http://poj.org/problem?id=2312
Battle City
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6903 | Accepted: 2336 |
Description
Many 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 walls and brick walls 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 walls, but it can destroy brick walls by shooting. A brick wall will be turned into empty spaces when you hit it, 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 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 should 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 4 YBEB EERE SSTE 0 0
Sample Output
8
Source
POJ Monthly,鲁小石
题意：就是从Y到T，S和R不能走，走B需要两步，问所需最少步数。
思路：比较简单的bfs，由于走B需要两步，所以第一次过B时停下来步数加一，下一次经过再加一。这样就保证bfs扩展的节点是最少的步数扩展到的。
也可以用优先队列直接bfs，很暴力。。
队列代码：
/**
* @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;
}
优先队列代码：
/**
* @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;
}