SOJ 1048.Inverso

解题技巧：

　　1.将3*3矩阵的状态压缩成一个整数来表示，使用二进制位运算来进行状态转移。

　　2.使用广度优先搜索求最短路径，从1-9种状态转移进行尝试，保证第一次得到的答案是字典序最小的。

　　3.如果题目给出的状态为目标状态，也就是wwwwwwwww，则应该输出11（表示先按1位置一次，再按1位置一次）。

代码如下：

#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <vector>
using namespace std;

// 一个代表区域状态的整数的二进制表示中，0表示白色，1表示黑色.

const int STATES = 1 << 9;
bool visited[STATES];

struct Elem {
    int state;
    int previous;
    int op_num;
    Elem(int s = 0, int p = 0, int on = 0) : state(s), previous(p), op_num(on) {}
};

int goal = 0;   // 目标状态，1-9全为白色.

vector<Elem> elems;
// Clicking field 1 inverts fields 1, 2, 4, 5
// Clicking field 2 inverts fields 1, 2, 3, 4, 5, 6
// Clicking field 3 inverts fields 2, 3, 5, 6
// Clicking 4 inverts fields 1, 2, 4, 5, 7, 8
// Clicking 5 inverts fields all fields
// Clicking 6 inverts fields 2, 3, 5, 6, 8, 9
// Clicking 7 inverts fields 4, 5, 7, 8
// Clicking 8 inverts fields 4, 5, 6, 7, 8, 9
// Clicking 9 inverts fields 5, 6, 8, 9
int click[10] = { 0, 27, 63, 54, 219, 511, 438, 216, 504, 432};

void init() {
    elems.clear();
    for (int i = 0; i < STATES; ++i) visited[i] = false;
}

int main() {
    int n; cin >> n;
    while (n--) {
        string s; cin >> s; init();
        elems.push_back(Elem()); Elem &elem = elems.back();
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == 'b') {
                elem.state += 1 << i;
            }
        }
        elem.previous = elem.op_num = -1;
        if (elem.state != goal) visited[elem.state] = true;
        int left = 0, right = 1;
        bool find = false;
        while (left < right && !find) {
            for (int i = 1; i <= 9 && !find; ++i) {
                Elem &e = elems[left];
                Elem next_e;
                next_e.state = e.state ^ click[i];
                next_e.previous = left;
                next_e.op_num = i;
                if (!visited[next_e.state]) {
                    visited[next_e.state] = true;
                    elems.push_back(next_e); ++right;
                }
                if (next_e.state == goal) {
                    find = true;
                }
            }
            ++left;
        }
        stack<int> st;
        int cur = right - 1;
        while (cur != 0) {
            st.push(elems[cur].op_num);
            cur = elems[cur].previous;
        }
        while (!st.empty()) {
            cout << st.top(); st.pop();
        }
        cout << endl;
    }

    return 0;
}