http://poj.org/problem?id=3669
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.
Also check http://www.cnblogs.com/7hat/p/3595630.html
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
* Lines 2..M+1: Line i+1 contains three space-separated integers: Xi, Yi, and Ti
4 0 0 2 2 1 2 1 1 2 0 3 5http://www.cnblogs.com/7hat/p/3595630.html
分析：依旧是迷宫问题。不同的是，需要自己构建出迷宫。首先将maze的所有格初始化为INF，表示这个格子被袭击的时间为INF（即永远不会被袭击）。对于每一个流星，将其影响反映到maze上，如果破坏范围由重叠，那么格子显示的是较早的破坏时间（因为一旦破坏了就不能进入），即maze[x][y] = min(maze[x][y], T)。迷宫构建起来后，回到问题本身。求最短时间，可以用BFS做到。使用d[x]][y] 来保存移动到该格时的最小时间。而对于约束条件，就是对于下一步能否移动到该地方，要看下一个时刻该地方是否会被破坏，若不会则可以，即可d[x][y] + 1 < maze[x][y]。另外，需要特别注意的是，若有流星在0时刻袭击（0, 0）位置，则无法逃生。
#define INDEX_MAX 512int map[INDEX_MAX][INDEX_MAX];bool visited[INDEX_MAX][INDEX_MAX];struct Meteor{ int x, y, t;};typedef Meteor P;Meteor m[50008];int n;const int direction[5][2] = { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 }, { 0, 0 },};bool cmp(const Meteor& a, const Meteor& b){ return a.t < b.t;}int bfs(){ memset(visited, 0, sizeof(visited)); queue<P> que; P current; current.x = 0; current.y = 0; // 当前花费时间 current.t = 0; que.push(current); while (que.size()) { // 做个备份 const P p = que.front(); que.pop(); for (int j = 0; j < 4; ++j) { current = p; current.x = current.x + direction[j][0]; current.y = current.y + direction[j][1]; ++current.t; if (current.x >= 0 && current.y >= 0 && map[current.x][current.y] > current.t && !visited[current.x][current.y]) { visited[current.x][current.y] = true; // 爆炸时间大于当前时间，是安全的 if (map[current.x][current.y] > m[n - 1].t) { // 当前位置爆炸时间大于流星雨最晚落下的时间，说明跑出了流星雨区域 return current.t; } que.push(current); } } } return -1;}int main(int argc, char *argv[]){#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout);#endif cin >> n; for (int i = 0; i < n; ++i) { cin >> m[i].x >> m[i].y >> m[i].t; } sort(m, m + n, cmp); // 地图中每个点的值表示最早在什么时候被炸毁 memset(map, 0x7F, sizeof(map)); for (int i = 0; i < n; ++i) { for (int j = 0; j < 5; ++j) { int nx = m[i].x + direction[j][0]; int ny = m[i].y + direction[j][1]; if (nx >= 0 && ny >= 0 && map[nx][ny] > m[i].t) { map[nx][ny] = m[i].t; } } } if (map[0][0] == 0) { cout << -1 << endl; } else { cout << bfs() << endl; }#ifndef ONLINE_JUDGE fclose(stdin); fclose(stdout); system("out.txt");#endif return 0;}Also check http://www.cnblogs.com/7hat/p/3595630.html
typedef pair<int, int> P; 9 10 const int MAX_M = 50000; 11 const int MAX_N = 400 + 1; 12 const int INF = 100000000; 13 14 //输入 15 int M; 16 int X[MAX_M], Y[MAX_M], T[MAX_M]; 17 18 int maze[MAX_N][MAX_N]; //保存地图 19 int d[MAX_N][MAX_N]; //保存最短步数 20 21 //4个方向 22 const int dx[4] = {-1, 1, 0, 0}; 23 const int dy[4] = {0, 0, -1, 1}; 24 25 int bfs(){ 26 //一开始就被炸 27 if(maze[0][0] == 0) return -1; 28 29 queue<P> que; 30 que.push(P(0, 0)); 31 d[0][0] = 0; 32 while(!que.empty()){ 33 P p = que.front(); 34 que.pop(); 35 //已到达安全位置 36 int x = p.first, y = p.second; 37 if(maze[x][y] == INF) return d[x][y]; 38 //4个方向走 39 for(int i = 0; i < 4; i ++){ 40 int nx = x + dx[i], ny = y + dy[i]; 41 //判断是否可移动，是否访问过，以及下一个时刻是否安全 42 if(0 <= nx && 0 <= ny && d[nx][ny] == INF && d[x][y] + 1 < maze[nx][ny]){ 43 que.push(P(nx, ny)); 44 d[nx][ny] = d[x][y] + 1; 45 } 46 } 47 } 48 return -1; 49 } 50 51 void solve(){ 52 //初始化地图 53 for(int i = 0; i < MAX_N; i ++) 54 fill(maze[i], maze[i] + MAX_N, INF); 55 //模拟轰炸场景 56 for(int i = 0; i < M; i ++){ 57 maze[X[i]][Y[i]] = min(maze[X[i]][Y[i]], T[i]); 58 for(int j = 0; j < 4; j ++){ 59 int nx = X[i] + dx[j], ny = Y[i] + dy[j]; 60 if(0 <= nx && 0 <= ny) 61 maze[nx][ny] = min(maze[nx][ny], T[i]); 62 } 63 } 64 //初始化地图最小步数 65 for(int i = 0; i < MAX_N; i ++) 66 fill(d[i], d[i] + MAX_N, INF); 67 //宽度优先搜索 68 int ans = bfs(); 69 printf("%d\n", ans); 70 }Read full article from 3669 -- Meteor Shower
