HDU3533:Escape(BFS)

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.

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.

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

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

9
Bad luck!


题意：
一个人从（0,0）跑到(n,m),只有k点能量，一秒消耗一点，在图中有k个炮塔，给出炮塔的射击方向c，射击间隔t，子弹速度v，坐标x,y
问这个人能不能安全到达终点
要求: 
	1.人不能到达炮塔所在的坐标
	2.炮塔会挡住子弹
	3.途中遇到子弹是安全的，但是人如果停在这个坐标，而子弹也刚好到这个坐标，人就被射死
	4.人可以选择停止不动

思路：其实不难，我们只需要看当人位于某个点的时候，其四个方向是否有炮塔，这个炮塔是都向人的方向射击，然后再看子弹是否刚好位于这个坐标即可。
而标记的话，vis[x][y][time]，对于time时刻，人位于x,y的情况只需要访问一次，这是唯一的

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
#define ads(x) (x<0?-x:x)
int n,m,k,life;
int to[5][2] = {0,1,1,0,0,-1,-1,0,0,0};//四个方向与停止不动的走法
int map[105][105];
bool vis[105][105][1005];

struct period
{
    char c;
    int t,v;
} s[105][105];

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

int check(int x,int y)
{
    if(x<0 || x>n || y<0 || y>m)
        return 1;
    return 0;
}

void bfs()
{
    node a,next;
    queue<node> Q;
    int i,j,flag,dis,tem;
    a.x = a.y = a.step = 0;
    Q.push(a);
    vis[0][0][0] = true;
    while(!Q.empty())
    {
        a = Q.front();
        Q.pop();
          if(a.step>life)
            break;
        if(a.x == n && a.y == m)
        {
            printf("%d\n",a.step);
            return ;
        }
        for(i = 0; i<5; i++)
        {
            next = a;
            next.x+=to[i][0];
            next.y+=to[i][1];
            next.step++;
            if(check(next.x,next.y)) continue;
            if(!s[next.x][next.y].t && !vis[next.x][next.y][next.step] && next.step<=life)//在符合条件的情况下，枚举四个方向
            {
                flag = 1;
                for(j = next.x-1; j>=0; j--)//当位于这点，我们往北向寻找是否有朝南方向射击的炮台
                {
                    if(s[j][next.y].t && s[j][next.y].c == 'S')//找到第一个炮台，且这个炮台是朝南射击的
                    {
                        dis = next.x-j;//看炮台与人的距离
                        if(dis%s[j][next.y].v) break;//因为不需要看子弹中途的点，子弹每一秒跑v，距离是dis，dis不能整除v的话，那么子弹是不可能停在这个点的
                        tem = next.step-dis/s[j][next.y].v;//人走的时间减去第一个子弹飞行到这个位置所需的时间
                        if(tem<0) break;//为负数就是第一个子弹都没有经过这个点，那么人绝对安全
                        if(tem%s[j][next.y].t==0)//看间隔，能整除，那么就是后续有子弹刚好到这个点，人死定了
                        {
                            flag = 0;
                            break;
                        }
                    }
                    if(s[j][next.y].t)//找到炮台但不是朝南射击，那么这个炮台会当下后面所有子弹，所以北方向安全我们不需要再找
                        break;
                }
                if(!flag)//这个方向都死定了，后面也就不需要看了
                    continue;
                //其他方向也是一样的道理，就不注释了
                for(j = next.x+1; j<=n; j++)
                {
                    if(s[j][next.y].t && s[j][next.y].c == 'N')
                    {
                        dis = j-next.x;
                        if(dis%s[j][next.y].v) break;
                        tem = next.step-dis/s[j][next.y].v;
                        if(tem<0) break;
                        if(tem%s[j][next.y].t==0)
                        {
                            flag = 0;
                            break;
                        }
                    }
                    if(s[j][next.y].t)
                        break;
                }
                if(!flag)
                    continue;
                for(j = next.y-1; j>=0; j--)
                {
                    if(s[next.x][j].t && s[next.x][j].c == 'E')
                    {
                        dis = next.y-j;
                        if(dis%s[next.x][j].v) break;
                        tem = next.step-dis/s[next.x][j].v;
                        if(tem<0) break;
                        if(tem%s[next.x][j].t==0)
                        {
                            flag = 0;
                            break;
                        }
                    }
                    if(s[next.x][j].t)
                        break;
                }
                if(!flag)
                    continue;
                for(j = next.y+1; j<=m; j++)
                {
                    if(s[next.x][j].t && s[next.x][j].c == 'W')
                    {
                        dis = j-next.y;
                        if(dis%s[next.x][j].v) break;
                        tem = next.step-dis/s[next.x][j].v;
                        if(tem<0) break;
                        if(tem%s[next.x][j].t==0)
                        {
                            flag = 0;
                            break;
                        }
                    }
                    if(s[next.x][j].t)
                        break;
                }
                if(!flag)
                    continue;
                vis[next.x][next.y][next.step] = true;
                Q.push(next);
            }
        }
    }
    printf("Bad luck!\n");
}

int main()
{
    int i,j,x,y,t,v;
    char str[3];
    while(~scanf("%d%d%d%d",&n,&m,&k,&life))
    {
        memset(s,0,sizeof(s));
        memset(vis,false,sizeof(vis));
        for(i = 0; i<k; i++)
        {
            scanf("%s%d%d%d%d",str,&t,&v,&x,&y);
            s[x][y].v = v;
            s[x][y].t = t;
            s[x][y].c = str[0];
        }
        bfs();
    }

    return 0;
}