《3432: Meteor Shower》

一开始以为只能在第一象限。。

这题主要就是可能0~300的点全都不是安全的，所以bfs的判断边界要到305才行。

然后，可能开始的点会被旁边的点炸到，这样开始就死了。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 2e4+5;
const int M = 1e6+5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
}
using namespace FASTIO;

int a[305][305],MI[305][305],b[4][2] = {1,0,-1,0,0,1,0,-1};
bool vis[305][305];
struct Node{int x,y,tim;};
int bfs(int x,int y)
{
    queue<Node> Q;
    Q.push(Node{x,y,0});
    vis[x][y] = 1;
    while(!Q.empty())
    {
        Node q = Q.front();
        Q.pop();
        if(a[q.x][q.y] != -1) return q.tim;
        for(int i = 0;i < 4;++i)
        {
            int px = q.x + b[i][0];
            int py = q.y + b[i][1];
            if(px >= 0 && px < 305 && py >= 0 && py < 305 && !vis[px][py] && q.tim + 1 < MI[px][py])
            {
                vis[px][py] = 1;
                Q.push(Node{px,py,q.tim + 1});
            }
        }
    }
    return -1;
}
int main()
{
    int m;
    while(~scanf("%d",&m))
    {
        memset(a,0,sizeof(a));
        memset(vis,0,sizeof(vis));
        memset(MI,0x3f,sizeof(MI));
        for(int i = 1;i <= m;++i)
        {
            int x,y,t;x = read(),y = read(),t = read();
            a[x][y] = -1;
            MI[x][y] = min(MI[x][y],t);
            for(int j = 0;j < 4;++j)
            {
                int px = x + b[j][0];
                int py = y + b[j][1];
                if(px >= 0 && py >= 0)
                {
                    a[px][py] = -1;
                    MI[px][py] = min(MI[px][py],t);
                }
            }
        }
        if(MI[0][0] <= 0) printf("-1\n");
        else printf("%d\n",bfs(0,0));
    }
    //system("pause");
    return 0;
}