POJ 369 Meteor Shower——bfs

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

40 0 22 1 21 1 20 3 5

Sample Output

5

先把图的每个位置标记为最小击中时间，安全位置特殊标记，然后bfs搜索，每次搜索严格限制当前时间小于搜索位置的时间，搜到特殊标记的安全位置返回当前时间，否则返回-1

注意：

1.输入的时间不一定从小到大；

2.我用的1010（即时间上限）特殊标记的安全位置，发现会出错，百度后得知是填充字节的问题，所以我这里采用了二重循环来初始化。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxn = 310;const int mint = 1000 + 10;const int dx[] = {-1, 1, 0, 0};const int dy[] = {0, 0, -1, 1};int graph[maxn][maxn];//搜索图int vis[maxn][maxn];//判断是否遍历过struct Node {    int x, y, t;}node;//bfs队列所需的节点int bfs(int x, int y){    memset(vis, 0, sizeof(vis));    queue<Node> q;    node.x = x, node.y = y, node.t = 0;    q.push(node);    vis[x][y] = 1;    while ( !q.empty() ) {        node = q.front();        q.pop();        int tempx = node.x, tempy = node.y, tempt = node.t;        if (graph[tempx][tempy] == mint) return tempt;        for (int i = 0; i < 4; i++) {            node.x = tempx + dx[i], node.y = tempy + dy[i], node.t = tempt + 1;            if (node.x >= 0 && node.y >= 0 && node.t < graph[node.x][node.y] && !vis[node.x][node.y]) {                q.push(node);                vis[node.x][node.y] = 1;            }        }    }    return -1;}int main(){    int N;    while (scanf("%d", &N) == 1) {        for (int i = 0; i < maxn; i++) {            for (int j = 0; j < maxn; j++) {                graph[i][j] = mint;            }        }        int x, y, t;        while (N--) {            scanf("%d %d %d", &x, &y, &t);            graph[x][y] = min(t, graph[x][y]);            for (int i = 0; i < 4; i++) {                int tempx = x + dx[i], tempy = y + dy[i];                if (tempx >= 0 && tempy >= 0) graph[tempx][tempy] = min(t, graph[tempx][tempy]);            }        }        //现在安全区域为-1，破坏区域为最小击中时间        int ans = bfs(0, 0);        printf("%d\n", ans);    }    return 0;}