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hdu1579 Function Run Fun(深搜+记忆化)
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1579
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.
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
代码一如下:
#include <cstdio> #include <cstring> int dp[57][57][57]; 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]; if(a < b && b < c) dp[a][b][c]=dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c); else 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); return dp[a][b][c]; } int main() { int a, b, c; while(~scanf("%d%d%d",&a,&b,&c)) { if(a == -1 && b == -1 && c == -1) break; int ans = dfs(a,b,c); printf("w(%d, %d, %d) = %d\n",a,b,c,ans); } return 0; }
代码二如下:
#include <cstdio> #define N 20 int a,b,c,w[N+1][N+1][N+1]; int dp(int a,int b,int c) { int i,j,k; if (a<=0 || b<=0 || c<=0) return 1; if (a>20 || b>20 || c>20) a=b=c=20; for (i=0;i<=N;i++) for (j=0;j<=N;j++) w[0][i][j]=w[i][0][j]=w[i][j][0]=1; for (i=1;i<=a;i++) for (j=1;j<=b;j++) for (k=1;k<=c;k++) if (i<j && j<k) w[i][j][k]=w[i][j][k-1]+w[i][j-1][k-1]-w[i][j-1][k]; else w[i][j][k]=w[i-1][j][k]+w[i-1][j-1][k]+w[i-1][j][k-1]-w[i-1][j-1][k-1]; return w[a][b][c]; } int main() { while (~scanf("%d %d %d",&a,&b,&c)) { if(a == -1 && b == -1 && c == -1) break; int ans = dp(a,b,c); printf("w(%d, %d, %d) = %d\n",a,b,c,ans); } return 0; }
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