POJ-1579 Function Run Fun

Function Run Fun Time Limit: 1000MS   Memory Limit: 10000K       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 Source Pacific Northwest 1999

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思路：最裸的记忆化搜索，裸就裸在直接跟你说，你在哪需要记忆化一下。
对于DFS，重复的计算是相当浪费时间的。这时候就需要利用参数和数组进行记忆化。对DFS来说，参数意味着状态。某一状态如果已经计算，直接返回值就行了。如果并不能确定0能否代表该状态是否已经计算，就要另开一个vis数组。
代码如下：
/****************************************/
#include <cstdio>
#include <cstring>
/****************************************/
bool vis[25][25][25];
int ans[25][25][25];
int w(int a, int b, int c) {
int ret;//小技巧，精简代码
if(a <= 0 || b <= 0 || c <= 0)
return 1;
else if(a > 20 || b > 20 || c > 20)
return w(20, 20, 20);
if(vis[a][b][c])//如果存在记忆，直接返回答案
return ans[a][b][c];
vis[a][b][c] = true;//记忆化
if(a < b && b < c) {
ret = w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c);
} else {
ret = w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1);
}
return ans[a][b][c] = ret;//返回答案值的同时储存进数组中
}
int main() {
int a, b, c;
while(~scanf("%d%d%d", &a, &b, &c), a!=-1||b!=-1||c!=-1) {
memset(vis, 0, sizeof(vis));
printf("w(%d, %d, %d) = %d\n", a, b, c, w(a, b, c));
}
return 0;
}