Sicily 1754. 逃离洞穴

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1754. 逃离洞穴

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

04级ZEH和YYR到一个洞穴探险，其中YYR进入洞穴探险，而ZEH在洞外附近利用自发研制的X-1探测器监测洞穴的喷气口是否开始喷射毒气。洞穴里面有若干个毒气喷射口，已知毒气的扩散速度是每单位时间向上下左右扩散一格，而YYR的逃逸速度也是每单位时间向上下左右移动一格。所有的毒气喷射口都是同时喷发的，当监测到喷射时，ZEH就马上通知YYR逃离，YYR必须在不遇到毒气的情况下尽快逃离洞穴。亲爱的师弟师妹，你们可以帮他们计算一下YYR最快逃离洞穴的时间吗？

Input

洞穴地图用矩阵表示，符号. 表示通路，符号X表示障碍物，符号E表示洞口，符号D表示毒气喷射口所在地，符号P表示YYR当前的位置。输入首先是两个整数m, n(2<=m,n<=1000)表示矩阵的大小。接下来是m行n列的矩阵图。矩阵描述之外的地方都是障碍物且只有一个逃离的洞口，但可能有多个毒气喷射口，且扩散到洞口的毒气立即消散。输入有多组数据，当m,n是0 0时表示输入结束。

Output

如果YYR能够顺利逃离洞穴，则输出最快逃离洞口的时间，否则输出“YYR is extremely dangerous!”

Sample Input

5 10XXXXX..XXXXXXE...XXXXX.......XX......P.XXD.....XXX5 10XXXXX..XXXXXXE...XXXXX.......XX......P.XX.D....XXX0 0

Sample Output

6YYR is extremely dangerous!
// Problem#: 1754// Submission#: 3326115// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <algorithm>#include <iostream>#include <string>#include <stdio.h>#include <queue>#include <string.h>#include <vector>#include <iomanip>#include <map>#include <stack>#include <functional>#include <list>#include <cmath>using namespace std;const int MAX_W = 1005;bool visGas[MAX_W][MAX_W];bool visP[MAX_W][MAX_W];char G[MAX_W][MAX_W];int H, W;int Ei, Ej;queue<pair<int, int> > gas;queue<pair<int, int> > q;int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};int step;bool isValidGas(int i, int j) {    return 0 <= i && i < H && 0 <= j && j < W && !visGas[i][j] && (G[i][j] == '.' || G[i][j] == 'P');}bool isValidP(int i, int j) {    return 0 <= i && i < H && 0 <= j && j < W && !visP[i][j] && G[i][j] == '.';}void GAS() {    int s = gas.size();    while (s--) {        pair<int, int> p = gas.front();        gas.pop();        for (int i = 0; i < 4; i++) {            int nextI = p.first + dir[i][0];            int nextJ = p.second + dir[i][1];            if (isValidGas(nextI, nextJ)) {                G[nextI][nextJ] = 'D';                visGas[nextI][nextJ] = true;                gas.push(make_pair(nextI, nextJ));            }        }    }}bool BFS() {    step = 0;    while (!q.empty()) {        int s = q.size();        GAS();        step++;        while (s--) {            pair<int, int> p = q.front();            q.pop();            //if (G[p.first][p.second] == 'D') continue;            for (int i = 0; i < 4; i++) {                int nextI = p.first + dir[i][0];                int nextJ = p.second + dir[i][1];                if (nextI == Ei && nextJ == Ej) return true;                if (isValidP(nextI, nextJ)) {                    visP[nextI][nextJ] = true;                    q.push(make_pair(nextI, nextJ));                }            }        }    }    return false;}int main() {    //std::ios::sync_with_stdio(false);    while (1) {        cin >> H >> W;        if (H == 0 && W == 0) break;        for (int i = 0; i < H; i++) {            cin >> G[i];            for (int j = 0; j < W; j++) {                if (G[i][j] == '.') visP[i][j] = visGas[i][j] = false;                if (G[i][j] == 'D') {                    gas.push(make_pair(i, j));                    visGas[i][j] = true;                }                if (G[i][j] == 'P') {                    q.push(make_pair(i, j));                    visP[i][j] = true;                }                if (G[i][j] == 'E') {                    Ei = i;                    Ej = j;                }            }        }        if (BFS()) cout << step << endl;        else cout << "YYR is extremely dangerous!" << endl;        while (!gas.empty()) gas.pop();        while (!q.empty()) q.pop();    }    //getchar();    //getchar();        return 0;}

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