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POJ2965 The Pilots Brothers' refrigerator 「BFS+状态压缩」

2014-08-14 20:57 537 查看

题目链接：http://poj.org/problem?id=2965

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18590		Accepted: 7116		Special Judge

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i,
j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is
initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions,
you may give any one of them.

Sample Input

-+--
----
----
-+--

Sample Output

Source

Northeastern
Europe 2004, Western Subregion

#include <iostream>
#include <queue>
#include <string.h>
#include <cstdio>
using namespace std;

const int maxn = 1 << 16;
int s[4][4];
char s1[4][4];
bool vis[maxn];
int pre[maxn];
int prex[maxn];
int prey[maxn];
int rex[maxn];
int rey[maxn];
int q[maxn];
int bits[16] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768};
int flip[4][4] = {4383, 8751, 17487, 34959, 4593, 8946, 17652, 35064, 7953, 12066, 20292, 36744, 61713, 61986, 62532, 63624};

int zip(int s[4][4])
{
int sum = 0;
for (int i = 0; i < 4; ++i)
for (int j = 0; j < 4; ++j)
if (s[i][j])
sum += bits[i * 4 + j];
return sum;
}

void unzip(int num)
{
memset(s, 0, sizeof(s));
for (int i = 0; i < 16; ++i)
if (num & bits[i])
s[i / 4][i % 4] = 1;
}

int main()
{
for (int i = 0; i < 4; ++i)
scanf("%s", s1[i]);
memset(s, 0, sizeof(s));
memset(vis, false, sizeof(vis));
for (int i = 0; i < 4; ++i)
for (int j = 0; j < 4; ++j)
if (s1[i][j] == '+')
s[i][j] = 1;
int num = zip(s);
int t;
int head = 0, tail = 0;
q[head] = num;
tail++;
vis[num] = true;
while (true) {
t = q[head++];
if (t == 0)
break;
unzip(t);
for (int i = 0; i < 4; ++i)
for (int j = 0; j < 4; ++j) {
int x = t ^ flip[i][j];
if (!vis[x]) {
q[tail++] = x;
vis[x] = true;
pre[x] = t;
prex[x] = i;
prey[x] = j;
}
}
}

int k = 0;
while (t != num) {
rex[k] = prex[t];
rey[k++] = prey[t];
t = pre[t];
}
printf("%d\n", k);
for (int i = k - 1; i >= 0; --i)
printf("%d %d\n", rex[i] + 1, rey[i] + 1);

}

G++ 594ms

比较暴力一点，网上有高效的方法。

这里主要是用数组记录了位移运算的值，避免重复进行位移运算;

另外，反转是可以用异或实现的，进行异或的值也用一个数组存起来。

异或数组生成代码：

#include <iostream>
#include <string.h>
using namespace std;

int main()
{
int s[4][4];
for (int i = 0; i < 4; ++i)
for (int j = 0; j < 4; ++j) {
memset(s, 0, sizeof(s));
for (int x = 0; x < 4; ++x) {
s[x][j] = 1;
s[i][x] = 1;
}
int sum = 0;
for (int i = 0; i < 4; ++i)
for (int j = 0; j < 4; ++j)
if (s[i][j])
sum += (1 << (i * 4 + j));
cout << sum << ", ";
}

}

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标签： poj bfs 状态压缩

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