P2895 [USACO08FEB] Meteor Shower S

[USACO08FEB] Meteor Shower S - 洛谷

这个题也是暴搜

将流星的时间填入地图,比对人能到达的时间即可

#include 
#include 
#include 
using namespace std;

typedef pair PII;
const int N = 1000;
int g[N][N];
int n;
queue q;
bool return0 = false;

void bfs()
{
    q.push({0, 0});
    g[0][0] = 0;
    
    int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
    
    while(q.size())
    {
        auto t = q.front();
        q.pop();
        
        for(int i = 0; i < 4; i ++ )
        {
            int a = t.first + dx[i], b = t.second + dy[i];
            
            if(a < 0 || b < 0) continue;
            if(g[a][b] == -1)
            {
                cout << g[t.first][t.second] + 1 << endl;
                return0 = true;
                return;
            }
            else if(g[t.first][t.second] + 1 < g[a][b])
            {
                q.push({a, b});
                g[a][b] = g[t.first][t.second] + 1;
            }
        }
    }
}

int main(){
    cin >> n;
    
    for(int i = 0; i < 1000; i ++ )
    {
        memset(g[i], -1, sizeof g[i]);
    }
    
    for(int i = 0; i < n; i ++ )
    {
        int a, b, t;
        cin >> a >> b >> t;
        
        if(g[a][b] == -1) g[a][b] = t;
        else if(t < g[a][b]) g[a][b] = t;
        
        if(g[a - 1][b] == -1) g[a - 1][b] = t;
        else if(t < g[a - 1][b]) g[a - 1][b] = t;
        
        if(g[a + 1][b] == -1) g[a + 1][b] = t;
        else if(t < g[a + 1][b]) g[a + 1][b] = t;
        
        if(g[a][b - 1] == -1) g[a][b - 1] = t;
        else if(t < g[a][b - 1]) g[a][b - 1] = t;
        
        if(g[a][b + 1] == -1) g[a][b + 1] = t;
        else if(t < g[a][b + 1]) g[a][b + 1] = t;
    }
    
    bfs();
    if(return0) return 0;
    printf("-1");
    
    return 0;
}

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