【HDU 5370】 Tree Maker（卡特兰数+dp）

时间：2016-09-21 23:12:43 阅读：283 评论：0 收藏：0 [点我收藏+]

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Tree Maker

Problem Description

Tree Lover loves trees crazily.

One day he invents an interesting game which is named Tree Maker.

In this game, all trees are binary trees.

Initially, there is a tree with only one vertex and a cursor on it. Tree Lover can control the cursor to apply 5 operations to build a tree, and their formats are following:

0 : Jump to the parent of the current vertex.

1 : Jump to the left child of the current vertex.

2 : Jump to the right child of the current vertex.

3 x : Generate a tree with x vertices arbitrarily and make it the left subtree of the current vertex.

4 x : Generate a tree with x vertices arbitrarily and make it the right subtree of the current vertex.

When applying an operation, the log system will log down a record of it.

Tree Lover played this game for a whole day yesterday. As a forgetful man, although Tree Lover knew the shape of the tree while playing, after a sleep he forgot it.

All he has now is the logs of operations.

Tree Lover wants to know: how many possible shapes of the tree can have yesterday according to the logs?

Can you answer this question?

Input

The input consists of multiple test cases.

For each test case:

The first line is an integer n (1 <= n <= 500), denoting the lines of logs.

Then follow n lines of logs. The formats of logs are as described above.

The integer x of operation 3 and 4 is positive.

In each case, the number of vertices of the tree will never exceed 500.

You can assume that the cursor will never jump to a non-existent vertex.

If the left child of a vertex exists, operation 3 will not be applied on this vertex, and operation 4 is similar.

Output

For each test case, ouput a single line “Case #x: y”, where x is the case number, starting from 1, and y is the answer to Tree Lover’s question.

Because the answer can be large, please output the answer mod 1000000007.

Sample Input

2

3 3

4 3

2

3 3

1

Sample Output

Case #1: 25

Case #2: 5

Hint
Because the tree is a binary tree, if left and right subtrees of a vertex are of different shapes, after swapping them, the new tree is considered different from the original one.

