POJ 1579-Function Run Fun（记忆化搜索）

Function Run Fun

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16503 Accepted: 8514

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 12 2 210 4 650 50 50-1 7 18-1 -1 -1

Sample Output

w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1

巨水。。

一开始wa了一组数据 （0,21,21）

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cctype>#include <vector>#include <cstdio>#include <cmath>#include <deque>#include <stack>#include <map>#include <set>#define ll long long#define maxn 116#define pp pair<int,int>#define INF 0x3f3f3f3f#define max(x,y) ( ((x) > (y)) ? (x) : (y) )#define min(x,y) ( ((x) > (y)) ? (y) : (x) )using namespace std;int n,m,dp[22][22][22];int dfs(int x,int y,int z){if(x<=0||y<=0||z<=0)return 1;if(x>20||y>20||z>20)return dfs(20,20,20);if(dp[x][y][z]!=INF)return dp[x][y][z];dp[x][y][z]=0;if(x<y&&y<z)dp[x][y][z]+=(dfs(x,y,z-1)+dfs(x,y-1,z-1)-dfs(x,y-1,z));else dp[x][y][z]+=(dfs(x-1,y,z)+dfs(x-1,y-1,z)+dfs(x-1,y,z-1)-dfs(x-1,y-1,z-1));return dp[x][y][z];}int main(){int a,b,c;    while(scanf("%d%d%d",&a,&b,&c)){if(a==-1&&b==-1&&c==-1)break;memset(dp,INF,sizeof(dp));printf("w(%d, %d, %d) = %d\n",a,b,c,dfs(a,b,c));}return 0;}