uva11624-fire! (BFs two times)

Topic links

Test instructions: Your task is to help J out of a maze of fires spreading. J can move up or down in four directions per minute, and all the burning squares will spread around. There are some obstacles in the maze, J and fire can not enter. When J went to the boundary lattice of a maze, we thought he had gone out of the maze.

Idea: This is a problem above the white, in fact, as long as each lattice what time on fire to deal with it. Two times BFS, the first time to deal with lattice fire, the second BFS is to determine whether in the shortest possible time to go out.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <

Algorithm> using namespace std;
const int MAXN = 1005;
const int INF = 0X3F3F3F3F;
const int dx[] = {-1, 0, 0, 1};

const int dy[] = {0,-1, 1, 0}; 
        struct node{node (int xx, int yy, int dd) {x = xx;
        y = yy;
    D = DD;
} int x, y, D;

};
Char G[MAXN][MAXN];
int VF[MAXN][MAXN], VM[MAXN][MAXN], T[MAXN][MAXN];

int R, c, SX, SY;

Queue<node> QF; 
        void Bfs_fire () {while (!qf.empty ()) {node u = qf.front (); 
        Qf.pop ();
        Node v = u; 
            for (int i = 0; i < 4; i++) {int tx = u.x + dx[i];      
            int ty = u.y + dy[i];
            if (TX < 0 | | TX >= R | | Ty < 0 | | ty >= c | | g[tx][ty] = = ' # ') continue; 
                if (!vf[tx][ty]) {Vf[tx][ty] = 1; 
                v.x = TX;
                v.y = Ty;
                V.D = u.d + 1; T[tx][tY] = V.D;
            Qf.push (v);
    }}}} int Bfs_man () {queue<node> QM; 
    while (!qm.empty ()) Qm.pop ();
    Node S (SX, SY, 0);
    Qm.push (s);
    memset (VM, 0, sizeof (VM));
    Vm[sx][sy] = 1; 
        while (!qm.empty ()) {node u = qm.front (); 
        Qm.pop ();
        if (u.x = = 0 | | u.x = = R-1 | | u.y = = 0 | | u.y = c-1) return U.D + 1;
        Node v = u; 
            for (int i = 0; i < 4; i++) {int tx = u.x + dx[i]; 
            int ty = u.y + dy[i];
            if (TX < 0 | | TX >= R | | Ty < 0 | | ty >= c | | g[tx][ty] = = ' # ') continue;
            if (u.d + 1 >= t[tx][ty]) continue; 
                if (!vm[tx][ty]) {Vm[tx][ty] = 1; 
                v.x = TX;
                v.y = Ty;
                V.D = u.d + 1;
            Qm.push (v);
}}} return-1; 
    } int main () {int cas;
    scanf ("%d", &cas); while (cas--) {scanf("%d%d", &r, &c);

        for (int i = 0; i < R; i++) scanf ("%s", G[i]);
        while (!qf.empty ()) Qf.pop ();
        memset (VF, 0, sizeof (VF));

        memset (t, INF, sizeof (t));
                    for (int i = 0, i < R; i++) {for (int j = 0; J < C; j + +) {if (g[i][j] = = ' J ') { 
                    SX = i; 
                sy = j;       
                    } if (g[i][j] = = ' F ') {node tmp (i, j, 0);
                    VF[I][J] = 1;
                    T[I][J] = 0;
                Qf.push (TMP);
        }}} bfs_fire ();    
        int ans = Bfs_man ();
        if (ans = =-1) printf ("impossible\n");
    else printf ("%d\n", ans);
} return 0; }