http://poj.org/problem?id=2965

? poj 1753擴展，dfs+枚舉，不過加了一個路徑。

The Pilots Brothers' refrigerator

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?19278	?	Accepted:?7384	?	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 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 row?i?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 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

6 1 1 1 3 1 4 4 1 4 3 4 4 #include<iostream> #include<cstdio> #include<cstring> using namespace std; int ans[6][6]; int step; int r[20],c[20]; bool flag=false; bool judge() {int i,j,sum=0;for(i=1;i<=4;i++)for(j=1;j<=4;j++)sum+=ans[i][j];if(sum==0)return true;elsereturn false; } void flip(int x,int y) {int i,j;for(j=1;j<=4;j++){if(ans[x][j]==1)ans[x][j]=0;elseans[x][j]=1;}for(i=1;i<=4;i++){if(i==x)continue;if(ans[i][y]==1)ans[i][y]=0;elseans[i][y]=1;} } void dfs(int x,int y,int d){if(d==step){flag=judge();return ;}if(flag||x>4){return ;}flip(x,y);r[d]=x;//保存路徑。c[d]=y;//保存路徑。if(y<4)dfs(x,y+1,d+1);elsedfs(x+1,1,d+1);flip(x,y);if(y<4)dfs(x,y+1,d);elsedfs(x+1,1,d);} int main() {int i,j;char temp;for(i=1;i<=4;i++){for(j=1;j<=4;j++){scanf("%c",&temp);if(temp=='+')ans[i][j]=1;elseans[i][j]=0;}getchar();}for(step=1;step<=16;step++){dfs(1,1,0);if(flag)break;}printf("%d\n",step);for(i=0;i<step;i++)printf("%d %d\n",r[i],c[i]);return 0; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/cancangood/p/4132078.html

總結(jié)

以上是生活随笔為你收集整理的poj 2965 The Pilots Brothers' refrigerator的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。