题目链接：
http://acm.hust.edu.cn/vjudge/problem/48421
Binary Tree
Time Limit: 3000MS
#### 问题描述
> Binary Tree is a tree data structure where each node has at most two children, usually they are
> distinguished as left and right child. And the node having the children are called parent of those
> children.
> An instruction string is a string consisting of the letters L, R and U. L stands for Left, R for Right
> and U for Up. Meaning of these will be clear shortly.
> One day I have drawn an infinitely large Binary Tree. In this tree each node has exactly two
> children (left child and right child) and each of them has a parent. For this problem, we will consider
> the parent of the root is root itself. I put a pen in the root and follow the instruction string S.
> That is, we look at the first character if it says L we go to left child, if it says R we go to right child
> and if it says U then to the parent. If we receive a U instruction at root we just end up at root since we
> assumed parent of the root is root itself.
> Now we have another instruction string T. Starting from the node where we are after following the
> instruction string S, we will follow the instruction string T. But this time, if we wish we may skip any
> instruction in the string T (possibly discarding all of them). You have to tell me how many different
> nodes I can end up after following instruction string T (skipping as many instructions as I wish).
> For example:
> Suppose: S = L and T = LU. Our answer is 3. Following S we will end up at the left child of the
> root. Now, when we follow T, there may be 4 cases:
> i Skipping all letters: we will be at the same node where we are.
> ii Skipping L and following U: we will be at the root.
> iii Following L and skipping U: we will be at the left child of current node.
> iv Following both L and U: we will be at the same node as in case i.
> Since 3 different nodes we can end up after following T, the answer is 3.
输入
First line of the test file contains an integer N (≤ 20) denoting number of test cases. Hence follow N
test cases. Each test case consists of two non empty strings. First line will contain instruction string S
and the second line will contain the instruction string T. You may assume that there will not be any
letter other than L, R or U in these strings. Length of the strings will not be greater than 100000.
输出
For each test case print the case number followed by the number of nodes we can end up finally. Since
the answer may be large, you have to give the answer modulo 21092013.
样例
sample input
2
L
LU
L
Lsample output
Case 1: 3
Case 2: 2
题解
首先我们考虑只有LR,没有U的情况：
我们定义三个计数变量：ans=1,l=1,r=1;
其中ans为答案，l表示当前局势下，往左走一步能够访问到的新节点；r表示当前局势往右走一步能够访问到的新节点。
如果现在的输入是'L'，那么显然会有ans+=l。并且不会影响r，且这些新节点还会创造出更多的能够往右走的新节点：r+=l。
那么，现在我们考虑有'U'的情况：
如果U没有上升到一个没有走过的点，显然是要亏一波的，还不如直接跳过。
因此我们只需要考虑往上走能够走到新节点的情况(用S串来获取信息）：
如果父亲在左边（它是右儿子），那么将会有l+=1; ans+=1;
如果父亲在右边：那么有r+=1,ans+=1;
代码
#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define pb(v) push_back(v)
#define sz() size()
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
using namespace std;
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const int maxn = 1e5+10;
const int mod=21092013;
char s1[maxn],s2[maxn];
vector<char> sta;
void add_mod(int &x,int y){ x=(x+y)%mod; }
void init(){
sta.clear();
}
int main() {
int tc,kase=0;
scanf("%d",&tc);
while(tc--){
init();
scanf("%s",s1);
int l1=strlen(s1);
rep(i,0,l1){
if(s1[i]=='U'){
if(sta.size()) sta.pop_back();
}else{
sta.pb(s1[i]);
}
}
scanf("%s",s2);
int l2=strlen(s2);
int l=1,r=1,ans=1;
rep(i,0,l2){
if(s2[i]=='U'){
if(sta.size()){
char last=sta[sta.sz()-1];
sta.pop_back();
if(last=='L'){
add_mod(r,1);
add_mod(ans,1);
}else{
add_mod(l,1);
add_mod(ans,1);
}
}
}else if(s2[i]=='L'){
add_mod(r,l);
add_mod(ans,l);
}else{
add_mod(l,r);
add_mod(ans,r);
}
}
printf("Case %d: %d\n",++kase,ans);
}
return 0;
}