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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 (Xi, Yi) (0 ≤ Xi ≤ 300; 0 ≤ Yi ≤ 300) at time Ti (0 ≤ Ti ≤ 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: Xi, Yi, and Ti
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
题意：一个2b要在下流星雨的地方玩大冒险。
数组maze记录该点不能走的时间。
注意：初始化的时候要初始化为-1，或inf
坑点写在注释了。
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 //因为这个地方wa了好几次，如果maze[0][0]=-1可以走的， 56 //这里却直接输出-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 //刚开始这次305是写成300的 87 //其实他是可以逃到301,302...的 88 //所以如果不把这些点放入队列的话， 89 //可能就找不到安全的地方了，也就错了 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 }
POJ 3669 Meteor shower 简单BFS， 有坑点
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原文地址：http://www.cnblogs.com/-maybe/p/4475076.html
