poj_3669_Meteor Shower(BFS+预处理)

http://poj.org/problem?id=3669

/*思路：BFS+预处理先预处理会爆炸的区域，BFS，遇到-1则结束*/#include <iostream>#include <queue>#include <algorithm>using namespace std;int visit[1001][1001];int dx[5] = {0,0,1,0,-1};int dy[5] = {0,1,0,-1,0};typedef struct{int x,y;int step; }point;queue<point> que;int bfs(int x,int y){int cnt;if(visit[0][0] == 0)  return -1;//无法逃离 if(visit[0][0] == -1) return 0;  //安全，无需逃离 point s,cur,next;s.x = x;s.y = y;s.step = 0;que.push(s);while(!que.empty()){cur = que.front();que.pop();for(int i = 0;i < 5;i++){next.x = cur.x + dx[i];next.y = cur.y + dy[i];next.step = cur.step + 1;if(next.x >= 0 && next.y >= 0 && next.y < 1001 && next.x < 1001){if(visit[next.x][next.y] == -1){return next.step;}if(next.step < visit[next.x][next.y]){visit[next.x][next.y] = next.step;//记录最小的步数 que.push(next);}} } } return -1;}int main(){int m;    while(cin >> m){        int x,y,t,xx,yy;    memset(visit,-1,sizeof(visit));    for(int i = 0;i < m;i++)    {        cin >> x >> y >> t;                  for(int i = 0;i <= 4;i++)        {            xx = x + dx[i];            yy = y + dy[i];            if(xx >= 0 && yy >= 0 && xx < 1001 && yy < 1001)//保证在有限区域内 {            if(visit[xx][yy] == -1)             visit[xx][yy] = t;            else            visit[xx][yy] = min(visit[xx][yy],t);}        }    }    cout << bfs(0,0) << "\n"; }    return 0;}