POJ 3669 Meteor shower 簡單BFS， 有坑點

　　　　　　　　　　　　　　　　Meteor Shower

 

Description

Bessie hears that an extraordinary meteor shower is coming; reports say that these meteors will crash into earth and destroy anything they hit. Anxious for her safety, she vows to find her way to a safe location (one that is never destroyed by a meteor) . She is currently grazing at the origin in the coordinate plane and wants to move to a new, safer location while avoiding being destroyed by meteors along her way.

The reports say that M meteors (1 ≤ M ≤ 50,000) will strike, with meteor i will striking point (X_i, Y_i) (0 ≤ X_i ≤ 300; 0 ≤ Y_i ≤ 300) at time T_i (0 ≤ T_i  ≤ 1,000). Each meteor destroys the point that it strikes and also the four rectilinearly adjacent lattice points.

Bessie leaves the origin at time 0 and can travel in the first quadrant and parallel to the axes at the rate of one distance unit per second to any of the (often 4) adjacent rectilinear points that are not yet destroyed by a meteor. She cannot be located on a point at any time greater than or equal to the time it is destroyed).

Determine the minimum time it takes Bessie to get to a safe place.

Input

* Line 1: A single integer: M
* Lines 2..M+1: Line i+1 contains three space-separated integers: X_i, Y_i, and T_i

Output

* Line 1: The minimum time it takes Bessie to get to a safe place or -1 if it is impossible.

Sample Input

4
0 0 2
2 1 2
1 1 2
0 3 5

Sample Output

5



題意：一個2b要在下流星雨的地方玩大冒險。

數組maze記錄該點不能走的時間。
注意：初始化的時候要初始化爲-1，或inf

坑點寫在註釋了。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<queue>
  5 
  6 using namespace std;
  7 
  8 const int maxn=310;
  9 
 10 int maze[maxn][maxn];
 11 bool vis[maxn][maxn];
 12 
 13 struct Node
 14 {
 15     int x,y,time;
 16 };
 17 
 18 int dx[4]={0,0,-1,1};
 19 int dy[4]={1,-1,0,0};
 20 
 21 void initMaze(int m)
 22 {
 23     memset(maze,-1,sizeof(maze));
 24 
 25     int u,v,w;
 26 
 27     for(int i=1;i<=m;i++)
 28     {
 29         scanf("%d %d %d",&u,&v,&w);
 30         if(maze[u][v]==-1||maze[u][v]>w)
 31         {
 32             maze[u][v]=w;
 33         }
 34         for(int j=0;j<4;j++)
 35         {
 36             int du=u+dx[j];
 37             int dv=v+dy[j];
 38             if(du<0||du>305||dv<0||dv>305)
 39                 continue;
 40             if(maze[du][dv]==-1||maze[du][dv]>w)
 41                 maze[du][dv]=w;
 42         }
 43 
 44     }
 45 }
 46 
 47 int bfs()
 48 {
 49     memset(vis,false,sizeof(vis));
 50 
 51     Node start;
 52     start.x=start.y=start.time=0;
 53     
 54     
 55     //由於這個地方wa了好幾回，若是maze[0][0]=-1能夠走的，
 56     //這裏卻直接輸出-1
 57     /*
 58     if(maze[0][0]<=1)
 59         return -1;
 60     */
 61 
 62     queue<Node>que;
 63     while(!que.empty())
 64         que.pop();
 65 
 66 
 67     que.push(start);
 68 
 69     vis[0][0]=true;
 70 
 71     while(!que.empty())
 72     {
 73         Node cur=que.front();
 74         que.pop();
 75 
 76         if(maze[cur.x][cur.y]==-1)
 77             return cur.time;
 78 
 79         for(int i=0;i<4;i++)
 80         {
 81             Node cnt;
 82             cnt.x=cur.x+dx[i];
 83             cnt.y=cur.y+dy[i];
 84             cnt.time=cur.time+1;
 85             
 86             //剛開始此次305是寫成300的
 87             //其實他是能夠逃到301,302...的
 88             //因此若是不把這些點放入隊列的話，
 89             //可能就找不到安全的地方了，也就錯了
 90             
 91             if(cnt.x<0||cnt.x>305||cnt.y<0||cnt.y>305)
 92                 continue;
 93 
 94             if(vis[cnt.x][cnt.y])
 95                 continue;
 96             if(maze[cnt.x][cnt.y]!=-1&&maze[cnt.x][cnt.y]<=cnt.time)
 97                 continue;
 98             que.push(cnt);
 99             vis[cnt.x][cnt.y]=true;
100         }
101     }
102     return -1;
103 }
104 
105 int main()
106 {
107     int m;
108     while(scanf("%d",&m)!=EOF)
109     {
110         initMaze(m);
111 
112         int ans=bfs();
113 
114         printf("%d\n",ans);
115         
116         //debug用
117         /*
118         for(int i=4;i>=0;i--)
119         {
120             for(int j=0;j<5;j++)
121                 printf("%d ",maze[i][j]);
122             printf("\n");
123         }
124         */
125     }
126 
127     return 0;
128 }

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