LightOJ-1165 Digit Dancing(BFS+康托展开)

B - Digit Dancing

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1165 uDebug

Description

Digits like to dance. One day, 1, 2, 3, 4, 5, 6, 7 and 8 stand in a line to have a wonderful party. Each time, a male digit can ask a female digit to dance with him, or a female digit can ask a male digit to dance with her, as long as their sum is a prime. Before every dance, exactly one digit goes to who he/she wants to dance with - either to its immediate left or immediate right.

For simplicity, we denote a male digit x by itself x, and denote a female digit x by -x. Suppose the digits are in order {1, 2, 4, 5, 6, -7, -3, 8}. If -3 wants to dance with 4, she must go either to 4's left, resulting {1, 2, -3, 4, 5, 6, -7, 8} or his right, resulting {1, 2, 4, -3, 5, 6, -7, 8}.

Note that -3 cannot dance with 5, since their sum 3+5 = 8 is not a prime; 2 cannot dance with 5, since they're both male.

Given the initial ordering of the digits, find the minimal number of dances needed for them to sort in increasing order (ignoring signs of course).

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains exactly eight integers in a single line. The absolute values of these integers form a permutation of {1, 2, 3, 4, 5, 6, 7, 8}.

Output

For each test case, print the case number and the minimal number of dances needed. If they can never be sorted in increasing order, print -1.

Sample Input

5

1 2 4 5 6 -7 -3 8

1 2 3 4 5 6 7 8

1 2 3 5 -4 6 7 8

1 2 3 5 4 6 7 8

2 -8 -4 5 6 7 3 -1

Sample Output

Case 1: 1

Case 2: 0

Case 3: 1

Case 4: -1

Case 5: 3

题意:如果一个正数和一个负数的绝对值之和是质数,那么可以把其中一个数移到另一个数的前面或者后面,问要最少移动多少次才能把这些数的绝对值按升序排序

题解:BFS,因为只有8个数,可以先用康拓展开压缩每个状态,搜索时注意剪枝即可

#include<cstdio>#include<algorithm>#include<string.h>#include<queue>using namespace std;const int INF = 0x6f6f6f6f;int fac[]={1,1,2,6,24,120,720,5040,40320};int prime[]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1},a[10],ans;int vis[5040*100];struct node{    int x[10],step;}now,nxt;bool check(node now){    for(int i=0;i<7;i++) if(abs(now.x[i])>abs(now.x[i+1])) return false;    return true;}int Cantor(node& now){    int cnt,ret=0;    for(int i=0;i<8;i++){        cnt=0;        for(int j=i+1;j<8;j++){            if(now.x[j]<now.x[i]) cnt++;        }        ret+=cnt*fac[7-i];    }    return ret;}void solve(int flag,int i,int j){    int index=0;    if(flag){        if(i==0) nxt.x[index++]=now.x[j];        for(int k=0;k<8;k++){            if(k==j) continue;            else if(k==i-1){                nxt.x[index++]=now.x[k];                nxt.x[index++]=now.x[j];            }            else nxt.x[index++]=now.x[k];        }    }    else{        if(i==7) nxt.x[7]=now.x[j];        for(int k=0;k<8;k++){            if(k==j) continue;            else if(k==i+1){                nxt.x[index++]=now.x[j];                nxt.x[index++]=now.x[k];            }            else nxt.x[index++]=now.x[k];        }    }   // printf("%d\n",index);}void bfs(){    ans=INF;    queue<node>q;    now.step=0;    for(int i=0;i<8;i++) now.x[i]=a[i];    q.push(now);    vis[Cantor(now)]=1;    while(!q.empty()){        now=q.front();q.pop();        if(check(now)) {ans=now.step;break;}        for(int i=0;i<8;i++){            for(int j=0;j<8;j++){                if(now.x[i]*now.x[j]<0&&prime[abs(now.x[i])+abs(now.x[j])]){                    nxt.step=now.step+1;                    if(j!=i+1) {                        solve(0,i,j);                        int tmp=Cantor(nxt);                        if(!vis[tmp]){                            vis[tmp]=1;                            q.push(nxt);                        }                    }                    if(j!=i-1) {                        solve(1,i,j);                        int tmp=Cantor(nxt);                        if(!vis[tmp]){                            vis[tmp]=1;                            q.push(nxt);                        }                    }                }            }        }    }    if(ans==INF) printf("-1\n");    else printf("%d\n",ans);}int main(){    int T;  //  freopen("in.txt","r",stdin);    scanf("%d",&T);    for(int cas=1;cas<=T;cas++){        printf("Case %d: ",cas);        memset(vis,0,sizeof(vis));        for(int i=0;i<8;i++) scanf("%d",&a[i]);        bfs();    }}