Poj-3669-Meteor shower-BFS

Think of the ground as the grid in the First quadrant. There is a meteor shower and m meteor. If a meteor falls into a certain grid on the ground, it will destroy the grid and the four grids above and below. These places cannot go any further. Now, Bessie starts from [0, 0], returns the coordinates (whole point) and time of each meteor falling to the ground, and asks Bessie how much time it will take to reach a safe point forever, if output-1 is not displayed.

Idea: Read point and time, pre-process on the [301,301] mesh, store the time at which the point was first destroyed, and then compare it during BFs. It cannot reach the mark as int_max.

AC code:

 1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <vector> 5 #include <queue> 6 #include <cstring> 7 #include <climits> 8 using namespace std; 9 10 int m;11 int const maxn = 302;12 int const mde = 50004;13 struct node14 {15     int x, y, t;16     bool operator < (const node that) const {17         return t < that.t;18     }19     node() {x = y = t = 0;}20 };21 int mp[maxn][maxn];22 int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};23 bool border(int x, int y)24 {25     if(x < 0 || y < 0 || y > 301 || x > 301) return false;26     return true;27 }28 void init()29 {30     for(int i = 0; i < maxn; i++) {31         for(int j = 0; j < maxn; j++) mp[i][j] = INT_MAX;32     }33     for(int i = 0; i < m; i++) {34         int a, b, c; scanf("%d%d%d", &a, &b, &c);35         if(c < mp[a][b]) mp[a][b] = c;36         for(int j = 0; j < 4; j++) {37             int dx = a + dir[j][0], dy = b + dir[j][1];38             if(border(dx, dy) && c < mp[dx][dy]) mp[dx][dy] = c;39         }40     }41 }42 void bfs(int &ans)43 {44     queue<node> q; node rec;45     q.push(rec);46     if(mp[0][0] == 0) return;47     bool inq[maxn][maxn]; memset(inq, 0, sizeof(inq)); inq[0][0] = 1;48     while(!q.empty()) {49         rec = q.front(); q.pop();50         if(mp[rec.x][rec.y] == INT_MAX) {ans = rec.t; return; }51         for(int i = 0; i < 4; i++) {52             node dn; dn.x = rec.x + dir[i][0]; dn.y = rec.y + dir[i][1]; dn.t = rec.t + 1;53             if(border(dn.x, dn.y) && mp[dn.x][dn.y] > dn.t && !inq[dn.x][dn.y]) {54                 q.push(dn);55                 inq[dn.x][dn.y] = 1;56             }57         }58     }59 }60 void work()61 {62     int ans = -1;63     init();64     bfs(ans);65     printf("%d\n", ans);66 }67 int main()68 {69     while(scanf("%d", &m) != EOF) work();70     return 0;71 }

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Poj-3669-Meteor shower-BFS