hdu 4771——Stealing Harry Potter's Precious

来源：互联网发布：店侦探类似软件编辑：程序博客网时间：2024/05/22 04:40
题意：遍历一个点格阵，方格中有一些宝物，要拿到所有的宝物，并且使时间最少
思路：bfs,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){}        bool operator == (const pos &a) const{                return x==a.x&&y==a.y;        }};struct sta{        pos s;        int k;        sta(pos s=pos(-1,-1),int k=0):s(s),k(k){}};const int dir[4][2]={-1,0,0,1,1,0,0,-1};char maze[105][105];int n,m;pos start;int key[105][105];int dp[105][105][1<<5];bool judge(int x,int y){        if(x<0||x>=n||y<0||y>=n)return false;        return true;}void bfs(){        queue<sta> q;        memset(dp,-1,sizeof(dp));        dp[start.x][start.y][0]=0;        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.x,to.y))continue;                        if(maze[to.x][to.y]=='#')continue;                        int sk=tmp.k;                        if(maze[to.x][to.y]=='k'){                                sk|=(1<<key[to.x][to.y]);                        }                         if(dp[to.x][to.y][sk]==-1||dp[to.x][to.y][sk]>dp[tmp.s.x][tmp.s.y][tmp.k]+1){                                q.push(sta(to,sk));                                dp[to.x][to.y][sk]=dp[tmp.s.x][tmp.s.y][tmp.k]+1;                        }                }        }}int main(){//        freopen("data.txt","r",stdin);        while(scanf("%d%d",&n,&m)!=EOF)        {                if(n==0&&m==0)break;                for(int i=0;i<n;++i){                        scanf("%s",maze[i]);                        for(int j=0;j<m;++j){                                if(maze[i][j]=='@'){                                        start.x=i;                                        start.y=j;                                }                        }                }                int K;                scanf("%d",&K);                for(int i=0;i<K;++i){                        int a,b;                        scanf("%d%d",&a,&b);                        a--;                        b--;                        maze[a][b]='k';                        key[a][b]=i;                }                for(int i=0;i<n;++i){                }                bfs();                int ans=-1;                for(int i=0;i<n;++i){                        for(int j=0;j<m;++j){                                if(maze[i][j]!='k')continue;                                if(dp[i][j][(1<<K)-1]==-1)continue;                                if(ans==-1||ans>dp[i][j][(1<<K)-1])ans=dp[i][j][(1<<K)-1];                        }                }                printf("%d\n",ans);        }        return 0;}
0 0