首页 > ACM题库 > HDU-杭电 > HDU 1579 Function Run Fun-动态规划-[解题报告] C++
2013
12-12

HDU 1579 Function Run Fun-动态规划-[解题报告] C++

Function Run Fun

问题描述 :

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.

输入:

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.

输出:

Print the value for w(a,b,c) for each triple.

样例输入:

1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1

样例输出:

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

递归会超时,用递推

#include<stdio.h>
int dp[25][25][25];
int main()
{
	int a,b,c,k;
	for(a=0;a<=20;a++)
		for(b=0;b<=20;b++)
			for(c=0;c<=20;c++)
			{
				if(a<=0||b<=0||c<=0)
					dp[a][b][c]=1;
				else if(a<b&&b<c)
						dp[a][b][c]=dp[a][b][c-1]+dp[a][b-1][c-1]-dp[a][b-1][c];
				else
					dp[a][b][c]=dp[a-1][b][c]+dp[a-1][b-1][c]+dp[a-1][b][c-1]-dp[a-1][b-1][c-1];
			}
			while(scanf("%d%d%d",&a,&b,&c),a!=-1||b!=-1||c!=-1)
			{
				if(a<=0||b<=0||c<=0)
                    k=dp[0][0][0];
				else if(a>20||b>20||c>20)
					k=dp[20][20][20];
				else k=dp[a][b][c];
				printf("w(%d, %d, %d) = %d\n",a,b,c,k);
			}
			return 0;
}

解题报告转自:http://blog.csdn.net/aixiaoling1314/article/details/8788502


  1. 第二块代码if(it != mp.end())应改为if(it != mp.end() && (i+1)!=(it->second +1));因为第二种解法如果数组有重复元素 就不正确