One day Leon finds a very classic game called Wumpus.The game is as follow.
Once an agent fell into a cave. The legend said that in this cave lived a kind of monster called Wumpus, and there were horrible pits which could lead to death everywhere. However, there were also a huge amount of gold in the cave. The agent must be careful and sensitive so that he could grab all of the gold and climb out of the cave safely.

The cave can be regarded as a n*n board. In each square there could be a Wumpus, a pit, a brick of gold, or nothing. The agent would be at position (0,0) at first and headed right.(As the picture below)

If there is a pit at (0, 0), the agent dies immediately. There will not be a Wumpus at (0, 0).

Input

There are multiple cases. The first line will contain one integer k that indicates the number of cases.

For each case:
The first line will contain one integer n (n <= 20).
The following lines will contain three integers, each line shows a position of an object. The first one indicates the type of the object. 1 for Wumpus, 2 for pit and 3 for gold. Then the next two integers show the x and y coordinates of the object.
The input end with -1 -1 -1. (It is guaranteed that no two things appear in one position.)

Output

The output contains one line with one integer, which is the highest point Leon could possibly get. If he cannot finish the game with a non-negative score, print "-1".

Sample Input

2
3
1 1 1
2 2 0
3 2 2
-1 -1 -1
3
1 1 1
3 2 2
-1 -1 -1

Sample Output

850
870

Hint

For the sample 1, the following steps are taken:
turn left, forward, forward, turn right, forward, forward, grab, turn left, turn left, forward, forward, turn left, forward, forward, climb.
There are in all 15 steps, so the final score is 840. For the sample 2 , the path is as follow:

大致題意：人在迷宮中，有一個寶藏，須要找到這個寶藏然後離開出口，求最短花費的時間，每秒能夠做這幾件事：

拿寶藏，離開出口，向左轉。向右轉，向前走

大致思路：

BFS，每一個格子相應了四個狀態，方向狀態，然後就是bfs辣。

acm半年了。又寫了一次基礎的搜索，莫名感概

//Accepted	3890	C++	0	280
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
typedef pair<ll,ll> pii;

const int N = 25;
int mp[N][N];
int n;

bool vis[N][N][8];
struct node{
        int x,y,dir,t;
        node(int x = 0,int y = 0,int dir = 0,int t= 0):x(x),y(y),dir(dir),t(t){}
};
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
queue<node>que;
int bfs(){
        que.push(node(1,1,0,0));
        vis[1][1][0] = 1;
        while(!que.empty()){
                node cur = que.front();
                que.pop();
                int d = 0;
                if( mp[cur.x][cur.y] == 3 && cur.dir < 4){
                        if(vis[cur.x][cur.y][cur.dir+4] == 0){
                                que.push(node(cur.x,cur.y,cur.dir+4,cur.t+1));
                                vis[cur.x][cur.y][cur.dir+4] = 1;
                        }
                        continue;
                }
                if( cur.x == 1 && cur.y == 1 && cur.dir > 3) return 1000-(cur.t+1)*10;
                if( cur.x < 1 || cur.y < 1 || cur.x > n || cur.y > n) continue;
                if( mp[cur.x][cur.y] == 1 || mp[cur.x][cur.y] == 2) continue;
                if(cur.dir > 3){
                        d += 4;
                        cur.dir -= 4;
                }
                if(vis[cur.x][cur.y][(cur.dir+1+4)%4+d] == 0){
                        que.push(node(cur.x,cur.y,(cur.dir+1+4)%4+d,cur.t+1));
                        vis[cur.x][cur.y][(cur.dir+1+4)%4+d] = 1;
                }
                if(vis[cur.x][cur.y][(cur.dir-1+4)%4+d] == 0){
                        que.push(node(cur.x,cur.y,(cur.dir-1+4)%4+d,cur.t+1));
                        vis[cur.x][cur.y][(cur.dir-1+4)%4+d] = 1;
                }
                if(vis[cur.x+dx[cur.dir]][cur.y+dy[cur.dir]][cur.dir+d] == 0){
                        que.push(node(cur.x+dx[cur.dir],cur.y+dy[cur.dir],cur.dir+d,cur.t+1));
                        vis[cur.x+dx[cur.dir]][cur.y+dy[cur.dir]][cur.dir+d] = 1;
                }
        }
        return -1;
}
void ini(){
        memset(vis,0,sizeof(vis));
        memset(mp,0,sizeof(mp));
        while(!que.empty()) que.pop();
}
int main(){

        int T;
        cin>>T;
        while(T--){
                scanf("%d",&n);
                ini();
                while(true){
                        int a,x,y;
                        scanf("%d%d%d",&a,&x,&y);
                        if( a == -1 && x == -1 && y == -1) break;
                        mp[++x][++y] = a;
                }
                if(mp[1][1] == 2) {
                        puts("-1");
                        continue;
                }
                int ans = bfs();
                if(ans < 0) puts("-1");
                else printf("%d\n",ans);
        }
}

ZOJ - 3890 Wumpus（BFS基礎題）