POJ 2312 Battle City (priority queue + BFS), pojbfs

POJ 2312 Battle City (priority queue + BFS), pojbfs

Question link: http://poj.org/problem? Id = 2312

Battle City

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:7085		Accepted:2390

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

Question: How long does it take at least from Y to T.

Problem: first, calculate the shortest time and use BFS. Because the weights of each point may be different, priority queue is required ~~

PS: it is convenient to use the priority queue in STL for the first time !!!

AC code:

#include<iostream>#include<cstring>#include<queue>#define N 305using namespace std;int n,m,sx,sy,ex,ey,visit[N][N];int dir[][2]={    {0,1},{0,-1},{1,0},{-1,0}};char chess[N][N];struct Node{    int x,y,s;    friend bool  operator <(Node a,Node b ){        return a.s>b.s;    }};bool ok(int x,int y){    if(x>=0&&x<m&&y>=0&&y<n&&chess[x][y]!='S'&&chess[x][y]!='R')        return true;    return false;}int bfs(){    priority_queue<Node> q;    memset(visit,-1,sizeof(visit));    visit[sx][sy]=0;    Node head={sx,sy,0};    q.push(head);    while(!q.empty())    {        Node f=q.top();        q.pop();        if(f.x==ex&&f.y==ey)return f.s;        for(int i=0;i<4;i++){            int dx=f.x+dir[i][0],dy=f.y+dir[i][1];            if(ok(dx,dy)&&visit[dx][dy]){                visit[dx][dy]=0;                int temp=0;                if(chess[dx][dy]=='B')temp=2;                else temp=1;                Node tmp={dx,dy,f.s+temp};                q.push(tmp);            }        }    }    return -1;}int main(){    cin.sync_with_stdio(false);    while(cin>>m>>n&&(m||n))    {        for(int i=0;i<m;i++)            for(int j=0;j<n;j++){                cin>>chess[i][j];                if(chess[i][j]=='Y'){                    sx=i;sy=j;                    chess[i][j]='S';                }                else if(chess[i][j]=='T'){                    ex=i;ey=j;                }            }        cout<<bfs()<<endl;    }    return 0;}

[Reprinted with the source]

Author: MummyDing

Source: http://blog.csdn.net/mummyding/article/details/43676441