题目链接:https://cn.vjudge.net/contest/312165#problem/D 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. 翻译： 给定a，b，c的值，按照上述规则，计算最终w(a,b,c); 分析： 将搜过的结果，存到dp数组中，遇到相同的a，b，c，直接返回对应dp的值。 Sample Input
1 1 1 2 2 2 10 4 6 50 50 50 -1 7 18 -1 -1 -1Sample 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代码:
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int dp[100][100][100]; int dfs(int a,int b,int c){ if(a<=0||b<=0||c<=0) return 1; if(a>20||b>20||c>20) return dfs(20,20,20); if(dp[a][b][c]) return dp[a][b][c]; else{ if(a<b&&b<c) return dp[a][b][c]=dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c); else return dp[a][b][c]=dfs(a-1,b,c)+dfs(a-1,b-1,c)+dfs(a-1,b,c-1)-dfs(a-1,b-1,c-1); } } int main(){ int a,b,c; while(~scanf("%d%d%d",&a,&b,&c)){ memset(dp,0,sizeof(dp)); if(a==-1&&b==-1&&c==-1)break; printf("w(%d, %d, %d) = %d\n",a,b,c,dfs(a,b,c)); } return 0; }记忆化搜索，博客中有相关练习。
