UVA 11624 Fire！【两点BFS】

Joe works in a maze. Unfortunately, portions of the maze have
caught on fire, and the owner of the maze neglected to create a fire
escape plan. Help Joe escape the maze.
Given Joe’s location in the maze and which squares of the maze
are on fire, you must determine whether Joe can exit the maze before
the fire reaches him, and how fast he can do it.
Joe and the fire each move one square per minute, vertically or
horizontally (not diagonally). The fire spreads all four directions
from each square that is on fire. Joe may exit the maze from any
square that borders the edge of the maze. Neither Joe nor the fire
may enter a square that is occupied by a wall.
Input
The first line of input contains a single integer, the number of test
cases to follow. The first line of each test case contains the two
integers R and C, separated by spaces, with 1 ≤ R, C ≤ 1000. The
following R lines of the test case each contain one row of the maze. Each of these lines contains exactly
C characters, and each of these characters is one of:
• #, a wall
• ., a passable square
• J, Joe’s initial position in the maze, which is a passable square
• F, a square that is on fire
There will be exactly one J in each test case.
Output
For each test case, output a single line containing ‘IMPOSSIBLE’ if Joe cannot exit the maze before the
fire reaches him, or an integer giving the earliest time Joe can safely exit the maze, in minutes.
Sample Input
2
4 4
####
#JF#
#..#
#..#
3 3
###
#J.
#.F
Sample Output
3
IMPOSSIBLE

【题意】：一个人要逃离迷宫，迷宫中有多处起火了。问能否逃出迷宫，能输出最小步数，不能输出"IMPOSSIBLE"。迷宫的边缘都是出口。

【分析】：双点BFS，火和人同时进行BFS即可。注意首先火源不只一处，可以有多处，那么我们就要把每处火都数组记录下来，然后bfs搜索前让火源全部入队，还有就是不需要逃出地图，只要跑到边界就ok。

【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 1e3 + 20;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n,m,t;
char a[1005][1005];
int v[1005][1005];
int step[1005][1005];
struct node
{
    int x,y,s;
    node(){}
    node(int xx,int yy,int ss)
    {
        x=xx;
        y=yy;
        s=ss;
    }
};
bool ok(int x,int y)
{
    return x>=0 && y>=0 && x<n && y<m && !v[x][y] && a[x][y]!='#';
}
queue<node>q;
void bfs1() //火
{

    while(!q.empty())
    {
        node st = q.front();
        q.pop();
        node ed;
        for(int i=0; i<4; i++)
        {
            ed.x = st.x + dir[i][0];
            ed.y = st.y + dir[i][1];
            if(!ok(ed.x,ed.y)) continue;
            step[ed.x][ed.y] = st.s + 1;
            v[ed.x][ed.y]=1;
            q.push(node(ed.x,ed.y,st.s+1));
        }
    }
}

int bfs2(int x,int y) //人
{
    ms(v,0);
    v[x][y]=1;
    q.push(node(x,y,0));
    while(!q.empty())
    {
        node st = q.front();
        q.pop();
        node ed;
        if(st.x==0||st.x==n-1||st.y==0||st.y==m-1)
        {
            return st.s+1;
        }
        for(int i=0;i<4;i++)
        {
            ed.x = st.x + dir[i][0];
            ed.y = st.y + dir[i][1];
            if(!ok(ed.x,ed.y)) continue;
            if(st.s + 1 < step[ed.x][ed.y]) //人逃跑花的时间比火烧的时间短才行
            {
                v[ed.x][ed.y]=1;
                q.push(node(ed.x,ed.y,st.s+1));
            }
        }
    }
    return -1;
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        while(!q.empty()) q.pop();
        ms(v,0);
        ms(step,INF);
        int x,y;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%s",a[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(a[i][j] == 'F') //多个用数组存
                {
                    q.push(node(i,j,0));
                    v[i][j]=1;
                    step[i][j]=0;
                }
                if(a[i][j] == 'J')
                {
                    x = i;
                    y = j;
                }
            }
        }
        bfs1();
        int ans=bfs2(x,y);
        if(ans==-1) puts("IMPOSSIBLE");
        else printf("%d\n",ans);
    }
}