hdu4912 LCA+贪心

题意:
      给你一棵树和m条边,问你在这些边里面最多能够挑出多少条边,使得这些边之间不能相互交叉。

思路:

     lca+贪心,首先对于给的每个条边,我们用lca求出他们的公共节点,然后在公共节点的深度排序,排序之后我们先从最深的开始,每次判断当前的这条边的两点是否用过,如果没用过,那么就把以当前两点的公共点为树根的子树全部标记上,然后答案+1,就这样一直遍历到最后就行了,一开始敲了一个,TLE了,各种优化之后还是TLE了,最后没办法了,用了一下自己存的一个LCA的模板,这个比我自己写的LCA快,所以就AC了,思路就是贪心+LCA。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>

#define MAXN 110000
#define MAXM 210000

using namespace std;
//*********************************************************
int n;
struct EDGE
{
    int v, next, w;
}edge[MAXM];
int head[MAXN], e;
int index, tmpdfn;
int f[2 * MAXN], id[MAXN], vis[MAXN], pos[MAXN], dis[MAXN];
int mi[2 * MAXN][18];
void init()
{
    memset(head, -1, sizeof(head));
    e = 0;
    index = tmpdfn = 0;
    memset(vis, 0, sizeof(vis));
    dis[1] = 0;
}
void add(int u, int v, int w)
{
    edge[e].v = v;
    edge[e].w = w;
    edge[e].next = head[u];
    head[u] = e++;
}
void dfs(int u)
{
    vis[u] = 1;
    int tmp = ++tmpdfn;
    f[++index] = tmp;
    id[tmp] = u;
    pos[u] = index;
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if(!vis[v])
        {
            dis[v] = dis[u] + edge[i].w;
            dfs(v);
            f[++index] = tmp;
        }
    }
}
void rmqinit(int n, int *w)
{
    for(int i = 1; i <= n; i++) mi[i][0] = w[i];
    int m = (int)(log(n * 1.0) / log(2.0));
    for(int i = 1; i <= m; i++)
        for(int j = 1; j <= n; j++)
        {
            mi[j][i] = mi[j][i - 1];
            if(j + (1 << (i - 1)) <= n) mi[j][i] = min(mi[j][i], mi[j + (1 << (i - 1))][i - 1]);
        }
}
int rmqmin(int l,int r)
{
    int m = (int)(log((r - l + 1) * 1.0) / log(2.0));
    return min(mi[l][m] , mi[r - (1 << m) + 1][m]);
}
int LCA(int l, int r)
{
    if(pos[l] > pos[r]) swap(l, r);
    int ans = rmqmin(pos[l], pos[r]);
    return id[ans];
}
//**********************************************************


typedef struct
{
   int a ,b ,ggdeep ,lca;
}EDGEE;

EDGEE edgee[110000];

bool camp(EDGEE a ,EDGEE b)
{
   return a.ggdeep > b.ggdeep;
}

int use[110000];

void mk_dfs(int x)
{
   use[x] = 1;
   for(int k = head[x] ;k + 1 ;k = edge[k].next)
   {
      int to =  edge[k].v;
      if(use[to] || dis[x] + 1 != dis[to]) continue;
      mk_dfs(to);
   }
}
   


int main()
{    
    int u, v, w, l, r ,m;
    while(~scanf("%d%d", &n, &m))
    {
       
       init();
       for(int i = 1 ;i < n ;i ++)
       {
            scanf("%d %d" ,&u ,&v);
            add(u ,v ,1);
            add(v ,u ,1);
       }
       for(int i = 1 ;i <= m ;i ++)
       scanf("%d %d" ,&edgee[i].a ,&edgee[i].b);
       dfs(1);
       rmqinit(index, f);
       for(int i = 1 ;i <= m ;i ++)
       {
            int lca = LCA(edgee[i].a ,edgee[i].b);
            edgee[i].ggdeep = dis[lca];
            edgee[i].lca = lca;
       }
       
       sort(edgee + 1 ,edgee + m + 1 ,camp);
       memset(use ,0 ,sizeof(use));
       int ans = 0;
       for(int i = 1 ;i <= m ;i ++)
       {
           
            if(use[edgee[i].a] || use[edgee[i].b])
            continue;
            ans ++;
            mk_dfs(edgee[i].lca);
       }
       printf("%d
" ,ans);        
   }
   return 0;
}
    

原文地址:https://www.cnblogs.com/csnd/p/12062898.html