hdu 5025 bfs+状压

http://acm.hdu.edu.cn/showproblem.php?pid=5025

N*N矩阵 M个钥匙

K起点，T终点，S点需多花费1点且只需要一次，1-9表示9把钥匙，只有当前有I号钥匙才能拿I+1号钥匙，可以不拿钥匙只从上面走过

BFS+优先队列。蛇最多只有5条，状压即可。

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;
typedef pair<int,int> pii;
const int INF = 2000000007;
map <pii,int> hash;
int n,m,sx,sy,tx,ty,cnt_snake;
char s[105][105];
int dx[] = {1,-1,0,0},
    dy[] = {0,0,1,-1};
int _key[105][105][10];
int anss;
bool in(int x,int y)
{
    return 0 <= x && x < n && 0 <= y && y < n;
}
struct node{
    int x,y,key,snak,t;
    bool operator < (const node b )const
    {
        return t > b.t;
    }
};
//class cmp
//{
//public:
//    cmp(){};
//    bool operator ()( const node a, const node b )
//    {
//        if(a.x!=b.x)
//            return a.x<b.x;
//        if(a.y!=b.y)
//            return a.y<b.y;
//        if(a.key != b.key)
//            return a.key < b.key;
//        if(a.snak != b.snak)
//            return a.snak < b.snak;
//        return false;
//    }
//};
void bfs()
{
    //map <node,int,cmp> vis;
    priority_queue<node> q;
    q.push((node){sx,sy,0,0,0});
    while(!q.empty()){
        node now = q.top(),to;
        int x = now.x,y = now.y,key = now.key,snak = now.snak,t = now.t;
        q.pop();
        //vis[now] = 0;
        if(s[x][y] == 'T'  && key == m){//cout<<snak;
            anss = min(anss,now.t);
            return ;
        }
        for(int i = 0;i < 4;++i){
            int mx = x+dx[i],my = y + dy[i];
            if(!in(mx,my) || s[mx][my] == '#')continue;
            if(s[mx][my] == 'S'){
                pii tt = make_pair(mx,my);
                if(snak & (1<<(hash[tt] - 1))){
                    to = (node){mx,my,key,snak,t+1};
                }
                else{
                    int tosnak = snak | (1<<(hash[tt] - 1));
                    to = (node){mx,my,key,tosnak,t+2};
                }
            }
            else if(s[mx][my] - '0' == key + 1){
                to = (node){mx,my,key+1,snak,t+1};
            }
            else{
                to = (node){mx,my,key,snak,t+1};
            }
            if(to.t < _key[mx][my][to.key]){
                q.push(to);
                _key[mx][my][to.key] = to.t;
            }
        }
    }
}
int main(){
    while(~scanf("%d%d",&n,&m),n|m){
        hash.clear();
        cnt_snake = 0;
        for(int i = 0;i < n;++i){
            scanf("%s",s[i]);
            for(int j = 0;j < n;++j){
                if(s[i][j] == 'K')
                    sx = i,sy = j;
                else if(s[i][j] == 'T')
                    tx = i,ty = j;
                else if(s[i][j] == 'S'){
                    pii tt = make_pair(i,j);
                    hash[tt] = ++cnt_snake;
                }
                for(int k = 0;k <= m;++k)
                    _key[i][j][k] = INF;
            }
        }
        //int mm = 1<<cnt_snake;
        //memset(dp,0x3f,sizeof(dp));
        _key[sx][sy][0] = 0;
        anss = INF;
        bfs();
        if(anss >= INF)
            puts("impossible");
        else
            printf("%d\n",anss);
     }
     return 0;
 }