钥匙是必须有序，蛇是不要求有序的。所以一个需要状压一个不用

因为时间计算和步数计算不同。所以要遍历整个空间，或者使用优先队列。优先时间短的。

风格就这样了.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#define inf 0x3f3f3f3f
#define maxn 110
using namespace std;
int N,M,T;
int D[maxn][maxn][10][32];
char G[maxn][maxn];
int dr[]={-1,0,1,0},dc[]={0,1,0,-1};
int sr[5],sc[5],scnt;
struct P
{
    int r,c,s,k;
}s,e,use,in;
int bfs()
{
    queue<struct P> q;
    q.push(s);int res=inf;
    while(q.size()){
        use=q.front();q.pop();
        if(use.k==M && use.r==e.r && use.c==e.c)
            res=min(res,D[use.r][use.c][use.k][use.s]);
        for(int i=0;i<4;i++){
            int t=1;
            in=use;
            in.r=use.r+dr[i];
            in.c=use.c+dc[i];
            if(D[in.r][in.c][in.k][in.s]!=inf || G[in.r][in.c]=='#' || in.r<1 || in.r>N || in.c<1 || in.c>N) continue;
            if(G[in.r][in.c]=='S'){
                for(int j=0;j<scnt;j++)
                    if(in.r==sr[j] && in.c==sc[j]){
                        if( (in.s & (1<<j)) ==0)
                            t++,in.s|=1<<j;
                        break;
                    }
            }
            else if(G[in.r][in.c]<='9' && G[in.r][in.c]>='1' && G[in.r][in.c]-'0'==in.k+1)
                in.k++;
            D[in.r][in.c][in.k][in.s]=D[use.r][use.c][use.k][use.s]+t;
            q.push(in);
        }
    }
    return res==inf?-1:res;
}
int main()
{
    while(~scanf("%d%d",&N,&M) && (N+M)){
        scnt=0;
        for(int i=1;i<=N;i++){
            getchar();
            for(int j=1;j<=N;j++){
                scanf("%c",G[i]+j);
                if(G[i][j]=='K')
                    s.r=i,s.c=j,s.k=0,s.s=0;
                else if(G[i][j]=='T')
                    e.r=i,e.c=j,e.k=0,e.s=0;
                else if(G[i][j]=='S')
                    sr[scnt]=i,sc[scnt++]=j;
            }
        }
        memset(D,0x3f,sizeof(D));
        D[s.r][s.c][0][0]=0;
        int ans=bfs();
        if(ans==-1) cout<<"impossible"<<endl;
        else        cout<<ans<<endl;
    }
	return 0;
}

HDU 5025 Saving Tang Monk(状压搜索)