2014多校5（1002）hdu4912(贪心+LCA)

Paths on the tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 424    Accepted Submission(s): 127

Problem Description

bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.

There are m paths on the tree. bobo would like to pick some paths while any two paths do not share common vertices.

Find the maximum number of paths bobo can pick.

 

Input

The input consists of several tests. For each tests:

The first line contains n,m (1≤n,m≤10 ⁵). Each of the following (n - 1) lines contain 2 integers a _i,b _i denoting an edge between vertices a _i and b _i (1≤a _i,b _i≤n). Each of the following m lines contain 2 integers u _i,v _i denoting a path between vertices u _i and v _i (1≤u _i,v _i≤n).

 

Output

For each tests:

A single integer, the maximum number of paths.

 

Sample Input

   
   

    
    3 2
1 2
1 3
1 2
1 3
7 3
1 2
1 3
2 4
2 5
3 6
3 7
2 3
4 5
6 7
   
   

 

Sample Output

   
   

    
    1
2
   
   

题意：给出了m条路径，需要你选择尽量多的路径使得两两之间没有公共点

思路：以任意一点为根dfs树，然后优先选择路径中最小深度的节点（可以用LCA求）最深的路径，每次选完一条路径以后将该路径中以最小深度节点为根的子树标记（之后不能再选），然后就没了。

#include 
   
   
    
    
#include 
    
    
     
     
#include 
     
     
      
      
#include 
      
      
       
       
#include 
       
       
         #pragma comment(linker, "/STACK:16777216") using namespace std; #define maxn 100010 int m,n; int head[maxn]; int edge[maxn<<1]; int next[maxn<<1]; int d; int cent[maxn]; int dp[maxn][20]; int fa[maxn]; int vis[maxn]; struct node{ int i; int p; int u; int v; }; node vex[maxn]; int maxi(int x,int y) { return x>y?x:y; } void add(int u,int v) { edge[d]=v; next[d]=head[u]; head[u]=d++; } void dfs(int u,int pre,int ce) { int i; cent[u]=ce; fa[u]=pre; for(i=head[u];i!=-1;i=next[i])if(edge[i]!=pre)dfs(edge[i],u,ce+1); } void dfs2(int u,int pre) { int i; vis[u]=1; for(i=head[u];i!=-1;i=next[i])if(!vis[edge[i]]&&edge[i]!=pre)dfs2(edge[i],u); } void LCA() { int i,j; memset(dp,0,sizeof(dp)); for(i=1;i<=n;i++)dp[i][0]=fa[i]; for(j=1;j<=19;j++) for(i=1;i<=n;i++)dp[i][j]=dp[dp[i][j-1]][j-1]; } int query(int u,int v) { int temp,i; if(u==v)return u; if(cent[u] 
        
          =0;i--)if(dp[u][i]!=dp[v][i]){ u=dp[u][i]; v=dp[v][i]; } return fa[u]; } bool cmp(node a,node b) { return a.i>b.i; } int main() { while(scanf("%d%d",&n,&m)!=EOF) { int i,j,u,v; d=0; memset(head,-1,sizeof(head)); memset(vis,0,sizeof(vis)); for(i=1;i