The 15th UESTC Programming Contest Preliminary B - B0n0 Path cdoj1559

地址：http://acm.uestc.edu.cn/#/problem/show/1559

题目：

B0n0 Path

Time Limit: 1500/500MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

There is a country with NN cities, there are roads between some pairs of cities.

For every pair of cities, there is exactly one path between them, on which there are no cities passed more than once.

So it is obvious that there are N−1N−1 roads and N(N−1)2N(N−1)2 paths.

The cities number from 11 to NN, and the ithith city has one number AiAi.

A path is Bono if and only if the numbers AiAi on the path are non-strictly increasing or decreasing.

i.e., if the numbers are Ab1,Ab2,…,AbmAb1,Ab2,…,Abm,

Abi≤Abi+1 ,∀1≤i<mAbi≤Abi+1 ,∀1≤i<m

or

Abi≥Abi+1 ,∀1≤i<mAbi≥Abi+1 ,∀1≤i<m

should be satisfied.

How many Bono paths in the country?

Input

The first line contains one integer NN.

The second line contains NN integers, where the ithith indicates AiAi.

For the next N−1N−1 lines, each line contains two integers u,vu,v, meaning that there is a road between uthuth city and vthvth city.

1≤N≤105,1≤u,v≤N,1≤Ai≤1091≤N≤105,1≤u,v≤N,1≤Ai≤109

Output

One integer indicating the number of bono paths in the country.

Sample input and output

Sample Input	Sample Output
4 1 7 1 9 1 3 1 4 2 1	5
6 1 1 2 2 3 3 1 2 2 3 3 4 4 5 5 6	15

Hint

For sample 1:

The format of the text on ithith node is (i:Ai)(i:Ai).

There are 5 bono paths: (1,2),(1,3),(1,4),(2,1,3),(3,1,4)(1,2),(1,3),(1,4),(2,1,3),(3,1,4), while path (2,1,4)(2,1,4) is not bono path.

For sample 2:

All path are bono paths.

Source

The 15th UESTC Programming Contest Preliminary

思路：树形dp

　　dp[x][0]：表示从x的子孙节点出发的路径，并且路径上所有点权值都相等的路径数

　　dp[x][0]：表示从x的子孙节点出发的路径，并且路径上所有点权值是严格递增的路径数

　　dp[x][0]：表示从x的子孙节点出发的路径，并且路径上所有点权值是严格递减的路径数

具体转移过程见代码吧：

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;
 vector<int>mp[K];
 int n,val[K];
 LL dp[K][];
 LL ans;

 void dfs(int x,int f)
 {
     for(int i=;i<mp[x].size();i++)
     {
         int v=mp[x][i];
         if(v==f) continue;
         dfs(v,x);
         if(val[v]<val[x])
         {
             ans+=(dp[x][]+dp[x][])*(dp[v][]+dp[v][]+);
             dp[x][]+=dp[v][]+dp[v][]+;
         }
         else if(val[v]>val[x])
         {
             ans+=(dp[x][]+dp[x][])*(dp[v][]+dp[v][]+);
             dp[x][]+=dp[v][]+dp[v][]+;
         }
         else
         {
             ans+=dp[x][]*(dp[v][]+dp[v][]+dp[v][]+)+dp[x][]*(dp[v][]+dp[v][]+)+dp[x][]*(dp[v][]+dp[v][]+);
             dp[x][]+=dp[v][]+;
             dp[x][]+=dp[v][];
             dp[x][]+=dp[v][];
         }
     }
     ans+=dp[x][]+dp[x][]+dp[x][];
     //cout<<"x="<<x<<" "<<dp[x][0]<<" "<<dp[x][1]<<" "<<dp[x][2]<<" "<<ans<<endl;
     //printf("x=%I64d %I64d %I64d %I64d\n",x,dp[x][0],dp[x][1],dp[x][2]);
 }

 int main(void)
 {
     cin>>n;
     for(int i=;i<=n;i++)
         scanf("%d",val+i);
     for(int i=,x,y;i<n;i++)
         scanf("%d%d",&x,&y),mp[x].PB(y),mp[y].PB(x);
     dfs(,);
     cout<<ans<<endl;
     return ;
 }