广搜1—— Meteor Shower（流星雨）

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：给你一个数字n，然后n行，每行有三个数（x, y, t）表示在t时刻流星雨落在（x， y）坐标，并且把四个方向全炸毁，让你求能逃到一个安全的地方用到的最短时间。
思路：先初始化，将（x， y）及其四个方向的点初始化成时间t，然后再搜索，从（0， 0）开始，能逃到的地方可以在300以外，这是一个坑。
初始化时候要的是时间小的。
给一下代码：

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <queue>#include <algorithm>using namespace std;int vis[400][400], vim[400][400];int dis[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};struct xx{    int x, y;}vv;int check(int x, int y){    if((x >= 0 && x <= 500) && (y >= 0 && y <= 500)) return 1;    else return 0;}void bfs(){    int ans = 0, fl = 0;    queue <xx> qq;    vv.x = vv.y = 0;    vim[0][0] = 0;    qq.push(vv);    while(!qq.empty())    {        vv = qq.front();        qq.pop();        int x1 = vv.x;        int y1 = vv.y;        if(vis[x1][y1] == -1)        {            ans = vim[x1][y1];            fl = 2;            printf("%d\n", ans);            break;        }        for(int i = 0; i < 4; i++)        {            int x0 = x1 + dis[i][0];            int y0 = y1 + dis[i][1];            if(check(x0, y0) && !vim[x0][y0] && (vim[x1][y1] + 1 < vis[x0][y0] || vis[x0][y0] == -1))            {                vim[x0][y0] += vim[x1][y1] + 1;                vv.x = x0;                vv.y = y0;                qq.push(vv);            }        }    }    if(!fl) puts("-1");}int main(){    int n, a, b, time;    while(~scanf("%d", &n))    {        memset(vis, -1, sizeof(vis));        memset(vim, 0, sizeof(vim));        for(int i = 0; i < n; i++)        {            scanf("%d %d %d", &a, &b, &time);            if(vis[a][b] == -1) vis[a][b] = time;            else vis[a][b] = min(vis[a][b], time);            for(int j = 0; j < 4; j++)            {                int x0 = a + dis[j][0];                int y0 = b + dis[j][1];                if(check(x0, y0))                {                    if(vis[x0][y0] == -1) vis[x0][y0] = time;                    else vis[x0][y0] = min(vis[x0][y0], time);                }            }        }        bfs();    }    return 0;}