孤岛营救问题(BFS+状压DP)

孤岛营救问题

https://www.luogu.org/problemnew/show/P4011

 

用状压DP标记拿到钥匙的数量

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
using namespace std;

int n,m,p;
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int door[15][15][15][15];
int map[15][15];
int book[15][15][1<<12];
struct sair{
    int x,y,step,key;
};

void bfs(){
    sair s,e;
    s.x=1,s.y=1,s.key=map[1][1],s.step=0;
    queue<sair>Q;
    book[1][1][0]=1;
    Q.push(s);
    while(!Q.empty()){
        s=Q.front();
        Q.pop();
        for(int i=0;i<4;i++){
            e.x=s.x+dir[i][0];
            e.y=s.y+dir[i][1];
            if(e.x>=1&&e.x<=n&&e.y>=1&&e.y<=m&&(door[s.x][s.y][e.x][e.y]!=0)){
                if(e.x==n&&e.y==m){
                    cout<<s.step+1<<endl;
                    return;
                }
                if((door[s.x][s.y][e.x][e.y]>0)&&((s.key&door[s.x][s.y][e.x][e.y])>0)){
                    int tmp=s.key;
                    if(!book[e.x][e.y][tmp]){
                        book[e.x][e.y][tmp]=1;
                        e.step=s.step+1;
                        e.key=s.key;
                        if(map[e.x][e.y]){
                            e.key|=map[e.x][e.y];
                        }
                        Q.push(e);
                    }
                }
                else if(door[s.x][s.y][e.x][e.y]==-1){
                    int tmp=s.key;
                    if(!book[e.x][e.y][tmp]){
                        book[e.x][e.y][tmp]=1;
                        e.step=s.step+1;
                        e.key=s.key;
                        if(map[e.x][e.y]){
                            e.key|=map[e.x][e.y];
                        }
                        Q.push(e);
                    }
                }
            }
        }
    }
    cout<<-1<<endl;
    return;
}

int main(){
    cin>>n>>m>>p;
    int k;
    cin>>k;
    int x1,x2,y1,y2,G;
    memset(door,-1,sizeof(door));
    for(int i=1;i<=k;i++){
        cin>>x1>>y1>>x2>>y2>>G;
        if(!G){
            door[x1][y1][x2][y2]=0;
            door[x2][y2][x1][y1]=0;
        }
        else{
            door[x1][y1][x2][y2]=(1<<G);
            door[x2][y2][x1][y1]=(1<<G);
        }
    }
    cin>>k;
    for(int i=1;i<=k;i++){
        cin>>x1>>y1>>G;
        map[x1][y1]|=(1<<G);
    }
    bfs();
}