HDU1429 胜利大逃亡 状压bfs

http://acm.hdu.edu.cn/viewcode.php?rid=22225154

因为总共a-j有10种钥匙，所以可以把有没有钥匙的状态压到一个int数里，然后dfs。

昨天状态特别不好写超时了好几次，但是这个题很简单的，算是水题。

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int maxn=200000;
const double eps=1e-8;
const int modn=998244353;
int vis[21][21][1<<11]={};
char a[21][21]={};
int id1[4]={0,0,1,-1};int id2[4]={1,-1,0,0};
int m,n,t;
struct nod{
    int x,y,k,t;
};
queue<nod>q;
int bfs(int xi,int yi){
    while(!q.empty()){
        q.pop();
    }
    nod c; int i,j,ff,ff1;
    c.x=xi,c.y=yi,c.k=0,c.t=0;
    vis[xi][yi][0]=1;
    q.push(c);
    while(!q.empty()){
        c=q.front();
        q.pop();
        nod b;
        if(c.t+1>=t){
            return -1;
        }
        for(int zz=0;zz<4;zz++){
            i=c.x+id1[zz]; j=c.y+id2[zz];
            if(i<0||i>=n||j<0||j>=m||a[i][j]=='*'){
                continue;
            }b.k=c.k;
            if(a[i][j]=='^'){
                return c.t+1;
            }
            if(a[i][j]>='A'&&a[i][j]<='Z'){
                ff1=a[i][j]-'A';
                ff=c.k&(1<<ff1);
                if(!ff)continue;
                else if(!vis[i][j][b.k]) { vis[i][j][b.k]=1; b.x=i;b.y=j;b.t=c.t+1; q.push(b);}
            }
            if(a[i][j]>='a'&&a[i][j]<='z'){
                ff=a[i][j]-'a';
                b.k=c.k|(1<<ff);
                if(!vis[i][j][b.k]) { vis[i][j][b.k]=1; b.x=i; b.y=j; b.t=c.t+1; q.push(b);}
            }
            if(a[i][j]=='.'&&!vis[i][j][b.k]){
                vis[i][j][b.k]=1;
                b.x=i;b.y=j;b.t=c.t+1;
                q.push(b);
            }
        }
    }
    return -1;
}
int main(){
    while(~scanf("%d%d%d",&n,&m,&t)){
        int x,y;memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++){
            scanf("%s",&a[i]);
            for(int j=0;j<m;j++){
                if(a[i][j]=='@'){
                    x=i;y=j;
                    a[i][j]='.';
                }
            }
        }
        printf("%d
",bfs(x,y));
    }
}