Escape （BFS + 模拟）

Problem Description

The students of the HEU are maneuvering for their military training.
The red army and the blue army are at war today. The blue army finds that Little A is the spy of the red army, so Little A has to escape from the headquarters of the blue army to that of the red army. The battle field is a rectangle of size m*n, and the headquarters of the blue army and the red army are placed at (0, 0) and (m, n), respectively, which means that Little A will go from (0, 0) to (m, n). The picture below denotes the shape of the battle field and the notation of directions that we will use later.



The blue army is eager to revenge, so it tries its best to kill Little A during his escape. The blue army places many castles, which will shoot to a fixed direction periodically. It costs Little A one unit of energy per second, whether he moves or not. If he uses up all his energy or gets shot at sometime, then he fails. Little A can move north, south, east or west, one unit per second. Note he may stay at times in order not to be shot.
To simplify the problem, let’s assume that Little A cannot stop in the middle of a second. He will neither get shot nor block the bullet during his move, which means that a bullet can only kill Little A at positions with integer coordinates. Consider the example below. The bullet moves from (0, 3) to (0, 0) at the speed of 3 units per second, and Little A moves from (0, 0) to (0, 1) at the speed of 1 unit per second. Then Little A is not killed. But if the bullet moves 2 units per second in the above example, Little A will be killed at (0, 1).
Now, please tell Little A whether he can escape.

Input

For every test case, the first line has four integers, m, n, k and d (2<=m, n<=100, 0<=k<=100, m+ n<=d<=1000). m and n are the size of the battle ground, k is the number of castles and d is the units of energy Little A initially has. The next k lines describe the castles each. Each line contains a character c and four integers, t, v, x and y. Here c is ‘N’, ‘S’, ‘E’ or ‘W’ giving the direction to which the castle shoots, t is the period, v is the velocity of the bullets shot (i.e. units passed per second), and (x, y) is the location of the castle. Here we suppose that if a castle is shot by other castles, it will block others’ shots but will NOT be destroyed. And two bullets will pass each other without affecting their directions and velocities.
All castles begin to shoot when Little A starts to escape.
Proceed to the end of file.

Output

If Little A can escape, print the minimum time required in seconds on a single line. Otherwise print “Bad luck!” without quotes.

SampleInput

4 4 3 10
N 1 1 1 1
W 1 1 3 2
W 2 1 2 4
4 4 3 10
N 1 1 1 1
W 1 1 3 2
W 1 1 2 4

SampleOutput

9
Bad luck!

题意就是要你从（0，0）跑到（n，m），只能在d时间内跑到，并且在图中有k个炮台，炮台只能往一个方向（W，E，N，S）发送子弹，子弹发射间隔为t，速度为v。

人不能跑进炮台，炮台会挡住另一炮台的子弹，当且仅当人到达某个点且子弹也到达当点才算被射中，人可以不动。

题意清楚了就是BFS搜吧，但是同样标记数组得开一个三维数组标记坐标和时间，因为每个点可能不止走一次，所以这道题唯一的难度就是判断是否被射杀。其实不是很难，当你处于某个位置时，往四个方向找炮台判断是否会被杀死就行了。
代码：

#include <iostream>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iomanip>
#include <map>
#include <stack>
#include <deque>
#include <queue>
#include <vector>
#include <set>
#include <list>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <iterator>
#include <cmath>
#include <bitset>
#include <ctime>
#include <fstream>
#include <limits.h>
#include <numeric>

using namespace std;

#define F first
#define S second
#define mian main
#define ture true

#define MAXN 1000000+5
#define MOD 1000000007
#define PI (acos(-1.0))
#define EPS 1e-6
#define MMT(s) memset(s, 0, sizeof s)
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef stringstream sstm;
const int INF = 0x3f3f3f3f;

int m,n,k,d;
bool mp[110][110];
bool vis[110][110][1010];

int fx[5][2]={{1,0},{-1,0},{0,1},{0,-1},{0,0}};     //可以不动

struct paotai{  //记录炮台数据
    char d;
    int t,v;
}cas[110][110];

struct node{
    int x,y;
    int step;
    node(int _x,int _y,int _step):x(_x),y(_y),step(_step){}
};

bool judge(int x,int y,int time){   //往四个方向找炮台

    for(int i = x-1; i >= 0; i--){
        if(mp[i][y]){
            if(cas[i][y].d == 'S' && time >= db(x-i) / cas[i][y].v){
                double c = db(x-i) / cas[i][y].v;
                if(time == c)
                    return false;
                while(time >= c){
                    if(time == c)
                        return false;
                    c += cas[i][y].t;
                }
            }
            break;  //只要找最近的就行了
        }
    }

    for(int i = x+1; i <= m; i++){
        if(mp[i][y]){
              if(cas[i][y].d == 'N' && time >= db(i-x) / cas[i][y].v){
                double c = db(i-x) / cas[i][y].v;
                if(time == c)
                    return false;
                while(time >= c){
                    if(time == c)
                        return false;
                    c += cas[i][y].t;
                }
            }
            break;
        }
    }

    for(int i = y-1; i >= 0; i--){
        if(mp[x][i]){
            if(cas[x][i].d == 'E' && time >= db(y-i) / cas[x][i].v){
                double c = db(y-i) / cas[x][i].v;
                if(time == c)
                    return false;
                while(time >= c){
                    if(time == c)
                        return false;
                    c += cas[x][i].t;
                }
            }
            break;
        }
    }

    for(int i = y+1; i <= n; i++){
        if(mp[x][i]){
            if(cas[x][i].d == 'W' && time >= db(i-y) / cas[x][i].v){
                double c = db(i-y) / cas[x][i].v;
                if(time == c)
                    return false;
                while(time >= c){
                    if(time == c)
                        return false;
                    c += cas[x][i].t;
                }
            }
            break;
        }
    }
    return true;
}


bool check(int x,int y,int time){
    if(x < 0 || x > m || y < 0 || y > n || time > d || mp[x][y]){
        return false;
    }
    if(judge(x,y,time)){
        return true;
    }
    return false;
}

void bfs(){
    MMT(vis);

    queue<node> s;
    s.push(node(0,0,0));
    vis[0][0][0] = 1;
    while(!s.empty()){
        node cnt = s.front();
        s.pop();
        if(cnt.step > d){
            cout << "Bad luck!" << endl;
            return ;
        }
        if(cnt.x == m && cnt.y == n && cnt.step <= d){
            cout << cnt.step << endl;
            return;
        }
        for(int i = 0; i < 5; i++){
            int x = cnt.x + fx[i][0];
            int y = cnt.y + fx[i][1];
            int time = cnt.step + 1;
            if(check(x,y,time) && !vis[x][y][time]){
                vis[x][y][time] = 1;
                s.push(node(x,y,time));
            }
        }
    }
    cout << "Bad luck!" << endl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    char a;
    int b,c,x,y;
    while(cin>>m>>n>>k>>d){
        MMT(mp);
        for(int i = 0; i < k; i++){
            cin>>a>>b>>c>>x>>y;
            cas[x][y].d = a, cas[x][y].t = b,cas[x][y].v = c;
            mp[x][y] = 1;
        }
        bfs();
    }
}