hdu 1429——胜利大逃亡(续)

来源：互联网发布：阿里大数据应用编辑：程序博客网时间：2024/05/19 14:38
思路：状态压缩
代码如下：
#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;struct pos{        int x,y;        pos(int x=-1,int y=-1):x(x),y(y){}};struct sta{        pos s;        int k;        sta(pos _s,int _k):s(_s),k(_k){}};const int dir[4][2]={-1,0,0,1,1,0,0,-1};int n,m,t;char maze[25][25];int key[15];pos start;pos stop;int dis[25][25][1<<11];inline bool judge(pos a){        if(a.x<0)return false;        if(a.x>=n)return false;        if(a.y<0)return false;        if(a.y>=m)return false;        return true;}int bfs(){        int ans=-1;        memset(dis,-1,sizeof(dis));        dis[start.x][start.y][0]=0;        queue<sta> q;        q.push(sta(start,0));        while(!q.empty())        {                sta tmp=q.front();                q.pop();                for(int d=0;d<4;++d){                        pos to=pos(tmp.s.x+dir[d][0],tmp.s.y+dir[d][1]);                        if(!judge(to))continue;                        if(maze[to.x][to.y]=='*')continue;                        int sk=tmp.k;                        char c=maze[to.x][to.y];                        if(c>='A'&&c<='J'){                                if(key[c-'A']==-1)continue;                                if(((int)(1<<key[c-'A'])&sk)==0)continue;                                else{                                        sta p=sta(to,sk);                                        if(dis[to.x][to.y][sk]==-1||dis[to.x][to.y][sk]>dis[tmp.s.x][tmp.s.y][tmp.k]+1)                                        {                                                q.push(p);                                                dis[to.x][to.y][sk]=dis[tmp.s.x][tmp.s.y][tmp.k]+1;                                        }                                }                        }                        else if(maze[to.x][to.y]>='a'&&maze[to.x][to.y]<='j')                        {                                sk=(int)(sk|(1<<key[c-'a']));                                sta p=sta(to,sk);                                if(dis[to.x][to.y][sk]==-1||dis[to.x][to.y][sk]>dis[tmp.s.x][tmp.s.y][tmp.k]+1){                                        q.push(p);                                        dis[to.x][to.y][sk]=dis[tmp.s.x][tmp.s.y][tmp.k]+1;                                }                        }                        else if(dis[to.x][to.y][sk]==-1||dis[to.x][to.y][sk]>dis[tmp.s.x][tmp.s.y][tmp.k]+1){                                q.push(sta(to,sk));                                dis[to.x][to.y][sk]=dis[tmp.s.x][tmp.s.y][tmp.k]+1;                                if(c=='^'){                                        if(ans==-1||ans>dis[to.x][to.y][sk])ans=dis[to.x][to.y][sk];                                }                        }                }        }        return ans;}int main(){//        freopen("data.txt","r",stdin);        while(scanf("%d%d%d",&n,&m,&t)!=EOF)        {                int totkey=0;                memset(key,-1,sizeof(key));                for(int i=0;i<n;++i)                {                        scanf("%s",maze[i]);                        for(int j=0;j<m;++j){                                if(maze[i][j]>='a'&&maze[i][j]<='j')key[maze[i][j]-'a']=totkey++;                                else if(maze[i][j]=='@'){                                        start.x=i;                                        start.y=j;                                }                                else if(maze[i][j]=='^'){                                        stop.x=i;                                        stop.y=j;                                }                        }                }                int ans=bfs();                printf("%d\n",ans<t?ans:-1);        }        return 0;}
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