Poj2965the pilots ' refrigerator dfs+ Enumeration

The pilots ' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20858		Accepted: 8048		Special Judge

Description

The game "The pilots brothers:following the stripy elephant" has a quest where a player needs to open a refrigerator.

There is handles on the refrigerator door. Every handle can is in one of the States:open or closed. The refrigerator is open if all handles is open. The handles is 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 rowI 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 are 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 is several solutions, you could give any one of the them.

Sample Input

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

Sample Output

61 11 31 44 14 34 4

#include <iostream> #include <cstring> #include <cstdio> #include <queue>using namespace std;    int num=0x3f3f3f3f;int a[10][10],b[10][10],flag;int Fanzhuan (int x,int y) {a[x][y]=!a[x][y];    for (int i=0; i<4; i++) a[x][i]=!a[x][i]; for (int j=0; j<4; j + +) A[j][y]=!a[j][y];}  int Panduan () {for (Int. i=0; i<4; i++) for (int j=0; j<4; J + +) if (!a[i][j]) return    0; return 1;} struct node{int A, p[20];void DFS (int x,int y,int ans)//will run all num step cases to determine if there is a compliant {if (Num==ans) {flag=p        Anduan ();    return; } if (Flag| | x>=4| |    y>=4) return;   int fy= (y+1)%4;    int fx=x+ (Y+1)/4 moved by line;    Fanzhuan (x, y);    DFS (fx,fy,ans+1);    P[ans].a=x;    P[ans].b=y; Fanzhuan (x, y);//Return to DFS (Fx,fy,ans);}    int main () {string s[4];        while (Cin>>s[0]) {for (int i=1; i<4; i++) cin>>s[i]; for (int i=0, i<4; i++) for (int j=0; j<4; J+ +)//Format conversion if (s[i][j]== ' + ') a[i][j]=0;        else a[i][j]=1;        flag=0;            for (int i=0; i<=16; i++)//enumeration {num=i;            DFS (0,0,0);        if (flag) break;        } cout<<num<<endl;    for (int i=0; i<num; i++) cout<<p[i].a+1<< "" <<p[i].b+1<<endl; }}//Mode 2 state compression +bfs//.

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Poj2965the pilots ' refrigerator dfs+ Enumeration