UVA 11624 Fire!

题意：在一个森林中有某些地方起火，有一个在森林中要逃出森林，到矩阵边缘即算逃出森林，火蔓延的速度和逃跑速度相同，判断这人是否能够逃出森林

链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=28833

思路：对所有起火点进行广搜，记录每个点的最短时间，第二次对人进行广搜，判断到底边界时是否满足条件

注意点：起火点可能有多个

以下为AC代码：

RunID	User	OJ All	Prob ID	Result All	Memory (KB)	Time (ms)	Language All	Length (Bytes)	Submit Time
3390675	luminous11	UVA	11624	Accepted		335	C++11 4.8.2	3092	2015-02-28 16:56:42

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
int dir[4][2] = { 0,1, 0,-1, 1,0, -1,0 };
int adj[1005][1005];
char str[1005][1005];
bool vis[1005][1005];
int m, n;
struct node{
    int x, y, cnt;
    node(){}
    node ( int _x, int _y ) : x(_x), y(_y) {}
    node ( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}
};
queue<node> q;
void init()
{
    clr( str, '@' );
    for ( int i = 1; i <= m; i ++ ){
        scanf ( "%s", &str[i][1] );
        str[i][n+1] = '@';
    }
    clr ( adj, 0x3f3f3f3f );
}
void bfs ( )
{
    while ( ! q.empty() ){
        node tmp = q.front();
        q.pop();
        for ( int i = 0; i < 4; i ++ ){
            int xi = tmp.x + dir[i][0];
            int yi = tmp.y + dir[i][1];
            if ( xi < 0 || yi < 0 || xi > m + 1 || yi > n + 1 )continue;
            if ( adj[xi][yi] > adj[tmp.x][tmp.y] + 1 ){
                adj[xi][yi] = adj[tmp.x][tmp.y] + 1;
                if ( str[xi][yi] != '#' && str[xi][yi] != '@' )
                    q.push ( node ( xi, yi ) );
            }
        }
    }
}
int bfs2 ( int x, int y )
{
    queue<node> q;
    clr ( vis, 0 );
    vis[x][y] = 1;
    q.push ( node ( x, y, 1 ) );
    while ( ! q.empty() ){
        node tmp = q.front();
        q.pop();
        for ( int i = 0; i < 4; i ++ ){
            node now = node( tmp.x + dir[i][0], tmp.y +dir[i][1], tmp.cnt + 1 );
            if ( vis[now.x][now.y] == 0 && adj[now.x][now.y] > now.cnt && str[now.x][now.y] != '#' ){
                if ( str[now.x][now.y] == '@' ){
                    return tmp.cnt;
                }
                vis[now.x][now.y] = 1;
                q.push ( now );
            }
        }
    }
    return -1;
}
void print()
{
    for ( int i = 0; i <= 5; i ++ ){
        for ( int j = 0; j <= 5; j ++ ){
            cout << str[i][j];
        }
        cout << endl;
    }
}
void solve()
{
    int sx, sy;
    for ( int i = 1; i <= m; i ++ ){
        for ( int j = 1; j <= n; j ++ ){
            if ( str[i][j] == 'J' ){
                sx = i;
                sy = j;
            }
            if ( str[i][j] == 'F' ){
                q.push ( node ( i, j ) );
                adj[i][j] = 1;
            }
        }
    }
    int ans = 1 << 29;
    bfs ();
    ans = bfs2 ( sx, sy );
    if ( ans == -1 )
        printf ("IMPOSSIBLE\n");
    else
        printf ( "%d\n", ans );
}
int main()
{
    int t;
    scanf ( "%d", &t );
    while ( t -- ){
        scanf ( "%d%d", &m, &n );
        init();
        solve();
    }
    return 0;
}