UVa 11198 - Dancing Digits，Rujia Liu的神人题（二）

UVa 11198 - Dancing Digits，Rujia Liu的神题（二）

题目链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2139

类型： 隐式图搜索，  BFS， 哈希判重，模拟

原题：

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

The input consists of at most 20 test cases. Each case contains exactly 8 integers in a single line. The absolute values of these integers form a permutation of {1, 2, 3, 4, 5, 6, 7, 8}. The last case is followed by a single zero, which should not be processed.

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

1 2 4 5 6 -7 -3 81 2 3 4 5 6 7 81 2 3 5 -4 6 7 81 2 3 5 4 6 7 82 -8 -4 5 6 7 3 -10

Output for the Sample Input

Case 1: 1Case 2: 0Case 3: 1Case 4: -1Case 5: 3

题目大意:

数字喜欢跳舞。 一天，1，2，3，4，5，6，7，8站在一排准备跳舞。数字也有男女之别， 正数代表男性， 负数代表女性。每一次，男性数字可以邀请任何一个女性数字和他跳舞， 当然女性数字也可以任意一个男性数字一起跳舞。 前提是男女的数字之和（都是绝对值）是一个素数。也可以和他（她）相邻的数字一起跳舞，不管对方是男还是女。

假设初始排列为 {1, 2, 4, 5, 6, -7, -3, 8}，   如果-3想和4一起跳舞，那么-3就必须走到4的左边和右边， 如果在左边那么排列变为：{1, 2, -3, 4, 5, 6, -7, 8} ，  如果在右边那么排列变为{1, 2, 4, -3, 5, 6, -7, 8}.

问题是给一个初始排列，找到一个最小的邀请跳舞次数，使得它们的排列变成是升序的（按绝对值）。如果不可能，输出-1.

分析与总结：

这题有点恶心了，应该是我的解法比较挫，直接模拟状态转换的过程，然后得到新的排列状态。模拟过程需要注意的是，邀请的对方是在自己位置的左边还是右边。

只要细心一点基本没问题。

就是因为错了一个小小的字母而RE了好几次，都是粗心惹的祸， 郁闷。。。

/* * UVa 11198 - Dancing Digits * 隐式图搜索，BFS * Time:  0.740s (UVa) *  Author: D_Double */#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;typedef int State[8];const int HashSize = 1000003;State start;State que[50000];int head[HashSize], next[HashSize], step[50000], ans;int prime[] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1};inline void init_lookup_table(){    ans = -1;  step[0] = 0;    memset(head, 0, sizeof(head));}inline int hash(State &s){    int v=0;    for(int i=0; i<8; ++i) v = v*10 + abs(s[i]);    return (v & 0x7FFFFFFF) % HashSize;}inline bool try_to_insert(int s){    int h = hash(que[s]);    int u = head[h];    while(u){        if(memcmp(que[u], que[s], sizeof(que[s]))==0) return false;        u = next[u];    }    next[s] = head[h];    head[h] = s;    return true;}bool is_ok(State &s){    for(int i=0; i<7; ++i) if(abs(s[i]) > abs(s[i+1])) return false;    return true;}// 邀请跳舞之后的排列转化void goto_dance(State &s, int u, int v, int dir){    if(dir==1){ //左边        if(u==v-1) return;  // 如果本来就在他的左边，不移动        if(u==v+1){ // 如果再它右边，直接交换            int tmp = s[u];  s[u]=s[v]; s[v] = tmp;        }        int t = s[u];        if(u>v){            for(int i=u; i>v; --i) s[i] = s[i-1];            s[v] = t;        }        else{            for(int i=u; i<v-1; ++i) s[i] = s[i+1];            s[v-1] = t;        }     }    else{ // 右边        if(u==v+1) return ; //如果本来就在他右边， 不移动        if(u==v-1){             int tmp = s[u];  s[u]=s[v]; s[v] = tmp;        }        int t = s[u];        if(u>v){            for(int i=u; i>v+1; --i) s[i] = s[i-1];            s[v+1] = t;        }        else{            for(int i=u; i<v; ++i) s[i] = s[i+1];            s[v] = t;        }     }}void bfs(){    init_lookup_table();    int front = 0, rear = 1;    memcpy(que[0], start, sizeof(start));    try_to_insert(0);    while(front < rear){        State &s = que[front];        if(is_ok(s)){            ans = step[front]; return ;        }        for(int i=0; i<8; ++i){            for(int j=0; j<8; ++j)if(i!=j && (s[i]>0&&s[j]<0 || s[i]<0&&s[j]>0) ){                int sum = abs(s[i])+abs(s[j]);                if(!prime[sum]) continue;                // 移动到目标左边或右边：                for(int k=1; k<=2; ++k){                    State &t = que[rear];                    memcpy(t, s, sizeof(s));                    goto_dance(t, i, j, k);                    if(try_to_insert(rear)){                        step[rear] = step[front]+1;  ++rear;                    }                 }            }        }        ++front;    }}int main(){        int cas = 1;    while(scanf("%d",&start[0]), start[0]){        for(int i=1; i<8; ++i) scanf("%d", &start[i]);        bfs();        printf("Case %d: %d\n",cas++, ans);    }    return 0;}

——  生命的意义，在于赋予它意义。
               原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)