hdu4771 Stealing Harry Potter's Precious （状压+bfs）

Problem Description

Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon’s home. But he can’t bring his precious with him. As you know, uncle Vernon never allows such magic things in his house. So Harry has to deposit his precious in the Gringotts Wizarding Bank which is owned by some goblins. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate. The coordinates of the upper-left room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)… A 3×4 bank grid is shown below:

　　Some rooms are indestructible and some rooms are vulnerable. Goblins always care more about their own safety than their customers’ properties, so they live in the indestructible rooms and put customers’ properties in vulnerable rooms. Harry Potter’s precious are also put in some vulnerable rooms. Dudely wants to steal Harry’s things this holiday. He gets the most advanced drilling machine from his father, uncle Vernon, and drills into the bank. But he can only pass though the vulnerable rooms. He can’t access the indestructible rooms. He starts from a certain vulnerable room, and then moves in four directions: north, east, south and west. Dudely knows where Harry’s precious are. He wants to collect all Harry’s precious by as less steps as possible. Moving from one room to another adjacent room is called a ‘step’. Dudely doesn’t want to get out of the bank before he collects all Harry’s things. Dudely is stupid.He pay you $1,000,000 to figure out at least how many steps he must take to get all Harry’s precious.

Input

There are several test cases.
　　In each test cases:
　　The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 100).
　　Then a N×M matrix follows. Each element is a letter standing for a room. ‘#’ means a indestructible room, ‘.’ means a vulnerable room, and the only ‘@’ means the vulnerable room from which Dudely starts to move.
　　The next line is an integer K ( 0 < K <= 4), indicating there are K Harry Potter’s precious in the bank.
　　In next K lines, each line describes the position of a Harry Potter’s precious by two integers X and Y, meaning that there is a precious in room (X,Y).
　　The input ends with N = 0 and M = 0

Output

For each test case, print the minimum number of steps Dudely must take. If Dudely can’t get all Harry’s things, print -1.

Sample Input

2 3
##@
#.#
1
2 2
4 4
#@##
…

…
2
2 1
2 4
0 0

Sample Output

-1
5

思路：

bfs或者dfs用二维数组标记的话，走过的路径就不能再走回去了，
而题目显然是找到宝藏之后还要找另一个宝藏，会重复走之前走过的路径

开三维数组标记，第三维记录当前找到宝藏的状态(状态压缩)，这样就能重复走了

这题也可以bfs几次求出任意两个宝藏之间的最短路，然后全排列枚举取宝藏的顺序取答案最小值

code：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#include<queue>
const int maxm=105;
char g[maxm][maxm];
int mark[maxm][maxm][maxm];
int n,m,k;
struct Node{int x,y,s;
}st,K[maxm];
queue<Node>q;
int dir[4][2]={0,1,0,-1,1,0,-1,0};
int bfs(){//    for(int i=0;i<k;i++){//实测起点不判也能a
//        if(st.x==K[i].x&&st.y==K[i].y){//            st.s|=(1<<i);
//        }
//    }while(!q.empty())q.pop();q.push(st);memset(mark,-1,sizeof mark);mark[st.x][st.y][st.s]=0;while(!q.empty()){Node x=q.front();q.pop();for(int i=0;i<4;i++){int xx=x.x+dir[i][0];int yy=x.y+dir[i][1];int ss=x.s;if(xx<0||xx>=n||yy<0||yy>=m)continue;//越界if(g[xx][yy]==-1)continue;//墙for(int j=0;j<k;j++){if(xx==K[j].x&&yy==K[j].y){ss|=(1<<j);}}if(mark[xx][yy][ss]!=-1)continue;//如果来过了mark[xx][yy][ss]=mark[x.x][x.y][x.s]+1;if(ss==(1<<k)-1){//如果全部收集到了return mark[xx][yy][ss];}q.push({xx,yy,ss});}}return -1;
}
signed main(){while(scanf("%d%d",&n,&m)!=EOF){if(n==0&&m==0)break;for(int i=0;i<n;i++){scanf("%s",g[i]);for(int j=0;j<m;j++){if(g[i][j]=='@'){st.x=i,st.y=j;st.s=0;}else if(g[i][j]=='#'){g[i][j]=-1;}}}scanf("%d",&k);for(int i=0;i<k;i++){scanf("%d%d",&K[i].x,&K[i].y);K[i].x--;K[i].y--;}printf("%d\n",bfs());}return 0;
}