标签：dp dfs poj
Description
We all love recursion! Don‘t we?
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.
Input
The input for your program will be a series of integer triples, one per line, until the end-of-file flag of -1 -1 -1. Using the above technique, you are to calculate w(a, b, c) efficiently and print the result.
Output
Print the value for w(a,b,c) for each triple.
Sample Input
1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1
Sample Output
w(1, 1, 1) = 2
w(2, 2, 2) = 4
w(10, 4, 6) = 523
w(50, 50, 50) = 1048576
w(-1, 7, 18) = 1
状态方程都给出来了，直接无脑敲代码就行了
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[25][25][25];
int dfs(int a,int b,int c)
{
if(a<=0 || b<=0 || c<=0)
return 1;
if(a>20 || b>20 || c>20)
return dfs(20,20,20);
if(dp[a][b][c])
return dp[a][b][c];
if(a<b && b<c)
dp[a][b][c] = dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c);
else
dp[a][b][c] = dfs(a-1,b,c)+dfs(a-1,b-1,c)+dfs(a-1,b,c-1)-dfs(a-1,b-1,c-1);
return dp[a][b][c];
}
int main()
{
int a,b,c;
memset(dp,0,sizeof(dp));
while(~scanf("%d%d%d",&a,&b,&c))
{
if(a == -1 && b == -1 && c == -1)
break;
printf("w(%d, %d, %d) = %d\n",a,b,c,dfs(a,b,c));
}
return 0;
}
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POJ1579：Function Run Fun(记忆化)
标签：dp dfs poj
原文地址：http://blog.csdn.net/libin56842/article/details/38234059
