HDU 5025 Saving Tang Monk(状态转移， 广搜)

Posted 2021-01-08 jadon97

#include<bits/stdc++.h>
using namespace std;
const int maxN = 107;
const int inf = 1e9 + 7;
char G[maxN][maxN], snake[maxN][maxN];
int times[maxN][maxN][15];
int n, m, sx, sy, ex, ey, ans;
int dir[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};

struct node
{
    int x, y, t, key, killed; //坐标， 步数， 已经拿到的钥匙， 有没有杀蛇
    bool operator < (const node& p) const
    {
        return t > p.t;
    }
    node(int x, int y, int t, int key, int killed):x(x),y(y), t(t), key(key), killed(killed) {};
};
void bfs()
{
    node t = (node)
    {
        sx, sy, 0, 0, 0
    };
    times[sx][sy][0] = 0;
    queue<node> q;
    q.push(t);
    while(!q.empty())
    {
        node u = q.front();
        q.pop();
        int x = u.x, y = u.y, key = u.key, t = u.t, killed = u.killed;
        if(x == ex && y == ey && key == m)
        {
            ans = min(ans, t);
            continue;
        }
        for(int d = 0; d < 4; d++)
        {
            int tx = x + dir[d][0];
            int ty = y + dir[d][1];
            if(tx < 0 || tx >= n || ty < 0 || ty >= n || G[tx][ty] == ‘#‘) //越界
                continue;


            if(G[tx][ty] >= ‘1‘ && G[tx][ty] <= ‘9‘)   //如果是钥匙
            {
                int num = G[tx][ty] - ‘0‘;
                if(key + 1 == num && (times[tx][ty][key + 1] > t + 1))   //可以拿
                {
                    q.push(node(tx,ty,t+1,key+1,killed));
                    times[tx][ty][key + 1] = t + 1;
                }
                else if(((key + 1) != num) && (times[tx][ty][key] > t + 1)) //不可以拿
                {
                    q.push(node(tx,ty,t+1,key,killed));
                    times[tx][ty][key] = t + 1;
                }
            }
            else if(G[tx][ty] >= ‘A‘ && G[tx][ty] <= ‘E‘) //蛇
            {
                int cnt = G[tx][ty] - ‘A‘; //压位判断

                if((killed & (1<<cnt)) == 0)   //未杀
                {
                    if(times[tx][ty][key] > t + 2)
                    {
                        q.push(node(tx,ty,t+2,key,(killed | (1<<cnt))));
                        times[tx][ty][key] = t + 2;
                    }
                }
                else     //已经杀了
                {
                    if(times[tx][ty][key] > t + 1)
                    {
                        q.push(node(tx,ty,t+1,key,killed));
                        times[tx][ty][key] = t + 1;
                    }
                }
            }
            else if((G[tx][ty] == ‘.‘ || G[tx][ty] == ‘T‘ || G[tx][ty] == ‘K‘)  && times[tx][ty][key] > t + 1)
            {
                q.push(node(tx,ty,t+1,key,killed));
                times[tx][ty][key] = t + 1;
            }
        }
    }
}
int main()
{
//    freopen("1.txt","r", stdin);
    while(~scanf("%d %d", &n, &m))
    {
        int cnt = 0;
        if(n == 0)
            break;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                for(int k = 0; k < 10; k++)
                    times[i][j][k] = inf;
        for(int i = 0; i < n; i++)
        {
            scanf("%s", G[i]);
        }
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(G[i][j] == ‘K‘)
                    sx = i, sy = j;
                if(G[i][j] == ‘T‘)
                    ex = i, ey = j;
                if(G[i][j] == ‘S‘)
                {
                    G[i][j] = (cnt++ + ‘A‘);
                }
            }
        }

        ans = inf;
        bfs();
        if(ans == inf)
        {
            printf("impossible
");
        }
        else
        {
            printf("%d
", ans);
        }
    }
    return 0;
}