# 记忆化搜索专题

HDU 1331 Function Run Fun

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.

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

#include <stdio.h>

int dp[21][21][21] = {0};

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 dp[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;
dp[20][20][20] = dfs(20, 20, 20);
while (~scanf("%d %d %d", &a, &b, &c) && (a != -1 || b != -1 || c != -1))
{
printf("w(%d, %d, %d) = %d\n", a, b, c, dfs(a, b, c));
}
return 0;
}


POJ 1088 滑雪

Problem Description
Michael喜欢滑雪，这并不奇怪，因为滑雪的确很刺激。可是为了获得速度，滑的区域必须向下倾斜，而且当你滑到坡底，你不得不再次走上坡或者等待升降机来载你。Michael想知道载一个区域中最长底滑坡。区域由一个二维数组给出。数组的每个数字代表点的高度。

1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9

Input

Output

Sample Input
5 5
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9

Sample Output
25

#include <stdio.h>
#include <algorithm>
using namespace std;
int map[101][101];
int dp[101][101];
int dir[4][2] = {1, 0, 0, -1, -1, 0, 0, 1};
int r, c;

int dfs(int x, int y)
{
if (dp[x][y])
return dp[x][y];
for (int i = 0; i < 4; i++)
{
int dx = x + dir[i][0], dy = y + dir[i][1];
if (dx > 0 && dx <= r && dy > 0 && dy <= c && map[dx][dy] < map[x][y])
{
dp[x][y] = max(dfs(dx, dy) + 1, dp[x][y]);
}
}
return dp[x][y];
}

int main()
{
scanf("%d %d", &r, &c);
for (int i = 1; i <= r; i++)
{
for (int j = 1; j <= c; j++)
{
scanf("%d", &map[i][j]);
}
}
int len = 0;
for (int i = 1; i <= r; i++)
{
for (int j = 1; j <= c; j++)
{
len = max(dfs(i, j), len);
}
}
printf("%d\n", len + 1);
return 0;
}


NITOJ 425 老袁的迷宫

Problem Description

Input

Output

Sample Input
2
3 3
9 9 4
6 6 8
2 1 1
3 3
3 4 5
3 2 6
2 2 1

Sample Output
4
4

posted @ 2019-04-30 16:05  redleaves  阅读(395)  评论(0编辑  收藏
Live2D