poj 3669 BFS

思路还是不明朗，其实这题和迷宫的最短路径思路差不多，但是这题没有给出明显的边界条件，导致半天没有想明白，写的程序还是半天调试不过去，基础不扎实......keep moving .....

粗略的想法：

        预处理，把每个会被陨石雨破坏的坐标都用破坏时间表示出来，类似迷宫的最短路径的障碍，方便接下来的bfs判断，比如，在原点不需要动的情况或往回走的情况。

注意：

          在oj上  cin流输入 比scanf输入效率差几倍，用cin就会TLE。

/*
* =====================================================================================
*
*       Filename:  poj_3669.cpp
*
*    Description:  poj 3669 Meteor shower
*
*        Version:  1.0
*        Created:  03/29/2015 03:38:35 AM
*       Revision:  none
*       Compiler:  gcc
*
*		   Author:  Kart, 786900722@qq.com
*           Blog:  http://blog.csdn.net/linux_kernel_fan/ *
* =====================================================================================
*/
#include<cstdio>
#include<iostream>
#include<queue>
#include<utility>
#include<cstring>

using namespace std;
const int maxsize=405;
int map[maxsize][maxsize];
int M;
pair<int, int> dir[5] = {{0, 0}, {-1, 0}, {0, 1}, {1, 0}, {0, -1}};

typedef struct
{
int x;
int y;
int time;
}Node;

int solve()
{
if(map[0][0] == 0) return -1;
if(map[0][0] == -1) return 0;

queue<Node> que;
Node now, tmp;
tmp.x = tmp.y = tmp.time = 0;

que.push(tmp);
while(que.size())
{
now = que.front();
que.pop();
for(int i = 1; i < 5; i++)
{
tmp.x = now.x + dir[i].first;
tmp.y = now.y + dir[i].second;
tmp.time = now.time + 1;
if(tmp.x < 0 || tmp.x >= maxsize || tmp.y < 0 || tmp.y >= maxsize)
continue;
if(map[tmp.x][tmp.y] == -1) return tmp.time;
if(tmp.time >= map[tmp.x][tmp.y]) continue;
map[tmp.x][tmp.y] = tmp.time;
que.push(tmp);
}
}
return -1;
}
int main()
{
while(scanf("%d", &M) != EOF)
{
memset(map, -1, sizeof(map));
for(int i = 0; i < M; i++)
{
int x, y,t;
scanf("%d%d%d", &x, &y, &t);
for(int k = 0; k < 5; k++)
{
int nx = x + dir[k].first;
int ny = y + dir[k].second;
if(nx >= 0 && nx < maxsize && ny >= 0 && ny < maxsize)
{
if(map[nx][ny] == -1)
map[nx][ny] = t;
else
map[nx][ny] = min(t, map[nx][ny]);
}
}
}
printf("%d\n", solve());
}
return 0;
}

参考引用：http://www.cnblogs.com/justforgl/archive/2012/12/19/2825422.html