1e-9;const double pi=acos(-1.0);const int INF=0x3f3f3f3f;const LL inf=(((LL)1)<<61)+5;const int N=305;char s[N][N];bool vis[N][N];int n,m;int sx,sy;int dir[4][2]={-1,0,0,-1,1,0,0,1};struct node{    int x,y,step;    bool operator < (const node &p) const{        return step>p.step;    }};bool inmap(int x,int y){    return x>=0&&x<n&&y>=0&&y<=m&&s[x][y]!='S'&&s[x][y]!='R'&&!vis[x][y];}int bfs(){    clr(vis);    priority_queue<node>q;    node cur,next;    cur.x=sx;    cur.y=sy;    cur.step=0;    vis[sx][sy]=true;    q.push(cur);    while(!q.empty())    {        cur=q.top();        if(s[cur.x][cur.y]=='T') return cur.step;        q.pop();        for(int i=0;i<4;i++)        {            int xx=cur.x+dir[i][0];            int yy=cur.y+dir[i][1];            if(inmap(xx,yy))            {                next.x=xx,next.y=yy;                if(s[xx][yy]=='B')                    next.step=cur.step+2;                else                    next.step=cur.step+1;                vis[xx][yy]=true;                q.push(next);            }        }    }    return -1;}int main(){    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0) break;        for(int i=0;i<n;i++)        {            scanf("%s",s[i]);            for(int j=0;j<m;j++)                if(s[i][j]=='Y')                    sx=i,sy=j;        }        printf("%d\n",bfs());    }    return 0;}

Poj2312 Battle City BFS