（hdu 1579）Function Run Fun (记忆化搜索)

Time limit1000 ms Memory limit32768 kB

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

分析：记忆化搜索
DP也能实现

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define mem(a,n) memset(a,n,sizeof(a))typedef long long LL;const int N=100+5;const int INF=0x3f3f3f3f;int dp[N][N][N];LL fun(int a,int b,int c){    if(a<=0||b<=0||c<=0) return 1;    int& ret=dp[a][b][c];///由于a,b,c存在为负值的情况，不能把这一步放在函数第一行    if(ret!=-1) return ret;    else if(a>20||b>20||c>20) return ret=fun(20,20,20);    else if(a<b&&b<c) return ret=fun(a,b-1,c-1)+fun(a,b-1,c-1)-fun(a,b-1,c);    else return ret=fun(a-1,b,c)+fun(a-1,b-1,c)+fun(a-1,b,c-1)-fun(a-1,b-1,c-1);}int main(){    int a,b,c;    while(~scanf("%d%d%d",&a,&b,&c))    {        mem(dp,-1);        if(a==-1&&b==-1&&c==-1) break;        int ans=fun(a,b,c);        printf("w(%d, %d, %d) = %d\n",a,b,c,ans);    }    return 0;}