大火蔓延的迷宫（uva 11624）

题目链接 https://vjudge.net/problem/UVA-11624

题意

    大白书307页例题，走迷宫问题。

思路

    先用一个bfs求出大火蔓延到各个位置的最短时间，再用一个bfs求迷宫的最短路，当走到边缘时判断当前时间是否小于大火蔓延到该位置的时间即可，bfs的基础应用题。

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1050;
const int dx[4] = { 0, 1, 0, -1 };
const int dy[4] = { 1, 0, -1, 0 };

int n, m;
char g[maxn][maxn];
int a[maxn][maxn];//火蔓延到i,j处的时间为a[i][j]
int b[maxn][maxn];//人跑到i,j处的时间为b[i][j]

struct node {
int r, c;
node(int rr = 0, int cc = 0) :r(rr), c(cc) {}
};

void bfs1() {
queue<node> que;
while (!que.empty()) que.pop();
memset(a, -1, sizeof(a));//-1表示无法蔓延到

for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if ('F' == g[i][j]) {
a[i][j] = 0;
que.push(node(i, j));
}
}
}

while (!que.empty()) {
int r = que.front().r;
int c = que.front().c;
que.pop();

for (int k = 0; k < 4; ++k) {
int x = r + dx[k];
int y = c + dy[k];

if (x < 0 || x >= n || y < 0 || y >= m) continue;
if ('#' == g[x][y]) continue;
if (a[x][y] != -1) continue;

a[x][y] = a[r][c] + 1;
que.push(node(x, y));
}
}
}

int bfs2() {
queue<node> que;
while (!que.empty()) que.pop();
memset(b, -1, sizeof(b));

for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if ('J' == g[i][j]) {
b[i][j] = 0;
que.push(node(i, j));
goto End;
}
}
}
End:
while (!que.empty()) {
int r = que.front().r;
int c = que.front().c;
que.pop();
if (r == 0 || r + 1 == n || c == 0 || c + 1 == m) {
return b[r][c] + 1;
}

for (int k = 0; k < 4; ++k) {
int x = r + dx[k];
int y = c + dy[k];

if (x < 0 || x >= n || y < 0 || y >= m) continue;
if ('#' == g[x][y]) continue;
if (-1 != b[x][y]) continue;
if (-1 != a[x][y] && a[x][y] <= b[r][c] + 1) continue;

b[x][y] = b[r][c] + 1;
que.push(node(x, y));
}
}
return -1;
}

int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i) scanf("%s", g[i]);
bfs1();
int ans = bfs2();
if (ans == -1) printf("IMPOSSIBLE\n");
else printf("%d\n", ans);
}
return 0;
}