題目連結

題意：有一個迷宮，迷宮裡有多個火堆和一個人，每一分鐘人可以移動一個單元格，同時每堆火焰可以同時向上下左右四個方向移動，每分鐘火堆先移動，且人不能移動到有火堆的單元格，求人能否逃離迷宮且若能逃離則最短用時是多少。

思路：典型的bfs，由於火堆和人都要擴充套件，且火堆每分鐘要比人先移動，則初始化時，我們可以先把火堆放入佇列中再放入人，這樣每分鐘總能先擴充套件完所有火堆的移動，再擴充套件人的移動。之後就是模板題了。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<vector>
#include<queue>
#include<deque>
using namespace std;
#define ll long long
#define PI acos(-1)
#define INF 0x3f3f3f3f
#define NUM 33010
#define debug true
#define lowbit(x) ((-x)&x)
#define ffor(i,d,u) for(int i=(d);i<=(u);++i)
#define _ffor(i,u,d) for(int i=(u);i>=(d);--i)
#define mst(array,Num,Kind,Count) memset(array,Num,sizeof(Kind)*(Count))
const int P = 1e9+7;
int T;
int n,m;
char ma[1005][1005];
int check[1005][1005];
int h[4]={0,0,1,-1},l[4]={-1,1,0,0};
struct node
{
    int flag;
    int posx,posy;
    int step;
};
queue < node > q;
template <typename T>
void read(T& x)
{
    x=0;
    char c;T t=1;
    while(((c=getchar())<'0'||c>'9')&&c!='-');
    if(c=='-'){t=-1;c=getchar();}
    do(x*=10)+=(c-'0');while((c=getchar())>='0'&&c<='9');
    x*=t;
}
template <typename T>
void write(T x)
{
    int len=0;char c[21];
    if(x<0)putchar('-'),x*=(-1);
    do{++len;c[len]=(x%10)+'0';}while(x/=10);
    _ffor(i,len,1)putchar(c[i]);
}
inline bool bfs()
{
    node t1,t2;
    int x,y;
    while(!q.empty())
    {
        t1 = q.front(),q.pop();
        if(t1.flag == 2&&(t1.posx == 1||t1.posx == n||t1.posy == 1||t1.posy == m))
        {
            write(t1.step+1);
            putchar('\n');
            return false;
        }
        t2 = t1;
        ++t2.step;
        ffor(i,0,3)
        {
            x = t1.posx+h[i],y = t1.posy+l[i];
            if(x < 1||x > n||y < 1||y > m||ma[x][y] == '#')continue;
            if(t1.flag == 1&&check[x][y] != 1)
            {
                check[x][y] = 1;
                t2.posx = x,t2.posy = y;
                q.push(t2);
            }
            if(t1.flag == 2&&check[x][y] != 1&&check[x][y] !=2)
            {
                check[x][y] = 2;
                t2.posx = x,t2.posy = y;
                q.push(t2);
            }
        }
    }
    return true;
}
void AC()
{
    node x;
    read(T);
    int pos1,pos2;
    while(T--)
    {
        read(n),read(m);
        x.step = 0;
        x.flag = 1;
        while(!q.empty())q.pop();
        ffor(i,1,n)
        {
            ffor(j,1,m)
            {
                ma[i][j] = getchar();
                check[i][j] = 0;
                if(ma[i][j] == 'F')
                {
                    x.posx = i,x.posy = j;
                    check[i][j] = 1;
                    q.push(x);
                }
                if(ma[i][j] == 'J')pos1 = i,pos2 = j,check[i][j] = 2;
            }
            getchar();
        }
        x.posx = pos1,x.posy = pos2;
        x.flag = 2;
        q.push(x);
        if(bfs())puts("IMPOSSIBLE");
    }
}
int main()
{
    AC();
    return 0;
}