Poj 2965 the pilots brothers 'refrigerator

The pilots brothers 'refrigerator Time limit:1000 ms   Memory limit:65536 K Total submissions:18109   Accepted:6871   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 4 records 4. 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 rowIAnd all handles in ColumnJ. 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 61 11 31 44 14 34 4 Source Northeastern Europe 2004, western subregion # Include <iostream> # include <cstdio> # include <cstring> using namespace STD; bool map [20] [20]; int best = 1000; int LJ [70000] [2]; int B [70000] [2]; int L = 0, T; int check () // check whether the door is fully open {int I, j, TOT = 0; for (I = 0; I <4; I ++) for (j = 0; j <4; j ++) TOT + = map [I] [J]; If (TOT = 16) return 1; return 0;} void change (INT X, int y) // change the door status {/* map [x] [Y] =! Map [x] [Y]; Map [x + 1] [Y] =! Map [x + 1] [Y]; If (x> 0) map [x-1] [Y] =! Map [x-1] [Y]; Map [x] [Y + 1] =! Map [x] [Y + 1]; If (Y> 0) map [x] [Y-1] =! Map [x] [Y-1]; */INT I, j; for (I = 0; I <4; I ++) map [x] [I] =! Map [x] [I]; for (j = 0; j <4; j ++) map [J] [Y] =! Map [J] [Y]; Map [x] [Y] =! Map [x] [Y]; // The point is changed back, so it must be changed once} void search (int K, int ans) {int X, Y; int I; if (k = 16) {If (check () & Ans <best) {best = ans; for (I = 0; I <best; I ++) {B [I] [1] = LJ [I] [1]; B [I] [0] = LJ [I] [0] ;}} else {// cout <ans <Endl; X = K/4; y = K % 4; search (k + 1, ANS ); LJ [l] [0] = x + 1; LJ [l] [1] = Y + 1; l ++; change (x, y ); search (k + 1, ANS + 1); change (x, y); // restore the door status l -- ;}} int main () {int I, J; char A; for (I = 0; I <4; I ++) {for (j = 0; j <4; j ++ ){ Cin> A; if (a = '-') map [I] [J] = 1; else map [I] [J] = 0 ;}} search (0, 0); If (best! = 1000) cout <best <Endl; else cout <"impossible" <Endl; for (I = 0; I <best; I ++) cout <B [I] [0] <"" <B [I] [1] <Endl; return 0 ;}