挑战程序设计竞赛 POJ - 3669

题目
坑点 1：流星撞击范围在0~300，人可以走到300之外，保存地图的数组要开大一点。
坑点2：人在起点也会被 T = 0 的流星砸死。

#include <iostream>
#include <cstring>
#include <queue>
#include <fstream>
using namespace std;

struct Coordinate{
    int row;
    int col;
};
Coordinate mov_dir[5] = {{0, 0}, {0, -1}, {-1, 0}, {0, 1}, {1, 0}};//五个方向，包含原点
Coordinate start = {0, 0};
int map[500][500], now_time[500][500];//分别保存地图的状态和到达改点所需的时间
queue<Coordinate> que;
int M, row, col, t；

void UpdateStrikenPoint(int row, int col, int t)//标记流星可以撞击到的位置，
												//保存该位置最早被撞击到的时间
{
    for (int i = 0; i < 5; i++) {
		int now_row = row + mov_dir[i].row;
		int now_col = col + mov_dir[i].col;
        if (now_row < 0 || now_col < 0)
            continue;
        if (map[now_row][now_col] == -1) { 
            map[now_row][now_col] = t;
        }
        else if (map[now_row][now_col] > t) {
            map[now_row][now_col] = t;
        }
    }
}
int bfs()
{
    while (que.size()) {
        Coordinate head = que.front();
		if (map[head.row][head.col] == -1) {
			return now_time[head.row][head.col];
		}
        que.pop();
        Coordinate now;
        for (int i = 0; i < 5; i++) {
            now.row = head.row + mov_dir[i].row;
            now.col = head.col + mov_dir[i].col;
			
            if (now.row < 0 || now.col < 0) {
                continue;
            }
			if (map[now.row][now.col] != -1 && (now_time[head.row][head.col] + 1 
			>= map[now.row][now.col])) {
                continue;
            }
            if (now_time[now.row][now.col] == -1){
                que.push(now);
                now_time[now.row][now.col] = now_time[head.row][head.col] + 1;
            }
        }
    }
    return -1;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

	// ifstream IN("IN.txt", ios::in);
	memset(map, -1, sizeof(map));//地图初始值为-1.
    memset(now_time, -1, sizeof(now_time));
	cin >> M;
    for (int i = 0; i < M; i++) {
        cin >> row >> col >> t;
        UpdateStrikenPoint(row, col, t);
    }
    que.push(start);
	now_time[0][0] = 0;
    int ans = bfs();
	cout << ans << endl;
}