【POJ 3669】Meteor Shower

Meteor Shower

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11661		Accepted: 3182

Description

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 (X_i, Y_i) (0 ≤ X_i ≤ 300; 0 ≤ Y_i ≤ 300) at time T_i (0 ≤ T_i  ≤ 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: X_i, Y_i, and T_i

Output

* Line 1: The minimum time it takes Bessie to get to a safe place or -1 if it is impossible.

Sample Input

4
0 0 2
2 1 2
1 1 2
0 3 5

Sample Output

5

Source

USACO 2008 February Silver

 

裸广搜、无优化。

真的没什么说。

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

struct Vertex
{
    int x;
    int y;
    
    Vertex() {}
    
    Vertex(int x0, int y0) : x(x0), y(y0) {}
};

const int LOCATION = 500;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

int g[LOCATION][LOCATION];
Vertex Q[LOCATION * LOCATION];
int dis[LOCATION][LOCATION];
int h;
int t;
int m;

bool test(int x1, int y1)
{
    return x1 >= 0 && x1 < LOCATION && y1 >= 0 && y1 < LOCATION;
}

void bfs()
{
    memset(dis, 0, sizeof(dis));
    h = t = 0;
    Q[t++] = Vertex(0, 0);
    while (h < t)
    {
        Vertex a = Q[h++];
        for (int i = 0; i < 4; ++i)
        {
            int x0 = a.x + dx[i];
            int y0 = a.y + dy[i];
            if (test(x0, y0) && dis[x0][y0] == 0 && dis[a.x][a.y] + 1 < g[x0][y0])
            {
                dis[x0][y0] = dis[a.x][a.y] + 1;
                Q[t++] = Vertex(x0, y0);
                if (g[x0][y0] > 1000)
                {
                    printf("%d
", dis[x0][y0]);
                    return;
                }
            }
        }
    }
    printf("-1
");
}

int main()
{
    scanf("%d", &m);
    memset(g, 0x7f, sizeof(g));
    for (int i = 0; i < m; ++i)
    {
        Vertex A;
        int time;
        scanf("%d%d%d", &A.x, &A.y, &time);
        g[A.x][A.y] = min(g[A.x][A.y], time);
        for (int j = 0; j < 4; ++j)
            if (test(A.x + dx[j], A.y + dy[j]))
                g[A.x + dx[j]][A.y + dy[j]] = min(g[A.x + dx[j]][A.y + dy[j]], time);
    }
    if (g[0][0] > 1000) printf("0
");
    else bfs();
    return 0;
}