hdu 4845 : 拯救大兵瑞恩 (bfs+状态压缩)

题目链接

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

int n,m,p,s,k;
int dx[]={,,,-};
int dy[]={,-,,};

bool vis[][][<<];    //vis[i][j][k]记录是否以k状态访问过(i,j)位置
int wall[][][][];    //记录墙和门的情况
int key[][];            //记录迷宫中钥匙的存在状态 

struct node
{
    int x,y;        //x,y记录坐标
    int own;        //own记录钥匙的拥有情况
    int dis;        //dis记录到原点的最短距离
};

bool in(int x,int y)
{
    return <=x&&x<=n&&<=y&&y<=m;
}

int bfs()
{
    vis[][][key[][]]=;
    queue<node> q;
    q.push((node){,,key[][],});
    while(!q.empty())
    {
        node cur=q.front();
        q.pop();
        int px=cur.x,py=cur.y,pdis=cur.dis;
        for(int i=;i<;i++)
        {
            int own=cur.own;
            int x=px+dx[i];
            int y=py+dy[i];
            if(!in(x,y)||wall[px][py][x][y]==) continue;    //不在边界内或是一堵墙
            int door= (wall[px][py][x][y]==-? :wall[px][py][x][y]);
            if(!(own&(<<door)))    continue;    //没有钥匙
            if(x==n&&y==m) return pdis+;        //到达终点
            own|=key[x][y];        //捡到该点的钥匙
            if(vis[x][y][own])    continue;
            vis[x][y][own]=;
            q.push((node){x,y,own,pdis+});
        }
    }
    return -;
}

int main()
{
    while(cin>>n>>m>>p)
    {
        memset(wall,-,sizeof(wall));
        memset(key,,sizeof(key));
        memset(vis,,sizeof(vis));
        cin>>k;
        while(k--)
        {
            int x1,y1,x2,y2,g;
            cin>>x1>>y1>>x2>>y2>>g;
            wall[x1][y1][x2][y2]=wall[x2][y2][x1][y1]=g;
        }
        cin>>s;
        while(s--)
        {
            int x,y,q;
            cin>>x>>y>>q;
            key[x][y]|=<<q;
        }
        key[][]|=;    //捡到原点的钥匙
        cout<<bfs()<<endl;
    }
}