UVA 1362 Exploring Pyramids 区间DP

Archaeologists have discovered a new set of hidden caves in one of the Egyptian pyramids. The decryption of ancient hieroglyphs on the walls nearby showed that the caves structure is as follows. There are n caves in a pyramid, connected by narrow passages, one of the caves is connected by a passage to the outer world. The system of the passages is organized in such a way, that there is exactly one way to get from outside to each cave along passages. All caves are located in the basement of the pyramid, so we can consider them being located in the same plane. Passages do not intersect. Each cave has its walls colored in one of several various colors.
The scientists have decided to create a more detailed description of the caves, so they decided to use an exploring robot. The robot they are planning to use has two types of memory - the output tape, which is used for writing down the description of the caves, and the operating memory organized as a stack.
 The robot first enters the cave connected to the outer world along the passage. When it travels along any passage for the first time, it puts its description on the top of its stack. When the robot enters any cave, it prints the color of its walls to its output tape. After that it chooses the leftmost passage among those that it has not yet travelled and goes along it. If there is no such passage, the robot takes the passage description from the top of its stack and travels along it in the reverse direction. The robot‘s task is over when it returns to the outside of the pyramid. It is easy to see that during its trip the robot visits each cave at least once and travels along each passage exactly once in each direction.
 The scientists have sent the robot to its mission. After it returned they started to study the output tape. What a great disappointment they have had after they have understood that the output tape does not describe the cave system uniquely. Now they have a new problem - they want to know how many different cave systems could have produced the output tape they have. Help them to find that out.
   Since the requested number can be quite large, you should output it modulo 1 000 000 000. Please note, that the absolute locations of the caves are not important, but their relative locations are important, so the caves (c) and (d) on the picture below are considered different.

Input 
The input file contains several test cases, and each of them consists of a single line with the output tape that the archaeologists have. The output tape is the sequence of colors of caves in order the robot visited them. The colors are denoted by capital letters of the English alphabet. The length of the tape does not exceed 300 characters. 
Output 
For each input case, write to the output a single line containing one integer number - the number of different cave systems (modulo 1 000 000 000) that could produce the output tape. 
Sample Input 
ABABABA 
AB 
Sample Output 
5 0

题意:给出一颗多叉树,各个节点用字母表示,从根节点开始,每次尽量往左走,走不通了就回溯,把遇到的字母顺序记下来,可以得到一个序列,现给出这个序列,问有多少棵树与之对应。 
题解
设输入序列为S,d(i,j)为子序列Si,Si+1,…,Sj对应的树的个数,则可以发现d(i,i)=1;并且Si不等于Sj时d(i,j)=0(因为起点和终点应该是同一点)。在其他情况下,设第一个分支在Sk时回到树根(必须有Si=Sk),则这个分支对应的序列是Si+1…Sk-1,方案数为d(i+1,k-1);其他分支对应的访问序列为Sk,…Sj,方案数为d(k.j)。这样,在非边界情况,递推关系为; 
d(i,j)=sigma{d(i+1,k-1)*d(k,j)|i+2<=k<=j,Si=Sk=Sj}.

时间: 2024-08-05 02:41:24

UVA 1362 Exploring Pyramids 区间DP的相关文章

UVA 1362 - Exploring Pyramids(计数问题+区间DP)

题目链接:1362 - Exploring Pyramids 白书上的例题,思路是对于每个结点,往后遍历分为左右两边子树来考虑,左边的子树为去掉根节点剩下的子树,而右边是要算上根节点的,这样就不会有重复的情况出现,然后根据乘法原理,左右两边情况相乘为总情况数,然后计算这些总和. f[i][j]表示[i,j]结点的情况种数,那么 f[i][j] = sum{f[i + 1][k - 1] * f[k][j]},并且s[i] == s[k] == s[j],因为i, j, k表示的是根节点必须相同.

uva live 3516 Exploring Pyramids 区间DP

// uva live 3516 Exploring Pyramids 区间DP // // 题目大意: // // 给你一个多叉树,每个节点是一个大写字母,从根节点走,按照先序遍历的 // 原则访问,不能访问则回溯,每次记录一下节点的字符,最后得到一个字符串.现 // 在给你一个字符串,问可能符合条件的多叉树的数量. // // 解题思路: // // 区间DP,我们注意到,从根节点出发,一定会再次回到根节点,那么我们可以设 // d(i,j) 是序列i到j段形成的符合条件的多叉树的数量,则

UVALive 3516 Exploring Pyramids 区间dp+计数原理

题目链接:点击打开链接 给定多叉树的先序遍历结果,求多叉树的同构数 思路:区间dp import java.io.PrintWriter; import java.util.ArrayList; import java.util.Scanner; public class Main { int min(int a,int b){return a>b?b:a;} int max(int a,int b){return a>b?a:b;} long min(long a,long b){retur

UVALive 3516 Exploring Pyramids (区间dp)

#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; #define ll long long const int maxn = 310; const ll mod = 1e9; char s[maxn]; ll dp[maxn][maxn]; ll solve(int i, int j) { if(i == j) retu

UVA 10003 Cutting Sticks(区间dp)

Description  Cutting Sticks  You have to cut a wood stick into pieces. The most affordable company, The Analog Cutting Machinery, Inc. (ACM), charges money according to the length of the stick being cut. Their procedure of work requires that they onl

UVA 10003 Cutting Sticks 区间DP+记忆化搜索

UVA 10003 Cutting Sticks+区间DP 纵有疾风起 题目大意 有一个长为L的木棍,木棍中间有n个切点.每次切割的费用为当前木棍的长度.求切割木棍的最小费用 输入输出 第一行是木棍的长度L,第二行是切割点的个数n,接下来的n行是切割点在木棍上的坐标. 输出切割木棍的最小费用 前话-区间dp简单入门 区间dp的入门下面博客写的非常好,我就是看的他们博客学会的,入门简单,以后的应用就得靠自己了. https://blog.csdn.net/qq_41661809/article/d

Uva 10003-Cutting Sticks(区间DP)

题目链接:点击打开链接 题意: 一根长度为L 的木棒,要求要切割n次,给出n次需要切割的位置,每次切割花费为当前段的总长度.求最小花费. 思路:..一开始想了好久也没往区间DP那方面想QAQ.设 dp[i][j] 为切割[i,j] 区间的最小花费,状态转移方程为 dp[i][j]=max(dp[i][k-1]+dp[k+1][j]+a[j+1]-a[i-1]); #include <algorithm> #include <iostream> #include <cstrin

UVA 1331 Minimax Triangulation 区间DP

区间DP: 将一个多边形三角剖分,让可以得到的最大三角形的面积最小 dp[i][j]表示从i点到j点的最优值,枚举中间点k dp[i][j]=min(dp[i][j],max(area(i,j,k),max(dp[i][k],dp[k][j]))); 注意如果中间三角形i-j-k中有其他的点,这样的三角形是不可以剖分的 Minimax Triangulation Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld

Uva 10617 Again Palindrome(区间dp)

Again Palindromes Input: Standard Input Output: Standard Output Time Limit: 2 Seconds A palindorme is a sequence of one or more characters that reads the same from the left as it does from the right. For example, Z, TOT and MADAM are palindromes, but