HDU--1331--Function Run Fun--記憶化搜索
Function Run Fun
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2376 Accepted Submission(s): 1187
Problem 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
題意：修改給你的這個遞歸，使他能夠處理更大的數據
解析，a,b,c被控制在1~20之間，所以用記憶化搜索的方式記憶那些數據防止重復就行了
#include
#include
#include
using namespace std;
int dp[22][22][22],s=0; //記錄防止重復搜索
int w(int a,int b,int c)
{
if(a<=0||b<=0||c<=0)return 1;
if(a>20||b>20||c>20)return w(20,20,20);
if(dp[a][b][c])return dp[a][b][c]; //使用記憶點防止重復
if(a
莫名其妙，就直接把記憶化的思想加進來就搞定了，當年真是白活了啊！！
