Function Run Fun POJ

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 题目：

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 12 2 210 4 650 50 50-1 7 18-1 -1 -1

Sample Output

w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1

大致题意：给了递归式，让你剪枝。

解法：用一个三位数组将走过的状态保存起来，记忆化搜索。

代码1：32ms过

#include<stdio.h>#include<string>#include<string.h>#include<algorithm>#include<iostream>#include<math.h>using namespace std;const int maxn = 100;int a, b, c, dp[maxn][maxn][maxn];int dfs(int x, int y, int z){    if(x <= 0 || y <= 0 || z <= 0)        return 1;    if(x > 20 || y > 20 || z > 20)        return dfs(20, 20, 20);    if(dp[x][y][z])//如果已经搜过，直接返回这个状态。注意这个判断条件不能放在上面，因为可能会数组越界。        return dp[x][y][z];    if(x < y && y < z)    {        dp[x][y][z] = dfs(x, y, z-1) + dfs(x, y-1, z-1) - dfs(x, y-1, z);        return dp[x][y][z];    }    else    {        dp[x][y][z] = dfs(x-1, y, z) + dfs(x-1, y-1, z) + dfs(x-1, y, z-1) - dfs(x-1, y-1, z-1);        return dp[x][y][z];    }}int main(){    while(scanf("%d%d%d",&a,&b,&c) != EOF)    {        if(a == -1 && b == -1 && c == -1)            break;        memset(dp, 0, sizeof(dp));        int ans = dfs(a, b, c);        printf("w(%d, %d, %d) = %d\n", a, b, c, ans);    }    return 0;}

代码2：参考了大神的代码，先打表，0ms过。

#include<stdio.h>#include<algorithm>#include<string>#include<string.h>#include<iostream>#include<math.h>using namespace std;const int maxn = 25;int dp[maxn][maxn][maxn];int main(){    for(int i = 0; i < 21; i ++)        for(int j = 0; j < 21; j ++)            for(int k = 0; k < 21; k ++)            {                dp[i][j][0] = 1;                dp[i][0][k] = 1;                dp[0][j][k] = 1;            }    for(int i = 1; i < 21; i ++)        for(int j = 1; j < 21; j ++)            for(int k = 1; k < 21; k ++)            {                if(i < j && j < k)                    dp[i][j][k] = dp[i][j][k-1] + dp[i][j-1][k-1] - dp[i][j-1][k];                else                    dp[i][j][k] = dp[i-1][j][k] + dp[i-1][j-1][k] + dp[i-1][j][k-1] - dp[i-1][j-1][k-1];            }    int a, b, c;    while(scanf("%d%d%d",&a,&b,&c) != EOF)    {        if(a == -1 && b == -1 && c == -1)            break;        if(a <=0 || b <= 0 || c <= 0)            printf("w(%d, %d, %d) = 1\n", a, b, c);        else if(a > 20 || b > 20 || c >20)            printf("w(%d, %d, %d) = %d\n", a, b, c, dp[20][20][20]);        else            printf("w(%d, %d, %d) = %d\n",a, b, c, dp[a][b][c]);    }    return 0;}

如有错误，欢迎指出~