HDU 5040 Instrusive(BFS+优先队列)

题意比较啰嗦。

就是搜索加上一些特殊的条件，比如可以在原地不动，也就是在原地呆一秒，如果有监控也可以花3秒的时间走过去。

这种类型的题目还是比较常见的。以下代码b[i][j][x]表示格子i行j列在x时刻有监控照的到。因为只有4个方向，所以只需要时间对4取模就行。具体细节见代码。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 550;
const int dx[5] = {0, 0, 1, -1, 0}, dy[5] = {1, -1, 0, 0, 0};
char mp[maxn][maxn];
int b[maxn][maxn][4];
int T, n, kase;
bool vis[maxn][maxn][4];
struct Node {
    int x, y, step;
    bool operator < (const Node &b) const {
        return step > b.step;
    }
};
Node tar;
int bfs(int x, int y)
{
    memset(vis, false, sizeof(vis));
    priority_queue<Node> Q;
    Node cur, nex;
    cur.x = x; cur.y = y; cur.step = 0;
    vis[x][y][0] = true;
    Q.push(cur);
    while (!Q.empty())
    {
        cur = Q.top(); Q.pop();
        if (tar.x == cur.x && tar.y == cur.y) return cur.step;
        for (int i = 0; i < 5; i++)//最后一个状态是可以呆在原地不动
        {
            int nx = cur.x + dx[i];
            int ny = cur.y + dy[i];
            if (nx < 0 || ny < 0 || nx >= n || ny >= n) continue;
            if (mp[nx][ny] == '#') continue;
            nex = cur;
            if (b[nx][ny][cur.step % 4] || b[cur.x][cur.y][cur.step % 4])//如果下一个位置有监控或者当前这个位置监控的到
            {
                if (nx == cur.x && ny == cur.y && !vis[nx][ny][(cur.step + 1) % 4])//如果当前这个位置监控的到,
                {
                    nex.step++;
                    vis[nx][ny][nex.step % 4] = true;
                    Q.push(nex);
                }
                else if (!vis[nx][ny][(cur.step + 3) % 4])//如果下一个监控的到
                {
                    nex.x = nx; nex.y = ny;
                    nex.step += 3;
                    vis[nx][ny][nex.step % 4] = true;
                    Q.push(nex);
                }
            }
            else if (!vis[nx][ny][(cur.step + 1) % 4])//否则直接走
            {
                nex.x = nx; nex.y = ny; nex.step++;
                vis[nx][ny][nex.step % 4] = true;
                Q.push(nex);
            }
        }
    }
    return -1;
}
int main()
{
    scanf("%d", &T);
    while (T--)
    {
        int x, y;
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf("%s", mp[i]);
        memset(b, 0, sizeof(b));
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (mp[i][j] == 'M')
                {
                    x = i; y = j;
                }
                else if (mp[i][j] == 'N')
                {
                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;//它本身这个位置不管任意时刻都被监控到
                    if (i - 1 >= 0) b[i - 1][j][0] = 1;//表示紧挨着它上面的那个在1秒的时候被监控到
                    if (j + 1 < n) b[i][j + 1][1] = 1;//紧挨着它右边的在第2秒的时候,意思就是下一秒
                    if (i + 1 < n) b[i + 1][j][2] = 1;//下两秒
                    if (j - 1 >= 0) b[i][j - 1][3] = 1;//下三秒在忘左边的位置被监控的到
                }
                else if (mp[i][j] == 'E')
                {
                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;
                    if (i - 1 >= 0) b[i - 1][j][3] = 1;
                    if (j + 1 < n) b[i][j + 1][0] = 1;
                    if (i + 1 < n) b[i + 1][j][1] = 1;
                    if (j - 1 >= 0) b[i][j - 1][2] = 1; 
                }
                else if (mp[i][j] == 'S')
                {
                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;
                    if (i - 1 >= 0) b[i - 1][j][2] = 1;
                    if (j + 1 < n) b[i][j + 1][3] = 1;
                    if (i + 1 < n) b[i + 1][j][0] = 1;
                    if (j - 1 >= 0) b[i][j - 1][1] = 1; 
                }
                else if (mp[i][j] == 'W')
                {
                    b[i][j][0] = b[i][j][1] = b[i][j][2] = b[i][j][3] = 1;
                    if (i - 1 >= 0) b[i - 1][j][1] = 1;
                    if (j + 1 < n) b[i][j + 1][2] = 1;
                    if (i + 1 < n) b[i + 1][j][3] = 1;
                    if (j - 1 >= 0) b[i][j - 1][0] = 1; 
                }
                else if (mp[i][j] == 'T')
                {
                    tar.x = i; tar.y = j;
                }
            }
        }
        printf("Case #%d: %d\n", ++kase, bfs(x, y));
    }
    return 0;
}