J - Fire! UVA - 11624

题意：火每次能烧到上下左右，人碰到非墙的边界则逃火成功，求最短的逃离时间。

由于bfs每个位置最多入队出队一次，所以复杂度为 1e6
一发bfs直接TLE，妙啊 -_-

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
const int maxn = 1e3+10;
using namespace std;
int n, m;
bool vis[maxn][maxn];
char s[maxn][maxn]; // 地图
int Jx, Jy;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};bool checkF(int x, int y) {if(x < 1 || x > n || y < 1 || y > m) return 0;return 1;
} bool checkJ(int x, int y) {if(s[x][y] == 'F' || s[x][y] == '#' || vis[x][y]) return 0;return 1;
} struct node{int x, y;int step;
};queue<node> FQ;int bfs() {node sta;sta.x = Jx; sta.y = Jy; sta.step = 0;queue<node> Q; Q.push(sta); while(!Q.empty()) {node now = Q.front(); Q.pop();if(now.x == 0 || now.x == n+1 || now.y == 0 || now.y == m+1) return now.step;while(!FQ.empty()) {node Fnow = FQ.front(); FQ.pop();for(int i = 0; i < 4; i++) { // 更新火烧到的地方 int nx = Fnow.x + dx[i], ny = Fnow.y + dy[i];if(checkF(nx,ny)) s[nx][ny] = '#';}}for(int i = 0; i < 4; i++) {int nx = now.x + dx[i], ny = now.y + dy[i];if(now.x == 0 || now.x == n+1 || now.y == 0 || now.y == m+1) return now.step+1;if(checkJ(nx,ny)){node tmp;tmp.x = nx, tmp.y = ny, tmp.step = now.step + 1;Q.push(tmp);}}}return 0;
}int main() {//  freopen("test.in", "r", stdin);int T; cin >> T;while(T--) {ios::sync_with_stdio(false);memset(vis, 0, sizeof(vis));while(!FQ.empty()) FQ.pop();int ans = 0;cin >> n >> m;for(int i = 1; i <= n; i++) {cin >> s[i]+1;for(int j = 1; j <= m; j++) {if(s[i][j] == 'J') Jx = i, Jy = j;if(s[i][j] == 'F') {node fire;fire.x = i; fire.y = j;FQ.push(fire);}}}ans = bfs();if(!ans) cout << "IMPOSSIBLE" << endl;else cout << ans << endl;}return 0;
}

注意：最开始可能不止一处起火

发现TLE是因为我在更新火蔓延的位置的时候出问题了，每次都更新，就出不来了，已经有火的地方就没必要再入队了，墙也不需要入队，修改一下就AC了

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
const int maxn = 1e3+10;
using namespace std;
int n, m;
bool vis[maxn][maxn];
char s[maxn][maxn]; // 地图
int Jx, Jy;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};bool checkF(int x, int y) {if(x < 1 || x > n || y < 1 || y > m) return 0;if(s[x][y] == '#') return 0;return 1;
} bool checkJ(int x, int y) {if(s[x][y] == 'F' || s[x][y] == '#' || vis[x][y]) return 0;return 1;
} struct node{int x, y;int step;int id; // 1是人   0是火
};queue<node> Q;int bfs() {node sta;sta.x = Jx; sta.y = Jy; sta.step = 0; sta.id = 1;vis[Jx][Jy] = 1;Q.push(sta); while(!Q.empty()) {int flag = 0;node now = Q.front(); Q.pop();if(now.id == 1 && (now.x == 0 || now.x == n+1 || now.y == 0 || now.y == m+1)) return now.step;for(int i = 0; i < 4; i++) {node tmp; tmp.step = now.step + 1;int nx = now.x + dx[i], ny = now.y + dy[i];if(now.id == 0) {if(checkF(nx,ny)) {s[nx][ny] = '#';tmp.x = nx; tmp.y = ny; tmp.id = 0;Q.push(tmp);}}else if(now.id == 1) {if(nx == 0 || nx == n+1 || ny == 0 || ny == m+1) return tmp.step;if(checkJ(nx,ny)){flag = 1; // 人还有路可走 vis[nx][ny] = 1;tmp.x = nx; tmp.y = ny; tmp.id = 1;Q.push(tmp);}}}}return 0;
}int main() {//  freopen("test.in", "r", stdin);ios::sync_with_stdio(false);int T; cin >> T;while(T--) {memset(vis, 0, sizeof(vis));while(!Q.empty()) Q.pop();int ans = 0;cin >> n >> m;for(int i = 1; i <= n; i++) {cin >> s[i]+1;for(int j = 1; j <= m; j++) {if(s[i][j] == 'J') Jx = i, Jy = j;if(s[i][j] == 'F') {node fire;fire.x = i; fire.y = j; fire.id = 0;Q.push(fire);}}}ans = bfs();if(!ans) cout << "IMPOSSIBLE" << endl;else cout << ans << endl;}return 0;
}

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