POJ 3669 Meteor Shower(bfs)

题意：某人在时刻0从原点出发，在第一象限范围内移动。已知每个炸弹爆炸的地点和时刻，炸弹爆炸可毁坏该点和它上下左右的点。不能经过已毁坏的点，也不能在某点毁坏的时候位于该点。不会被炸弹毁坏的地方是安全之处，问某人到达安全之处的最短时间。

分析：bfs即可。注意：1、某点多次爆炸或受爆炸点影响而毁坏，取最早时间为该点的毁坏时间。2、bfs走过的点不能再走，注意标记。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, -1, 0, 1, -1, -1, 1, 1};
const int dc[] = {-1, 0, 1, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 300 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int vis[MAXN][MAXN];
int a[MAXN][MAXN];
bool judge(int x, int y){
    return x >= 0 && y >= 0;
}
int bfs(int sx, int sy){
    queue<int> x, y, step;
    x.push(sx);
    y.push(sy);
    step.push(0);
    a[sx][sy] = 1;
    while(!x.empty()){
        int tmpx = x.front();
        x.pop();
        int tmpy = y.front();
        y.pop();
        int tmpstep = step.front();
        step.pop();
        if(vis[tmpx][tmpy] == INT_INF) return tmpstep;
        for(int i = 0; i < 4; ++i){
            int tx = tmpx + dr[i];
            int ty = tmpy + dc[i];
            if(judge(tx, ty)){
                if(vis[tx][ty] == INT_INF) return tmpstep + 1;
                if(a[tx][ty]) continue;
                if(vis[tx][ty] > tmpstep + 1){
                    x.push(tx);
                    y.push(ty);
                    step.push(tmpstep + 1);
                    a[tx][ty] = 1;
                }
            }
        }
    }
    return -1;
}
int main(){
    int M;
    scanf("%d", &M);
    memset(vis, INT_INF, sizeof vis);
    while(M--){
        int x, y, t;
        scanf("%d%d%d", &x, &y, &t);
        vis[x][y] = Min(vis[x][y], t);
        for(int i = 0; i < 4; ++i){
            int tmpx = x + dr[i];
            int tmpy = y + dc[i];
            if(judge(tmpx, tmpy)) vis[tmpx][tmpy] = Min(vis[tmpx][tmpy], t);
        }
    }
    int ans = bfs(0, 0);
    printf("%d\n", ans);
    return 0;
}