模板题：https://acm.sdut.edu.cn/onlinejudge3/problems/2176

记忆化搜索和普通搜索的明显差距在于时间上记忆化搜索要快许多，尤其在递归调用较多的情况下

记忆化搜索保存了之前尝试过的一种可能性，这样在到达不可能到达的点时将直接返回的原来点
另外记忆化搜索的思想是一种空间换时间的思路

普通递归代码（超时）：

#include<stdio.h>
int p[21][21][21] = {0};
int f(int a, int b, int c)
{
    if(a <= 0 || b <= 0 || c <= 0)
		return 1;
    else if(a > 20 || b > 20 || c > 20)
		return f(20, 20, 20);
    if(a < b && b < c)
		return f(a, b, c - 1) +f(a, b - 1, c - 1) - f(a, b - 1, c);
    else 
		return f(a-1, b, c) + f(a - 1, b - 1, c) + f(a - 1, b, c - 1) - f(a - 1, b - 1, c - 1);
}
int main()
{
    int a, b, c;
    while(~scanf("%d%d%d", &a, &b, &c)){
        printf("%d\n", f(a, b, c));
    }
    return 0;
}

AC代码：

#include<stdio.h>
int p[21][21][21] = {0};
int f(int a, int b, int c)
{
    if(a <= 0 || b <= 0 || c <= 0)
		return 1;
    else if(a > 20 || b > 20 || c > 20)
		return f(20, 20, 20);
    if(p[a][b][c])
    {
        return p[a][b][c];
    }
    if(a < b && b < c)
		return p[a][b][c] = f(a, b, c - 1) +f(a, b - 1, c - 1) - f(a, b - 1, c);
    else 
		return p[a][b][c] = f(a-1, b, c) + f(a - 1, b - 1, c) + f(a - 1, b, c - 1) - f(a - 1, b - 1, c - 1);
}
int main()
{
    int a, b, c;
    while(~scanf("%d%d%d", &a, &b, &c)){
        printf("%d\n", f(a, b, c));
    }
    return 0;
}