poj 3669搜索

Meteor Shower

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6111		Accepted: 1803

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

这是一道非常坑的搜索提，首先题目给出的数据范围是陨石的范围，我看成了地图，开小了，一开始wa了一次，后来第二次出错是忘记了，在第一次一出来的时候，人就有可能被砸死，后来在wiking大神的提醒下，终于发现了。下面是代码：

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>

using namespace std;

const int maxn=405;
const int inf=9999999;
int dy[]={0,1,-1,0,0};//方向数组得开5个，因为还有保持原地这个状态
int dx[]={0,0,0,1,-1};
int map[maxn][maxn];

struct node{
       int x,y;
       int time;
};

int bfs(){
    if(map[0][0]==0) return -1;
    if(map[0][0]==-1) return 0;//如果一开始未被击中
    node tmp,now;
    tmp.x=tmp.y=tmp.time=0;
    queue<node>q;
    q.push(tmp);
    while(!q.empty()){
         now=q.front();
         q.pop();
         for(int i=1;i<5;i++){//记住是从1开始的，因为要向4个方向搜索
            tmp.x=now.x+dx[i];
            tmp.y=now.y+dy[i];
            tmp.time=now.time+1;
            if(tmp.x<0||tmp.x>=maxn||tmp.y<0||tmp.y>=maxn)
                continue;
            if(map[tmp.x][tmp.y]==-1)
                return tmp.time;
            if(tmp.time>=map[tmp.x][tmp.y])//人到达的时间不能被此地被击中的时间大
                continue;
            map[tmp.x][tmp.y]=tmp.time;
            q.push(tmp);
         }
    }
    return -1;
}

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        int x,y,t;
        memset(map,-1,sizeof(map));
        while(n--){
            cin>>x>>y>>t;
            for(int k=0;k<5;k++){
                int tx=x+dx[k];
                int ty=y+dy[k];
                if(tx<0||tx>=maxn||ty<0||ty>=maxn)
                   continue;
                if(map[tx][ty]==-1)
                     map[tx][ty]=t;
                else
                    map[tx][ty]=min(map[tx][ty],t);
            }
        }
        printf("%d\n",bfs());
    }
    return 0;
}