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HDU 1495 非常可乐(最笨办法版)

2015-08-11 17:22 405 查看

原题链接:

http://acm.hdu.edu.cn/showproblem.php?pid=1495

题目大意:



思路:

3个杯子。有6种情况的道法。每种遍历就可以了

主要是比较复杂的原因是加法较多。

结构体只是为了方便一点。

退出时的判定比较坑。要两个相等的同时另一个等于0才满足条件

代码如下:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

int S,N,M;
int NUMBER;
bool Can;
int vis[101][101][101];

struct Tri
{
int s;
int n;
int m;
int number;
};

bool IsOver( int s,int n,int m )
{
if( (s == m && s != 0 && n == 0 )|| ( s == n && n != 0 && m==0 )|| ( m == n && m !=0 && s == 0 ) )
return true;
return false;
}

bool IsLeagal( int s,int n,int m )
{
if( s < 0 || s > S || n < 0 || n > N || m < 0 || m > M )
return false;
return true;
}

void bfs()
{
queue<Tri> Ways;
memset( vis,0,sizeof( vis));

vis[S][0][0] = 1;

Tri Now,Next1,Next2,Next3,Next4,Next5,Next6;
Now.s = S;
Now.n = 0;
Now.m = 0;
Now.number  = 0;
Ways.push ( Now );
while( !Ways.empty ())
{
Now.s = Ways.front ().s ;
Now.n = Ways.front ().n ;
Now.m = Ways.front ().m ;
Now.number = Ways.front ().number ;
Ways.pop ();

if( Can )
return;
/* 从第一个杯子到给第二个杯子 */
if( Now.s > ( N - Now.n) )
{   Next1.s = Now.s - ( N - Now.n);
Next1.n = N;
Next1.m = Now.m;
Next1.number = Now.number + 1;}
else
{   Next1.s = 0;
Next1.n = Now.n + Now.s ;
Next1.m = Now.m;
Next1.number = Now.number + 1;}
if( IsLeagal( Next1.s ,Next1.n ,Next1.m ) && !vis[Next1.s ][Next1.n ][Next1.m ])
{
vis[Next1.s ][Next1.n ][Next1.m ] = 1;
if( IsOver( Next1.s ,Next1.n ,Next1.m ))
{   Can = true;
NUMBER = Next1.number ;
return; }
Ways.push ( Next1 );
}

/* 从第一个杯子到给第三个杯子 */
if( Now.s > ( M - Now.m ) )
{   Next2.s = Now.s - ( M - Now.m );
Next2.n = Now.n;
Next2.m = M;
Next2.number = Now.number + 1;}
else
{   Next2.s = 0;
Next2.n = Now.n;
Next2.m = Now.s + Now.m;
Next2.number = Now.number + 1;}
if( IsLeagal( Next2.s ,Next2.n ,Next2.m ) && !vis[Next2.s ][Next2.n ][Next2.m ])
{
vis[Next2.s ][Next2.n ][Next2.m ] = 1;
if( IsOver( Next2.s ,Next2.n,Next2.m ))
{   Can = true;
NUMBER =  Next2.number  ;
return; }
Ways.push ( Next2 );
}

/* 从第二个杯子到给第一个杯子 */
if( Now.n > ( S - Now.s ) )
{   Next3.s = S;
Next3.n = Now.n - ( S - Now.s );
Next3.m = Now.m;
Next3.number = Now.number + 1;}
else
{   Next3.s = Now.s + Now.n ;
Next3.n = 0;
Next3.m = Now.m;
Next3.number = Now.number + 1;}
if( IsLeagal( Next3.s ,Next3.n ,Next3.m ) && !vis[Next3.s ][Next3.n ][Next3.m ])
{
vis[Next3.s ][Next3.n ][Next3.m ] = 1;
if( IsOver( Next3.s ,Next3.n ,Next3.m ))
{   Can = true;
NUMBER = Next3.number ;
return; }
Ways.push ( Next3 );
}

/* 从第二个杯子到给第三个杯子 */
if( Now.n > ( M - Now.m ))
{   Next4.s = Now.s ;
Next4.n = Now.n - ( M - Now.m );
Next4.m = M;
Next4.number = Now.number + 1;}
else
{   Next4.s = Now.s ;
Next4.n = 0;
Next4.m = Now.n + Now.m ;
Next4.number = Now.number + 1;}
if( IsLeagal( Next4.s ,Next4.n ,Next4.m ) && !vis[Next4.s ][Next4.n ][Next4.m ])
{
vis[Next4.s ][Next4.n ][Next4.m ] = 1;
if( IsOver( Next4.s ,Next4.n ,Next4.m ))
{   Can = true;
NUMBER = Next4.number ;
return; }
Ways.push ( Next4 );
}

/* 从第三个杯子到给第一个杯子 */
if( Now.m - ( S - Now.s ) )
{   Next5.s = S;
Next5.n = Now.n ;
Next5.m = Now.m - ( S - Now.s );
Next5.number = Now.number + 1;}
else
{   Next5.s = Now.s + Now.m ;
Next5.n = Now.n ;
Next5.m = 0;
Next5.number = Now.number + 1;}
if( IsLeagal( Next5.s ,Next5.n ,Next5.m ) && !vis[Next5.s ][Next5.n ][Next5.m ])
{
vis[Next5.s ][Next5.n ][Next5.m ] = 1;
if( IsOver( Next5.s ,Next5.n ,Next5.m ))
{   Can = true;
NUMBER = Next5.number ;
return; }
Ways.push ( Next5 );
}

/* 从第三个杯子到给第二个杯子 */
if( Now.m > ( N - Now.n ))
{   Next6.s = Now.s;
Next6.n = N;
Next6.m = Now.m - ( N - Now.n );
Next6.number = Now.number + 1;}
else
{   Next6.s = Now.s;
Next6.n = Now.m + Now.n ;
Next6.m = 0;
Next6.number = Now.number + 1;}
if( IsLeagal( Next6.s ,Next6.n ,Next6.m ) && !vis[Next6.s ][Next6.n ][Next6.m ])
{
vis[Next6.s ][Next6.n ][Next6.m ] = 1;
if( IsOver( Next6.s ,Next6.n ,Next6.m ))
{   Can = true;
NUMBER = Next6.number ;
return; }
Ways.push ( Next6 );
}

}
}
int main()
{
while( cin >> S >> N >> M )
{
if( S == 0 && N == 0 && M == 0 ) break;
if( N > M )
{
int tmp;
tmp = N;
N =M;
M = tmp;
}
if( S % 2 != 0 )
{
cout<<"NO"<<endl;
continue;
}
Can = false;
bfs();
if( Can )
cout<< NUMBER <<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
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