USACO算法系列十二——msquare

    题目：http://www.nocow.cn/index.php/Translate:USACO/msquare

    看到这道题我还以为有什么比较捷径的策略，毕竟广度优先遍历，是很耗内存空间的，而且需要去重的话，不断遍历数组也是很费时的。但是看了参考答案以后，好像也没什么特殊的解法。所以就开始写了。

    跟预期的一样，如果一遍遍的检索的话，在Test7的时候，会超时，但是如果用一个2^24次方的数组记录是否是否检索过，又很快超出了空间。思索了很久，终于想到了知道前7个数时最后个数也已经被确定，只需要将用来表示状态的二进制右移3位，达到2^21，空间就没问题了。

    广度优先遍历，代码思路很清晰的。

#include <iostream>#include <fstream>#define LENGTH 40320using namespace std;bool contain[2097152] = {false};int state[LENGTH];char step[LENGTH];int parent[LENGTH];int length = 0;int end[8];int T = 0;int S = 0;ifstream fin("msquare.in");ofstream fout("msquare.out");//是否存在bool isContain(int s){for (int i=0; i < length; i ++){if (state[i] == s){return true;}}return false;}//a变换int atransform(int source){int target = 0;for (int i= 0; i < 8; i ++){target = target << 3;target = target | (source & 7);source = source >> 3;}return target;}//B变换int btransform(int source){int temp[8];for (int i=7; i >=0; i --){temp[i] = source & 7;source = source >> 3;}int target = 0;target = target | temp[3];target = target << 3;target = target | temp[0];target = target << 3;target = target | temp[1];target = target << 3;target = target | temp[2];target = target << 3;target = target | temp[5];target = target << 3;target = target | temp[6];target = target << 3;target = target | temp[7];target = target << 3;target = target | temp[4];return target;}//C 变换int ctransform(int source){int temp[8];for (int i=7; i >=0; i --){temp[i] = source & 7;source = source >> 3;}int target = 0;target = target | temp[0];target = target << 3;target = target | temp[6];target = target << 3;target = target | temp[1];target = target << 3;target = target | temp[3];target = target << 3;target = target | temp[4];target = target << 3;target = target | temp[2];target = target << 3;target = target | temp[5];target = target << 3;target = target | temp[7];return target;}int main(){//输入数据for (int i=0; i < 8; i ++){fin >> end[i];}//转化为最终状态for (int i=0; i < 8; i ++){T = (T << 3);T = (T | (end[i]-1));S = (S << 3);S = (S | i);}state[0] = S;length ++;//广度遍历for (int i =0; i < length;i++){int temp = state[i];if (temp == T){break;}//end if temp == Telse{// A 变换int a = atransform(temp);if (!contain[(a>>3)]){state[length] = a;parent[length] = i;step[length] = 'A';contain[(a>>3)] = true;length ++;}if (a == T){break;}//B变换int b = btransform(temp);if (!contain[(b>>3)]){state[length] = b;parent[length] = i;step[length] = 'B';contain[(b>>3)] = true;length ++;}if (b == T){break;}//C变换int c = ctransform(temp);if (!contain[(c>>3)]){state[length] = c;parent[length] = i;step[length] = 'C';contain[(c>>3)] = true;length ++;}if (c == T){break;}}//end else}//打印结果char result[24];int rl = 0;for (int i=length-1; i >= 1; i = parent[i]){result[rl] = step[i];rl ++;}fout << rl << endl;for (int i=rl-1; i >=0; i --){fout << result[i] ;}fout << endl;return 0;}

    顺利通过了USACO测试，结果如下：

Executing... Test 1: TEST OK [0.022 secs, 5412 KB] Test 2: TEST OK [0.000 secs, 5412 KB] Test 3: TEST OK [0.011 secs, 5412 KB] Test 4: TEST OK [0.022 secs, 5412 KB] Test 5: TEST OK [0.000 secs, 5412 KB] Test 6: TEST OK [0.022 secs, 5412 KB] Test 7: TEST OK [0.011 secs, 5412 KB] Test 8: TEST OK [0.022 secs, 5412 KB]All tests OK.