题目不在描述，见计算机算法设计与分析习题解答（第二版）王晓东 第6页
/*
* CoinSort.cpp
*
* Created on: Dec 4, 2014
* Author: hujianjie
*/
#include<iostream>
using namespace std;
int Max = 10;
int n, m, count, best;
bool flag;
int arrOld[10][10];
int arrNew[10][10];
int arrMid[10][10];
void copy(int arr1[10][10], int arr2[10][10]) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
arr1[i][j] = arr2[i][j];
}
}
}
void transRow(int row) {
for (int i = 0; i < m; i++) {
// arrNew[row][i] = arrNew[row][i] ^ 1;
arrOld[row][i] = arrOld[row][i] ^ 1;
}
cout << "transRow" << endl;
count++;
}
void transColum(int mid, int tem) {
int temp;
for (int k = 0; k < n; k++) {
//temp = arrNew[k][mid];
//arrNew[k][mid] = arrNew[k][tem];
//arrNew[k][tem] = temp;
temp = arrOld[k][mid];
arrOld[k][mid] = arrOld[k][tem];
arrOld[k][tem] = temp;
}
cout << "transColum" << endl;
count++;
}
bool isSame(int mid, int tem) {
for (int k = 0; k < n; k++) {
if (arrOld[k][tem] != arrNew[k][mid]) {
return false;
}
}
return true;
}
int main() {
flag = false;
count = 0;
cout << "Please input row n and colum m !" << endl;
cin >> n >> m;
cout << "Please input old !" << endl;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> arrOld[i][j];
}
}
cout << "Please input New !" << endl;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> arrNew[i][j];
}
}
//copy(arrMid, arrNew);
copy(arrMid, arrOld);
best = n + m + 1;
for (int colum = 0; colum < m; colum++) {
cout << colum << endl;
//copy(arrNew,arrMid);
copy(arrOld, arrMid);
count = 0;
//将每一列转换成第一列
if (colum != 0)
transColum(0, colum);
//比较第一列的每一行，不同的则将其取反
for (int row = 0; row < n; row++) {
if (arrNew[row][0] != arrOld[row][0]) {
transRow(row);
}
}
//在第一行确定并翻转之后，比较剩下的列
for (int mid = 1; mid < m; mid++) {
for (int tem = mid; tem < m; tem++) {
if (isSame(mid, tem)) {
if (mid != tem)
transColum(mid, tem);
flag = true;
} else {
flag = false;
}
}
}
if (flag == true && count < best)
best = count;
}
if (best < (m + n + 1))
cout << "The best steps are : " << best << endl;
else
cout << "Can't finished! " << endl;
return 0;
}
/*
1 0 1
0 0 0
1 1 0
1 0 1
*
*
1 0 1
1 1 1
0 1 1
1 0 1
*
*
*
*
*/
C 语言写的不好，思路是：将初始矩阵的每一列尝试作为目标的第一列，按枚举的思路一列一列的比较