【题解】UVA-11624 Fire!

Fire! (UVA - 11624)

题面

img

思路

先跑一次bfs求出火蔓延到每一格的时间,再以此为限制对人跑bfs求解,注意有多个起火点

代码

查看代码
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using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 1e3+5;

char b[N][N] = {{0}};
bool vis[N][N] = {{0}};
int t[N][N] = {{0}};
int T, n, m, ans = 0;

int dx[] = { 1, -1, 0, 0};
int dy[] = { 0, 0, 1, -1};

struct F {
int x, y;
int step;
};

struct P {
int x, y;
int step;
};
queue<F> f;

void pre() {
memset(vis, 0, sizeof(vis));
while(!f.empty()) {
F tf = f.front();
f.pop();

F nf = tf;
for(int i=0; i<4; i++) {
int x = nf.x = tf.x+dx[i];
int y = nf.y = tf.y+dy[i];

if(x<0 || x>=n || y<0 || y>=m) {
continue;
}

if(!vis[x][y] && b[x][y]=='.') {
nf.step = tf.step+1;
f.push(nf);
vis[x][y] = 1;
t[x][y] = nf.step;
}
}
}
}

bool bfs(int x, int y) {
memset(vis, 0, sizeof(vis));

P sp;
sp.x = x;
sp.y = y;
sp.step = 0;

queue<P> q;
q.push(sp);
vis[x][y] = 1;

while(!q.empty()) {
P tp = q.front();
q.pop();

P np = tp;
for(int i=0; i<4; i++) {
int x = np.x = tp.x+dx[i];
int y = np.y = tp.y+dy[i];

if(x<0 || x>=n || y<0 || y>=m) {
ans = tp.step+1;
return true;
}

if(!vis[x][y] && b[x][y]=='.' && tp.step+1<t[x][y]) {
np.step = tp.step+1;
q.push(np);
vis[x][y] = 1;
}
}
}
return false;
}

int main(void) {
scanf("%d", &T);
for(int cs=1; cs<=T; cs++) {
memset(t, inf, sizeof(t));
scanf("%d%d", &n, &m);
int x, y;
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
scanf(" %c", &b[i][j]);
if(b[i][j]=='F') {
F sf;
sf.x = i;
sf.y = j;
sf.step = 0;
t[i][j] = 0;
f.push(sf);
} else if(b[i][j]=='J') {
x = i;
y = j;
}
}
}
pre();
if(bfs(x, y))
printf("%d\n", ans);
else
printf("IMPOSSIBLE\n");
}
return 0;
}
_/_/_/_/_/ EOF _/_/_/_/_/