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11198 - Dancing Digits(BFS + hash判重)


题目:11198 - Dancing Digits


题目大意:每组数据给出8个数字,可能正可能负。要求最后将这8个数字按照数字绝对值从小到大的排序。排序的规则是让某个数字a邀请另一个数字b跳舞,这样a就可以插到b的左边或是右边,a能邀请b跳舞,则a* b <0 ,且a+b要是素数。题目问给出一组数据问能否通过邀请跳舞来排序,能的话就输出最少的邀请次数,否则输出-1.


解题思路:这题一开始竟然想着dfs,但是后面发现,这样的判断树可以是无限大,因为可以a邀请完b,然后b在邀请a,这样一来一回有可能保持原样。所应该用隐式图遍历,因为这里的不能让相同的状态重复执行,并且要求的是最少次数,隐式图是bfs遍历,正好是先找到终态即路径最短。这里对于一个状态需要从头开始每个数字都考虑一下,能否和其他的数字跳舞,邀请成功后就要选择站到左边还是右边。

注意:插入到某个数的左边,右边情况还需要根据跳舞的两个数的位置来定,要细心。


代码:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
using namespace std;

const int MAXN = 1000005;
const int N = 8;
int state[MAXN][N], head[MAXN], next[MAXN], dist[MAXN];

bool cmp (const int& a, const int& b) {

	if (abs(a) < abs(b))
		return true;
	return false;
}

bool is_prime (int n) {

	for (int i = 2; i <= sqrt(n); i++)
		if (n % i == 0)
			return false;
	return true;
}

int hash (int rear) {

	int sum = 0;
	for (int i = 0; i < N; i++) 
		sum = sum  * 10 + abs(state[rear][i]);
	return sum % MAXN;
}

bool trytoinsert (int rear) {

	int p = hash (rear);
	int u = head[p];
	while (u) {

		if (memcmp(state[rear], state[u], sizeof (state[u])) == 0)
			return false;
		u = next[u];
	}

	next[rear] = head[p]; 
	head[p] = rear;
	return true;
}
//插入到某个数的左边或是右边 p1要插入的数的位置, p2被插队的那个数的位置,dir = 0插入到左边,dir = 1插入到右边
void change (int p1, int p2, int front, int& rear, int dir) {

	int temp = state[front][p1];
	memcpy (state[rear], state[front], sizeof (state[front]));
	if (p1 < p2) {

		if (dir == 0) {

			for (int i = p1 + 1; i < p2; i++)
				state[rear][i - 1] = state[rear][i];
			state[rear][p2 - 1] = temp;
		} else {

			for (int i = p1 + 1;i <= p2; i++)
				state[rear][i - 1] = state[rear][i];
			state[rear][p2] = temp;

		}
	}
	if (p1 > p2) {

		if (dir == 1) {

			for (int i = p1; i > p2; i--) 
				state[rear][i] = state[rear][i - 1];
			state[rear][p2 + 1] = temp;
		} else {

			for (int i = p1; i >= p2; i--)
				state[rear][i]= state[rear][i - 1];
			state[rear][p2] = temp;
		}
	}//判断该状态是否重复,不重复就入队
	if (trytoinsert(rear)) {

		dist[rear] = dist[front] + 1; 
		rear++;
	}
}

int bfs () {

	int front, rear;
	front = 1;
	rear = 2;
	memset (dist, 0, sizeof (dist));
	memset (head, 0, sizeof(head));
	while (front < rear) {

		if (memcmp (state[0], state[front], sizeof (state[0])) == 0) 
			return dist[front];
		for (int i = 0; i < N; i++) {	
			for (int j = 0; j < N; j++) {

				if (i != j && state[front][i] * state[front][j] < 0 && is_prime (abs (state[front][i]) + abs (state[front][j]))) {

					change (i, j, front, rear, 0);
					change (i, j, front, rear, 1);
				}
			}
		}
		front++;
	}
	return -1;
}

int main () {

	int t = 0;
	while (scanf ("%d", &state[1][0]), state[1][0]) {

		for (int i = 1; i < N; i++) 
			scanf ("%d", &state[1][i]);
		memcpy(state[0], state[1], sizeof (state[1]));
		sort (state[0], state[0] + N, cmp);	
		printf ("Case %d: %d\n", ++t, bfs());
	}
	return 0;
}