UVA-1533 Moving Pegs （BFS+位壓縮）

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Description

 

Venture MFG Company, Inc. has made a game board. This game board has 15 holes and these holes are filled with pegs except one hole. A peg can jump over one or more consecutive peg s to the nearest empty hole along the straight line. As a peg jump over the pegs you remove them from the board. In the following figure, the peg at the hole number 12 or the peg at the hole number 14 can jump to the empty hole number 5. If the peg at the hole number 12 is moved then the peg at the hole number 8 is removed. Instead, if the peg at the hole number 14 is moved then the peg at the hole number 9 is removed.

　　　　　　　　　　　　　　　　　　　　

 

Write a program which find a shortest sequence of moving pegs to leave the last peg in the hole that was initially empty. If such a sequence does not exist the program should write a message ``IMPOSSIBLE".

 

Input 

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case is a single integer which means an empty hole number.

 

Output 

For each test case, the first line of the output file contains an integer which is the number of jumps in a shortest sequence of moving pegs. In the second line of the output file, print a sequence of peg movements. A peg movement consists of a pair o f integers separated by a space. The first integer of the pair denotes the hole number of the peg that is moving, and the second integer denotes a destination (empty) hole number.

If there are multiple solutions, output the lexicographically smallest one.

 

Sample Input 

 

1                                                   5

 

Sample Output 

 

1012 5 3 8 15 12 6 13 7 9 1 7 10 8 7 9 11 14 14 5

題目大意：在中的棋盤（固定5行）上，每個棋子的走法類似於象棋中“炮”的走法，只能隔著棋子沿直線走，每走一步造成的效果是該棋子落到第一個空白處，並且沿途經過的棋子全部消失。求使最後一個棋子恰好落在第n個點上的最短、字典序最小的路徑。
題目分析：這道題說白了有15個位置，每個位置上可能有棋子也可能沒有棋子，棋子的狀況總共有2^15種。起點是(2^15)-1，終點是1<<(n-1)，BFS即可，狀態轉移也不難，但比較複雜。


代碼如下：

# include<iostream># include<cstdio># include<queue># include<map># include<cmath># include<string># include<cstring># include<algorithm>using namespace std;const int tot=(1<<15)-1;struct node{    int s,t;    string step;    node(int _s,int _t,string _step):s(_s),t(_t),step(_step){}    bool operator < (const node &a) const {        if(t==a.t)            return step>a.step;        return t>a.t;    }};int mark[1<<15];map<int,char>mp;int d[6][2]={{-1,-1},{-1,0},{0,-1},{0,1},{1,0},{1,1}};int get_pos(int x,int y){    return x*(x-1)/2+y;}void get_XY(int n,int &x,int &y){    x=1;    for(int i=1;i<=5&&n-i>0;++i)        ++x,n-=i;    y=n;}bool ok(int x,int y){    if(x>=1&&x<=5&&y>=1&&y<=x)        return true;    return false;}void print(string p){    for(int i=0;i<p.size();++i)        printf("%d%c",p[i]-‘A‘+1,(i==p.size()-1)?‘\n‘:‘ ‘);}void bfs(int goal){    priority_queue<node>q;    memset(mark,0,sizeof(mark));    mark[tot^goal]=1;    q.push(node(tot^goal,0,""));    while(!q.empty())    {        node u=q.top();        q.pop();        //cout<<u.t<<‘ ‘<<u.s<<‘ ‘<<u.step<<endl;        if(u.s==goal){            printf("%d\n",u.t);            print(u.step);            return ;        }        int x,y;        for(int i=1;i<=15;++i){            if(u.s&(1<<(i-1))){                get_XY(i,x,y);                for(int j=0;j<6;++j){                    int nx=x+d[j][0],ny=y+d[j][1];                    if(!ok(nx,ny))                        continue;                    int pos=get_pos(nx,ny);                    if(!(u.s&(1<<(pos-1))))                        continue;                    int s=u.s^(1<<(i-1));                    while(u.s&(1<<(pos-1)))                    {                        s^=(1<<(pos-1));                        nx=nx+d[j][0],ny=ny+d[j][1];                        if(!ok(nx,ny))                            break;                        pos=get_pos(nx,ny);                    }                    s^=(1<<(pos-1));                    string step=u.step+mp[i];                    step+=mp[get_pos(nx,ny)];                    if(ok(nx,ny)&&!mark[s]){                        mark[s]=1;                        q.push(node(s,u.t+1,step));                    }                }            }        }    }    printf("IMPOSSIBLE\n");}int main(){    for(int i=1;i<=15;++i)        mp[i]=i+‘A‘-1;    int T,n;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        bfs(1<<(n-1));    }    return 0;}

　　

UVA-1533 Moving Pegs （BFS+位壓縮）