poj 2965

题意:给一个4x4的方格,‘+’符号代表关闭,‘-’符号代表打开,当所有的手柄都为符号‘-’,冰箱门才可以打开,每次对一个点进行操作,与这个点在同一行同一列的点全                部进行状态转变,问将门开打的最少操作次数;

思路:意思就只需要把所有的‘+’符号变成‘-’符号就行了,创建一个bool数组,每次遇到‘+’符号的话将这一行这一列全部进行!操作,最后bool数组中所有为true的坐标            即为要操作的点。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cmath>
 4 using namespace std;
 5 const int sasuke=4;
 6 bool num[5][5];
 7 int main()
 8 {
 9     char str[5][5];
10     for(int i=1;i<=sasuke;++i){
11         for(int j=1;j<=sasuke;++j)
12             str[i][j]=cin.get();
13             cin.get();
14     }
15
16
17     for(int j,i=1;i<=sasuke;++i)
18         for(j=1;j<=sasuke;++j){
19             if(str[i][j]==‘+‘){
20                 num[i][j]=!num[i][j];
21                 for(int k=1;k<=sasuke;++k){
22                     num[i][k]=!num[i][k];
23                     num[k][j]=!num[k][j];
24                 }
25
26             }
27         }
28     int a[20],b[20];int t=0;
29     int count=0;
30     for(int i=1;i<=sasuke;++i)
31         for(int j=1;j<=sasuke;++j)
32             if(num[i][j]){
33                 ++count;a[t]=i;b[t]=j;++t;
34             }
35     cout << count << endl;
36     for(int i=0;i<t;++i)
37         cout << a[i] << " " << b[i] << endl;
38 }
时间: 2024-10-13 15:12:09

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