POJ2965解题报告（BFS，位压缩）

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因为和之前的翻转问题十分类似，思想就不列举了，贴上代码，供以后温故：

#include <iostream>#include<fstream>#include<cstdlib>#include<cstring>#include<cstdio>#include<queue>#include<stack>using namespace std;const int MAX = 16;int final_step = MAX;const int MAX_STATE = 65536;int flips[4][4] = {4383, 8751, 17487, 34959, 4593, 8946, 17652, 35064, 7953, 12066, 20292, 36744, 61713, 61986, 62532, 63624};  struct POINT{int x;int y;};POINT pts[MAX_STATE];int states[MAX_STATE];int steps[MAX_STATE];int pres[MAX_STATE];int qu[MAX_STATE];int init(char str[]){int current_state = 0;for(int i=0;i<MAX;i++){if(str[i] == '+'){current_state += (1 << i);}}return current_state;}int flip(int state,int x,int y){state ^= flips[x][y];return state;}void ans(int current_state){memset(steps,0,sizeof(steps));memset(states,0,sizeof(states));memset(pres,0,sizeof(pres));states[current_state] = 1;steps[current_state] = 0;pres[current_state] = 0;pts[current_state].x = -1;pts[current_state].y = -1;int front =0;int tail = 0;tail ++;qu[front] = current_state;int final_step = 0;while(true){int cur = qu[front++];if(cur == 0){break;}for(int i=0;i<4;i++){for(int j=0;j<4;j++){int next  = flip(cur,i,j);if(!states[next]){states[next] = 1;steps[next] = steps[cur] + 1;pres[next] = cur;pts[next].x = i;pts[next].y = j;qu[tail++] = next;}}}}cout << steps[0] <<endl;stack<int> st;int tmp = 0;while(pres[tmp] != current_state){st.push(tmp);tmp = pres[tmp];}st.push(tmp);while(st.size()>0){cout << pts[st.top()].x + 1 <<' ' << pts[st.top()].y + 1 <<endl;st.pop();}}int main(int argc, char** argv) {//ifstream cin("L:/test.txt");char str[MAX];int number = 0;char c;while(cin >> c){if(c==' ' || c == '\n') continue; str[number++] = c; if(number == 16) break;}int current_state = init(str);ans(current_state);//system("pause");return 0;}

主要的点在于翻转控制，毕竟BFS太简单了，而且用queue的时候会有问题，自己定义队列最好！

对于位压缩的这些常量值，整理了一下思路，写出了下面的代码，可以对照，在表中出现的值，和相应的执行代码：

/*** 位分布图（每一位对应于原题中该位置）* 0  1  2  3* 4  5  6  7* 8  9  10 11* 12 13 14 15*/#include<iostream>#include<cstdlib>#include<vector>using namespace std;const int row[] = {0xF,0XF0,0XF00,0XF000};void binary(int num){vector<int> v;int yu = num % 2;while(num / 2){yu = num % 2;v.push_back(yu);num = num / 2;}v.push_back(1);while(v.size() <16){v.push_back(0);}for(int i=1;i<=16;i++){cout << v[i-1];if(i % 4 == 0)cout <<endl;}cout <<endl;}int main(){int state = 0;int count = -1;for(int i=0;i<4;i++){for(int j=0;j<4;j++){count ++;state = 0;state ^=  row[i];state ^= (1 <<count);//行列交汇处异或之后复原了，要补上这一步！！！switch(j){case 0:state ^= ((1<<0) + (1<<4) + (1<<8) +( 1<< 12));break;case 1:state ^= ((1<<1) + (1<<5) + (1<<9) + ( 1<< 13));break;case 2:state ^= ((1<<2) + (1<<6) + (1<<10) +( 1<< 14));break;case 3:state ^= ((1<<3) + (1<<7) + (1<<11) +( 1<< 15));break;}cout << count <<" : " <<state <<endl;binary(state);}}//system("pause");return 0;}

binary 把每个数字对应在原图中对应的位置标识了出来，非常直观！

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