20-2-4-状圧枚举-POJ1753

POJ1753

Flip Game

Description

Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it’s black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules:

1. Choose any one of the 16 pieces.
2. Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).

[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-Gjtek2MZ-1580839282810)(D:AFilesTyporaPictures1753_1-1580838869620.jpg)]Consider the following position as an example:

bwbw
wwww
bbwb
bwwb
Here “b” denotes pieces lying their black side up and “w” denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:

bwbw
bwww
wwwb
wwwb
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal.

Input

The input consists of 4 lines with 4 characters “w” or “b” each that denote game field position.

Output

Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it’s impossible to achieve the goal, then write the word “Impossible” (without quotes).

Sample Input

bwwb
bbwb
bwwb
bwww

Sample Output

4

Source

Northeastern Europe 2000

• 状圧枚举

• 每个地方的字符有改和不改的两种情况, 故一共2**16种状态, 每个地方的字符改一次和奇数次的效果相同, 不改和偶数次的效果相同.

• 一共16个位置, 2**16种状态, 考虑暴力枚举, 状态压缩

• 暴力枚举:

  • 按照翻一个棋子, 翻两个棋子, 翻三个棋子的顺序向下枚举, 这样可不重不漏的枚举出所有的状况.

• 状态压缩很简单, 只有把16个位置的棋子的状态按照二进制放入一个int中就好

#pragma warning(disable:4996)
#include <cstring>
#include<iostream>

#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 10;
const int MAXN = 100000;

char g[maxn][maxn];

int path[17];
int cntt;

int cnt;
//当前的值cur, 需要翻转的个数times, 当前光标所在的位置pos, 当前已翻转的个数cur_times
bool traverse(int cur, int times, int pos, int cur_times) {
	if (cur_times == times) {
		if (cur == 0 || cur == 0xffff) {
			return true;
		}

		return false;
	}
	if (pos >= 16) {
		return false;
	}
	int curr;
	for (int i = pos; i < 16; i++) {
		curr = cur ^ (1 << i);
		path[i] = 1;
		if (i - 4 >= 0) {
			curr ^= (1 << (i - 4));
		}
		if (i + 4 < 16) {
			curr ^= (1 << (i + 4));
		}
		if (i >= 1 && i <= 3 || i >= 5 && i <= 7 || i >= 9 && i <= 11 || i >= 13 && i <= 15) {
			curr ^= (1 << (i - 1));
		}
		if (i >= 0 && i <= 2 || i >= 4 && i <= 6 || i >= 8 && i <= 10 || i >= 12 && i <= 14) {
			curr ^= (1 << (i + 1));
		}
		if (traverse(curr, times, i + 1, cur_times + 1)) {
			return true;
		}
		path[i] = 0;
	}
	return false;
}

//白色用1
int solve() {
	int cur = g[0][0] == 'w' ? 1 : 0;
	for (int i = 0; i < 4; i++) {
		for (int j = 0; j < 4; j++) {
			if (i + j == 0) {
				continue;
			}
			//左移一位
			cur <<= 1;
			if (g[i][j] == 'w') {
				cur += 1;
			}
		}
	}
	if (cur == 0 || cur == 0xffff) {
		return 0;
	}
	//
	int last = cur;
	for (int i = 1; i <= 16; i++) {
		if (traverse(last, i, 0, 0)) {
			return i;
		}
	}
	return -1;
}
int main() {
	for (int i = 0; i < 4; i++) {
		for (int j = 0; j < 4; j++) {
			g[i][j] = getchar();
		}
		getchar();
	}
	int ans = solve();
	if (ans == -1) {
		cout << "Impossiblen";
	}
	else {
		cout << ans << "n";
	}
	return 0;
}

最后

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