HDU 4634 Swipe Bo （BFS+状压）

最新推荐文章于 2025-05-11 11:13:04 发布

紫杉丶

最新推荐文章于 2025-05-11 11:13:04 发布

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分类专栏： DFS/BFS HDU

本文链接：https://blog.youkuaiyun.com/secretbase_/article/details/47378487

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本文介绍了如何解决HDU 4634题目‘Swipe Bo’，重点探讨了使用宽度优先搜索（BFS）结合状态压缩（位运算）的解题策略。

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//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<set>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
const int M = 60005;
const int INF = 1e10;
//

struct node
{
    int x, y;
    int step;
    int key;

} st, nx;

int dir[4][2] = { 1,0,-1,0,0,1,0,-1 };
int n, m;
char mp[205][205];
bool vis[205][205][1 << 7];     
int ky[205][205];
int ans, cnt, cur, sx, sy;
bool bol[205][205][1 << 7][4];  //标记状态

int bfs()
{
    queue<node>q;
    memset(vis, 0, sizeof(vis));
    st.x = sx;
    st.y = sy;
    st.step = 0;
    st.key = 0;
    cur = 0;
    cnt = 0;
    vis[st.x][st.y][st.key] = 1;
    q.push(st);
    memset(bol, 0, sizeof(bol));
    while (!q.empty())
    {
        nx = q.front();
        q.pop();
        for (int i = 0; i < 4; ++i)
        {
            int xx = dir[i][0];
            int yy = dir[i][1];
            int s = nx.step;
            int k = nx.key;
            int x = nx.x;
            int y = nx.y;
            while (1)
            {
                if (mp[x][y] == 'L')
                {
                    xx = 0;
                    yy = -1;
                }
                if (mp[x][y] == 'U')
                {
                    xx = -1;
                    yy = 0;
                }
                if (mp[x][y] == 'D')
                {
                    xx = 1;
                    yy = 0;
                }
                if (mp[x][y] == 'R')
                {
                    xx = 0;
                    yy = 1;
                }
//                if(mp[nx.x][nx.y]=='K')
//                {
//                    k|=ky[nx.x][nx.y];
//                }
//                cout<<"dg"<<endl;
                if (xx == -1 && yy == 0)
                {
                    cnt = 0;
                }
                else if (xx == 1 && yy == 0)
                {
                    cnt = 1;
                }
                else if (xx == 0 && yy == 1)
                {
                    cnt = 2;
                }
                else
                {
                    cnt = 3;
                }
                if (bol[x][y][k][cnt])
                {
                    break;
                }
                bol[x][y][k][cnt] = 1;
                x += xx;
                y += yy;
                if (mp[x][y] == 'E' && k == (1 << ans) - 1)
                {
                    //puts("asd");
                    cur = nx.step + 1;
                    return 1;
                }
                if (x < 0 || x >= n || y < 0 || y >= m || mp[x][y] == '#')
                    break;
                if (mp[x][y] == 'L')
                {
                    xx = 0;
                    yy = -1;
                }
                if (mp[x][y] == 'U')
                {
                    xx = -1;
                    yy = 0;
                }
                if (mp[x][y] == 'D')
                {
                    xx = 1;
                    yy = 0;
                }
                if (mp[x][y] == 'R')
                {
                    xx = 0;
                    yy = 1;
                }
                if (mp[x][y] == 'K')
                {
                    k |= ky[x][y];      //拾取钥匙
                }
                if (mp[x + xx][y + yy] == '#' && x + xx >= 0 && x + xx < n && y + yy >= 0 && y + yy < m)
                {
                    if (vis[x][y][k])
                        break;
                    vis[x][y][k] = 1;
                    node nxx;
                    nxx.x = x;
                    nxx.y = y;
                    nxx.key = k;
                    nxx.step = nx.step + 1;
                    q.push(nxx);
                    break;
                }


            }



        }



    }
    return 0;
}


int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        ans = 0;
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < m; ++j)
            {
                cin >> mp[i][j];
                if (mp[i][j] == 'S')
                {
                    sx = i;
                    sy = j;

                }
                if (mp[i][j] == 'K')
                {
                    ky[i][j] = (1 << ans);  //储存钥匙
                    ans++;

                }
            }

        }
        if (bfs())
        {
            printf("%d\n", cur);
        }
        else
        {
            puts("-1");
        }


    }

    return 0;

}