Instrusive
Time Limit: 3000/1500 MS (Java/Others)
Memory Limit: 262144/262144 K (Java/Others)
Problem Description
The legendary mercenary Solid Matt gets a classic mission: infiltrate a military base.
The military base can be seen as an N * N grid. Matt’s target is in one of the grids and Matt is now in another grid.
In normal case, Matt can move from a grid to one of the four neighbor grids in a second. But this mission is not easy.
Around the military base there are fences, Matt can’t get out of the base.
There are some grids filled with obstacles and Matt can’t move into these grids.
There are also some surveillance cameras in the grids. Every camera is facing one of the four direction at first, but for every second, they will rotate 90 degree clockwisely. Every camera’s sight range is 2, which means that if Matt is in the same grid as the camera, or in the grid that the camera is facing, he will be seen immediately and the mission will fail.
Matt has a special equipment to sneak: a cardbox. Matt can hide himself in the card box and move without being noticed. But In this situation, Matt will have to use 3 seconds to move 1 grid. Matt can also just hide in the cardbox without moving. The time to hide and the time to get out of the cardbox can be ignored.
Matt can’t take the risk of being noticed, so he can’t move without cardbox into a grid which is now insight of cameras or from a grid which is now insight of cameras. What’s more, Matt may be in the cardbox at the beginning.
As a live legend, Matt wants to complete the mission in the shortest time.
Input
The first line of the input contains an integer T, denoting the number of testcases. Then T test cases follow.
For each test cases, the first line contains one integer:N(1<=N<=500)
In the following N lines, each line contains N characters, indicating the grids.
There will be the following characters:
● ‘.’ for empty
● ‘#’ for obstacle
● ‘N’ for camera facing north
● ‘W’ for camera facing west
● ‘S’ for camera facing south
● ‘E’ for camera facing east
● ‘T’ for target
● ‘M’ for Matt
Output
For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the answer.
If Matt cannot complete the mission, output ‘-1’.
Sample Input
2
3
M..
.N.
..T
3
M..
###
..T
Sample Output
Case #1: 5
Case #2: -1
解题心得：
- 题意就是一个骑士去救公主，但是迷宫中有监控，每个监控可以监控自己的位置和自己面对的前面一个位置，监控每1秒钟顺时针旋转90度，骑士有一个神奇的盒子，骑士在盒子下可以不被监控看到，在盒子下面可以选择不走也可以顶着盒子走路，每三秒走一格。问骑士最少要花多少时间才可以救到公主。
- 其实看似很多个状态，但是地图每四秒形成一个循环，所以在记录监控的时候就很简单了，每四个一个循环，而骑士也有三种选择，顶着盒子不走（在任意一个点都可以选择不走），自身位置没有被监控看到，要走的位置也没被监控看到，就直接花费一秒走过去，或者要被监控看到，可以选择顶着盒子每三秒走一格。
- 其实在标记已经走过的状态的时候用一个三维的数组就行了，前两维记录位置，第三维记录时间，时间每四秒循环一次，所以在记录时间也很简单。因为每一秒走的速度是不一样的（可以顶着盒子），所以选择优先队列，每次取出的时间都是花费的最少的时间，到达目标之后就可以直接跳出。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 510;
char maps[maxn][maxn];
int n,dir[4][2] = {-1,0,0,1,1,0,0,-1};
bool vis[maxn][maxn][10],oversee[maxn][maxn][10];
struct Tar
{
int x,y;
int dir;
} tar;
struct NODE
{
int x,y;
int step;
} now,Next;
priority_queue <NODE> qu;
vector <Tar> ve;
bool operator < (const NODE& a,const NODE& b)
{
return a.step > b.step;
}
void deal(int x,int y)
{
char ch = maps[x][y];
int i = x,j = y;
for(int i=0;i<4;i++)
oversee[x][y][i] = 1;
//写的很弱智
if(ch=='N')
{
oversee[i-1][j][0]=1;
oversee[i][j+1][1]=1;
oversee[i+1][j][2]=1;
oversee[i][j-1][3]=1;
}
if(ch=='W')
{
oversee[i][j-1][0]=1;
oversee[i-1][j][1]=1;
oversee[i][j+1][2]=1;
oversee[i+1][j][3]=1;
}
if(ch=='S')
{
oversee[i+1][j][0]=1;
oversee[i][j-1][1]=1;
oversee[i-1][j][2]=1;
oversee[i][j+1][3]=1;
}
if(ch=='E')
{
oversee[i][j+1][0]=1;
oversee[i+1][j][1]=1;
oversee[i][j-1][2]=1;
oversee[i-1][j][3]=1;
}
}
void deal_oversee()//处理监控在各个时间的状态
{
for(int i=0; i<ve.size(); i++)
{
tar = ve[i];
int x = tar.x;
int y = tar.y;
deal(x,y);
}
}
bool check(int x,int y,int step)
{
if(vis[x][y][step] || x<1 || y<1 || x>n || y>n || maps[x][y] == '#')
return false;
return true;
}
void check_maps()
{
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
tar.x = i;
tar.y = j;
if(maps[i][j] == 'M')
{
now.x = i;
now.y = j;
now.step = 0;
}
//把有监控的位置给记录下来
if(maps[i][j] == 'N')
{
tar.dir = 0;
ve.push_back(tar);
}
if(maps[i][j] == 'E')
{
tar.dir = 1;
ve.push_back(tar);
}
if(maps[i][j] == 'S')
{
tar.dir = 2;
ve.push_back(tar);
}
if(maps[i][j] == 'W')
{
tar.dir = 3;
ve.push_back(tar);
}
}
}
void init()
{
while(!qu.empty())//队列要记得清空
qu.pop();
scanf("%d",&n);
memset(vis,0,sizeof(vis));
memset(oversee,0,sizeof(oversee));
ve.clear();
for(int i=1; i<=n; i++)
scanf("%s",maps[i]+1);
check_maps();
deal_oversee();
}
int bfs()
{
qu.push(now);
vis[now.x][now.y][now.step] = true;
while(!qu.empty())
{
now = qu.top();
qu.pop();
if(maps[now.x][now.y] == 'T')
return now.step;
Next = now;
Next.step++;
if(!vis[Next.x][Next.y][Next.step%4])//骑士选择不走
{
qu.push(Next);
vis[Next.x][Next.y][Next.step%4] = true;
}
for(int i=0; i<4; i++)
{
int step = now.step + 1;
int x = now.x + dir[i][0];
int y = now.y + dir[i][1];
if(!oversee[x][y][now.step%4] && !oversee[now.x][now.y][now.step%4])//当前位置和将要求的位置都没被监控看到
{
if(check(x,y,step%4))
{
Next.x = x;
Next.y = y;
Next.step = step;
vis[x][y][step%4] = true;
qu.push(Next);
}
}
else
{
Next.x = x;
Next.y = y;
Next.step = now.step + 3;//骑士选择顶着盒子走
if(check(x,y,Next.step%4))
{
vis[x][y][Next.step%4] = true;
qu.push(Next);
}
}
}
}
return -1;
}
int main()
{
int t,T = 1;
scanf("%d",&t);
while(t--)
{
init();
int ans = bfs();
printf("Case #%d: ",T++);
printf("%d
",ans);
}
return 0;
}