POJ 3669 Meteor Shower

Meteor Shower

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16022		Accepted: 4217

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

Source

USACO 2008 February Silver

 

    用一个数组保存每个位置最早的损毁时刻。将数组每个元素的值初始化位INF。对于每一个流星，更新它下落位置、四个相邻的位置的损毁时刻：如果这个流星损毁位置的时刻小于这个位置已记录的损害时刻，就更改该位置的损毁时刻。使用宽度优先搜索，得到Bessie第一次到达一个安全位置的时间，也就是到达一个损毁时间为INF的位置所用的时间，若不能到达，则输出-1。

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
#include <string>
#include <cstring>
#include <queue>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
#define IOS ios_base::sync_with_stdio(false)
#define TIE std::cin.tie(0)
#define MIN2(a,b) (a<b?a:b)
#define MIN3(a,b) (a<b?(a<c?a:c):(b<c?b:c))
#define MAX2(a,b) (a>b?a:b)
#define MAX3(a,b,c)  (a>b?(a>c?a:c):(b>c?b:c))
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const double PI = 4.0*atan(1.0);

const int MAX_L = 305;
int n, ans, x, y, t;
int board[MAX_L][MAX_L];
bool vis[MAX_L][MAX_L];
int dx[4] = { 1, -1, 0, 0 }, dy[4] = { 0, 0, 1, -1 };
typedef struct Point{
    int x, y, t;
    Point(int x = 0, int y = 0, int t = 0) :x(x), y(y), t(t){}
}P;
bool solve()
{
    memset(vis, 0, sizeof(vis));
    queue<P> que;
    que.push(P(0, 0));
    vis[0][0] = true;
    while (que.size()){
        P p = que.front(); que.pop();
        if (board[p.x][p.y]==INF){
            ans = p.t; return true;
        }
        for (int i = 0; i < 4; i++){
            int nx = p.x + dx[i], ny = p.y + dy[i];
            if (0 <= nx && nx < MAX_L && 0 <= ny && ny<MAX_L
                &&!vis[nx][ny] && board[nx][ny]>p.t + 1){
                que.push(P(nx, ny, p.t + 1));
                vis[nx][ny] = true;
            }
        }
    }
    return false;
}
int main()
{
    while (~scanf("%d", &n)){
        memset(board, 0x3f, sizeof(board));
        for (int i = 0; i < n; i++){
            scanf("%d%d%d", &x, &y, &t);
            board[x][y] = MIN2(board[x][y],t);
            for (int i = 0; i < 4; i++){
                int nx = x + dx[i], ny = y + dy[i];
                if (0 <= nx && 0 <= ny){
                    board[nx][ny] = MIN2(board[nx][ny], t);
                }
            }
        }
        if (board[0][0] == 0){ puts("-1"); continue;}
        printf("%d\n", solve() ? ans : -1);
    }
}