HDU 5379 Mahjong Tree dp Getting Started

Description Little Sun is an artist. Today He is playing mahjong alone. He suddenly feels that the tree in the yard doesn ' t look good. So he wants to decorate the tree (the tree had n Vertexs, indexed from 1 to n.) Thought for a long time, and finally he decides the mahjong to decorate the tree. His mahjong are strange because all of the mahjong tiles had a distinct index. (Little Sun have only n mahjong tiles, and the Mahjong tiles indexed from 1 to n.) He put the Mahjong tiles on the vertexs of the tree. As is known to all, Little sun was an artist. So he want to decorate the tree as beautiful as possible. His decoration rules is as follows: (1) Place exact one mahjong tiles on each vertex. (2) The Mahjong Tiles ' index must be continues which is placed on the son Vertexs of a vertex. (3) The Mahjong Tiles ' index must be continues which is placed on the vertexs of any subtrees. Now he want to know so he can obtain how many different beautiful mahjong tree using these rules, because of the answer Can is very large, you need output the answer modulo 1e9 + 7. Input The first line of the input was a single integer T, indicates the number of test cases. For each test case, the first line contains an integers n. (1 <= n <= 100000) And the next n-1 lines, each line contains the integers UI and VI, which describes a edge of the tree, and Vertex 1 is The root of the tree. Output For each test case, the output one line. The output format is ' case #x: ans ' (without quotes), X is the case number, starting from 1. Sample Input Sample Output

The main topic: there is a tree of n points, the number of 1~n in these n points, to meet the following conditions of the maximum number of fill

1. All sons nodes of any node should be filled with successive numbers.

2. All nodes in any subtree should be filled with successive numbers.

Idea: Define DP[U] to indicate the number of successive numbers that are filled with the subtree of the root of U.

For the current node U, discuss the classification of its son node:

1, you son node if there are more than two nodes is not a leaf node, then dp[u]=0

2, u only one son node v,dp[u] = dp[v]*2;

3, u have two sons node V1,v2, and these two sons nodes are not leaf nodes, dp[u] = Dp[v1]*dp[v2]

...... The general idea is this, the rest of the classification is omitted.

#include <vector>#include<cstdio>#include<cstring>using namespacestd;Const intMAXN =100005;Const intMoD = 1e9 +7;intN;Long LongDP[MAXN];BOOLVis[maxn];vector<int>ADJ[MAXN];Long LongSolveintu) {    if(dp[u]!=-1)returnDp[u]; Vis[u]=true; intx=0, v1=-1, v2=-1; Dp[u]=1; intlen=adj[u].size (); intson=0;  for(intI=0; i<len; i++)    {        intv =Adj[u][i]; if(Vis[v])Continue;        Solve (v); Dp[u]*=Dp[v]; Dp[u]%=MoD; if(dp[v]==1) x + +; Son++; }    if(son-x>2|| dp[u]==0)returndp[u]=0; Else if(son==1) {Dp[u]*=2; returndp[u]%=MoD; }    if(x) {if(son==x| | son-x==1) dp[u]*=2; Dp[u]%=MoD;  while(x) {Dp[u]*=x; Dp[u]%=MoD; X--; }    }    returndp[u]%=MoD;}intMain () {intT; scanf ("%d",&T); intKase=1;  while(t--) {scanf ("%d",&N);  for(intI=1; i<n; i++)        {            intx, y; scanf ("%d%d",&x,&X);            Adj[x].push_back (y);        Adj[y].push_back (x); }        intRoot =1; Memset (DP,-1,sizeof(DP)); memset (Vis,0,sizeof(VIS)); printf ("Case #%d:%i64d\n", kase++, solve (root));  for(intI=1; i<=n; i++) adj[i].clear (); //for (int i=1;i<=n;i++) printf ("%d%d\n", I,dp[i]);    }    return 0;}

HDU 5379 Mahjong Tree dp Getting Started