【题意】

　　有一个造二叉树的程序,初始时只有一个节点并且光标指向它,然后进行了

光标指向当前节点的父亲节点.
光标指向当前节点的左儿子.
光标指向当前节点的右儿子.
随机造一棵包含
随机造一棵包含

　　给出这

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<queue>
  7 #include<cmath>
  8 using namespace std;
  9 #define LL long long
 10 #define Mod 1000000007
 11 #define Maxn 510
 12 
 13 LL p[2*Maxn],c[Maxn];
 14 LL f[Maxn][Maxn];
 15 
 16 LL qpow(LL x,LL b)
 17 {
 18     LL ans=1;
 19     while(b)
 20     {
 21         if(b&1) ans=(ans*x)%Mod;
 22         x=(x*x)%Mod;
 23         b>>=1;
 24     }
 25     return ans;
 26 }
 27 
 28 LL get_c(LL n)
 29 {
 30     LL k=qpow(p[n+1],Mod-2),ans=1;
 31     ans=(p[2*n]*k)%Mod;
 32     k=qpow(p[n],Mod-2);
 33     ans=(ans*k)%Mod;
 34     return ans;
 35 }
 36 
 37 void init()
 38 {
 39     p[0]=1;
 40     for(LL i=1;i<=2*Maxn-10;i++) p[i]=(p[i-1]*i)%Mod;
 41     for(LL i=0;i<=Maxn-10;i++) c[i]=get_c(i);
 42     memset(f,0,sizeof(f));
 43     f[0][0]=1;
 44     for(LL i=0;i<=Maxn-10;i++) f[1][i]=c[i];
 45     
 46     for(LL i=2;i<=Maxn-10;i++)
 47      for(LL j=0;j<=Maxn-10;j++)
 48      {
 49          // f[i][j]=0;
 50          for(LL k=0;k<=j;k++)
 51          f[i][j]=(f[i][j]+f[i-1][j-k]*c[k])%Mod;
 52      }
 53       
 54 }
 55 
 56 struct node
 57 {
 58     int x,lc,rc,f;
 59     int sum,is;
 60     bool al,ar;
 61 }t[Maxn];int cnt;
 62 
 63 struct hp
 64 {
 65     int x,id;
 66     bool q;
 67 }qy[Maxn];int ql;
 68 
 69 void upd(int x)
 70 {
 71     t[x].is=0; t[x].sum=1;
 72     t[x].sum+=t[x].al?0:t[t[x].lc].sum;
 73     t[x].sum+=t[x].ar?0:t[t[x].rc].sum;
 74     
 75     int y;
 76     y=t[x].lc?t[t[x].lc].is:1;
 77     if(t[x].al) y=0;
 78     t[x].is+=y;
 79     
 80     y=t[x].rc?t[t[x].rc].is:1;
 81     if(t[x].ar) y=0;
 82     t[x].is+=y;
 83 }
 84 
 85 int n;
 86 bool ffind()
 87 {
 88     int now=1;cnt=1;ql=0;
 89     t[1].lc=t[1].rc=0;
 90     t[1].al=t[1].ar=0;
 91     upd(1);
 92     bool ok=1;
 93     for(int i=1;i<=n;i++)
 94     {
 95         int opt;
 96         scanf("%d",&opt);opt++;
 97         if(opt==1)
 98         {
 99             if(now==1) ok=0;
100             now=t[now].f;
101         }
102         else if(opt==2)
103         {
104             if(!t[now].lc)
105             {
106                 t[++cnt].f=now;
107                 t[cnt].lc=t[cnt].rc=0;
108                 t[cnt].al=t[cnt].ar=0;
109                 upd(cnt);
110                 t[now].lc=cnt;
111                 upd(t[cnt].f);
112             }
113             now=t[now].lc;
114         }
115         else if(opt==3)
116         {
117             if(!t[now].rc)
118             {
119                 t[++cnt].f=now;
120                 t[cnt].lc=t[cnt].rc=0;
121                 t[cnt].al=t[cnt].ar=0;
122                 t[now].rc=cnt;
123             }
124             now=t[now].rc;
125         }
126         else if(opt==4)
127         {
128             if(t[now].al||t[now].lc) ok=0;
129             int x;
130             scanf("%d",&x);
131             qy[++ql].id=now;qy[ql].x=x;qy[ql].q=0;
132             t[now].al=1;
133         }
134         else
135         {
136             if(t[now].ar||t[now].rc) ok=0;
137             int x;
138             scanf("%d",&x);
139             qy[++ql].id=now;qy[ql].x=x;qy[ql].q=1;
140             t[now].ar=1;
141         }
142     }
143     return ok;
144 }
145 
146 void dfs(int x)
147 {
148     if(t[x].lc) dfs(t[x].lc);
149     if(t[x].rc) dfs(t[x].rc);
150     upd(x);
151 }
152 
153 LL get_ans()
154 {
155     LL ans=1;
156     for(int i=1;i<=ql;i++)
157     {
158         int now,y=qy[i].x;
159         if(qy[i].q==0)
160         {
161             if(t[qy[i].id].lc) now=t[t[qy[i].id].lc].is,y-=t[t[qy[i].id].lc].sum;
162             else now=1;
163         }
164         else
165         {
166             if(t[qy[i].id].rc) now=t[t[qy[i].id].rc].is,y-=t[t[qy[i].id].rc].sum;
167             else now=1;
168         }
169         if(y<0) return 0;
170         ans=(ans*f[now][y])%Mod;
171     }
172     return ans;
173 }
174 
175 int main()
176 {
177     int T,kase=0;
178     init();
179     while(scanf("%d",&n)!=EOF)
180     {
181         printf("Case #%d: ",++kase);
182         if(!ffind()) {printf("0\n");continue;}
183         dfs(1);
184         printf("%lld\n",get_ans());
185     }
186     return 0;
187 }

[HDU 5370]

2016-09-21 22:12:01

【HDU 5370】 Tree Maker（卡特兰数+dp）

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原文地址：http://www.cnblogs.com/Konjakmoyu/p/5894381.html

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