Consider a three-parameter recursive function w(a, b, c):
if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns:
1
if a > 20 or b > 20 or c > 20, then w(a, b, c) returns:
w(20, 20, 20)
if a < b and b < c, then w(a, b, c) returns:
w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c)
otherwise it returns:
w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1)
This is an easy function to implement. The problem is, if implemented directly, for moderate values of a, b and c (for example, a = 15, b = 15, c = 15), the program takes hours to run because of the massive recursion.
#include<stdio.h>
int w(int a,int b,int c)
{
if(a<=||b<=||c<=)return ;
else if(a>||b>||c>)return w(,,);
else if(a<b&&b<c)return w(a,b,c-)+w(a,b-,c-)-w(a,b-,c);
else return w(a-,b,c)+w(a-,b-,c)+w(a-,b,c-)-w(a-,b-,c-);
}
int main()
{
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
if(a==-&&b==-&&c==-)break;
printf("w(%d, %d, %d) = %d\n",a,b,c,w(a,b,c));
}
}
运行结果是这样的。
然后我发现在递归的过程中有很多重复的部分,导致超时,所以要用记忆化搜索来解决。
也就是把已经计算出来的结果放在一个数组里保存,下次计算到这里的时候直接读取结果就可以了。
下面是正确代码。
#include<stdio.h>
int dp[+][+][+]={};
int w(int a,int b,int c)
{
if(a<=||b<=||c<=)return ;
if(dp[a][b][c]!=)return dp[a][b][c];//直接读取
else if(a<b&&b<c)
{
dp[a][b][c]=w(a,b,c-)+w(a,b-,c-)-w(a,b-,c);
return dp[a][b][c];
}
else
{
dp[a][b][c]= w(a-,b,c)+w(a-,b-,c)+w(a-,b,c-)-w(a-,b-,c-);
return dp[a][b][c];
}
}
int main()
{
int a,b,c,res;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
if(a==-&&b==-&&c==-)break;
if(a<=||b<=||c<=)res=;
else if(a>||b>||c>)res=w(,,);
else res=w(a,b,c);
printf("w(%d, %d, %d) = %d\n",a,b,c,res);
}
}