编程之美: 24点游戏的解法

分治法:

把A={e,f,a,b}分治成为B={e,f,a+b}, C={e,f,a-b}, D={e,f,b-a}, E={e,f,a*b}, F={e,f,a/b}, G={e,f,b/a} 这样5个元素个数等于3的集合.

这样就把A中元素运算等于24, 转化为B或C或D或E或F或G中元素运算等于24.

这展示了有关集合运算的一种方法: 元素合并, 转化为元素更少的集合.

Code:

#include <iostream>
#include <exception>
#include <string>
#include <stdexcept>
#include <cmath>
using namespace std;

const int N = 4;
const double RESULT = 24.0;
const double DEGREE = 1e-6;
double num
;
string strRet
;

bool pointGame(int n)
{
if(n == 1)
{
if(fabs(num[0] - RESULT) < DEGREE)
{
cout << strRet[0] << endl;
return true;
}
else
{
return false;
}
}

for(int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
{
double tempA = num[i];
double tempB = num[j];
num[j] = num[n-1];

string expA = strRet[i];
string expB = strRet[j];
strRet[j] = strRet[n-1];

strRet[i] = '(' + expA + '+' + expB + ')';
num[i] = tempA + tempB;
if(pointGame(n-1))
return true;

strRet[i] = '(' + expA + '-' + expB + ')';
num[i] = tempA - tempB;
if(pointGame(n-1))
return true;

strRet[i] = '(' + expB + '-' + expA + ')';
num[i] = tempB - tempA;
if(pointGame(n-1))
return true;

strRet[i] = '(' + expA + '*' + expB + ')';
num[i]  = tempA * tempB;
if(pointGame(n-1))
return true;

if(tempB)
{
strRet[i] = '(' + expA + '/' + expB + ')';
num[i]  = tempA/tempB;
if(pointGame(n-1))
return true;
}

if(tempA)
{
strRet[i] = '(' + expB + '/' + expA + ')';
num[i] = tempB/tempA;
if(pointGame(n-1))
return true;
}

num[i] = tempA;
num[j] = tempB;
strRet[i] = expA;
strRet[j] = expB;
}
return false;
}

int main()
{
while (scanf("%lf%lf%lf%lf", &num[0], &num[1], &num[2], &num[3]) != EOF)
{
for (int i = 0; i < N; ++i)
{
char buffer[20];
itoa((int)num[i], buffer, 10);
strRet[i] = buffer;
}
if(!pointGame(N))
cout << "No result" << endl;
}
return 0;
}

集合的二进制表示法:

可以将元素个数为N集合的集合, 用0 ~ (2<<N)-1表示出所有子集.

从而可以利用二进制表示方便得到子集等操作.

对于本题, 还可以利用重复子问题, 减少重复计算.

Code:

#include <iostream>
#include <exception>
#include <string>
#include <stdexcept>
#include <cmath>
#include <set>
using namespace std;

const int N = 4;
const double RESULT = 24.0;
const double DEGREE = 1e-6;
double num
;
set<double> s[(1<<N)];

void f(int n)
{
if(s
.size() != 0)
return;

for(int i = 1; i < n; ++i)
if((i&n) == i)
{
f(i);
f(n-i);
for (set<double>::iterator itr1 = s[i].begin(); itr1 != s[i].end(); ++itr1)
for (set<double>::iterator itr2 = s[n-i].begin(); itr2 != s[n-i].end(); ++itr2)
{
double a = *itr1;
double b = *itr2;
s
.insert(a + b);
s
.insert(a - b);
s
.insert(b - a);
s
.insert(a * b);
if(b)
s
.insert(a / b);
if(a)
s
.insert(b / a);
}
}
}

int main()
{
for (int i = 0; i < N; ++i)
{
cin >> num[i];
s[(1<<i)].insert((double)num[i]);
}

for(int i = 1; i <= (1<<N)-1; ++i)
f(i);

for(set<double>::iterator itr = s[(1<<N)-1].begin(); itr != s[(1<<N)-1].end(); ++itr)
if(fabs(*itr - RESULT) < DEGREE)
cout << "Has an result" << endl;
return 0;
}