hdu 3078 Network (LCA)
题意:
一棵树,给出点权,问一条树链上第k大的点权,点权可以动态修改。
思路:
暴力即可orz(数据是真的水啊。
求路径上的点的时候需要用到LCA
/* ***********************************************
Author :111qqz
Created Time :2017年07月31日 星期一 01时12分54秒
File Name :3078.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define PB push_back
14#define fst first
15#define sec second
16#define lson l,m,rt<<1
17#define rson m+1,r,rt<<1|1
18#define ms(a,x) memset(a,x,sizeof(a))
19typedef long long LL;
20#define pi pair < int ,int >
21#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=8E4+7;
7int n,q;
8int val[N];
9vector < pi > edge[N];
10int in[N];
11int E[2*N],R[2*N],dis[N],depth[2*N];
12int p;
13int fa[N];
14int dp[2*N][20];
15void dfs( int u,int dep,int d,int pre)
16{
1 // cout<<"u:"<<u<<" dep:"<<dep<<" d:"<<d<<endl;
2 fa[u] = pre;
3 p++;
4 E[p] = u;
5 depth[p] = dep;
6 R[u] = p ;
7 dis[u] = d;
1 int siz = edge[u].size();
2 for ( int i = 0 ; i < siz ; i++)
3 {
4 int v = edge[u][i].fst;
5 if (v==pre) continue;
6 dfs(v,dep+1,d+edge[u][i].sec,u);
1 p++;
2 E[p] = u;
3 depth[p] = dep;
4 }
5}
1int _min( int l,int r)
2{
3 if (depth[l]<depth[r]) return l;
4 return r;
5}
6void rmq_init()
7{
8 for ( int i = 1 ; i <= 2*n+2 ; i++) dp[i][0] = i;
1 for ( int j = 1 ; (1<<j) <= 2*n+2 ; j++)
2 for ( int i = 1 ; i + (1<<j)-1 <= 2*n+2 ; i++)
3 dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
4}
1int rmq_min( int l,int r)
2{
3 if (l>r) swap(l,r);
4 int k = 0 ;
5 while (1<<(k+1)<=r-l+1) k++;
6 return _min(dp[l][k],dp[r-(1<<k)+1][k]);
7}
8int D(int u,int v) //计算u,v点的距离
9{
10 int LCA = E[rmq_min(R[u],R[v])];
11 int res = dis[u] + dis[v] - 2*dis[LCA];
12 return res;
13}
14void get_path(vector<int> &v,int s,int t)
15{
16 while (s!=t)
17 {
18 v.PB(val[s]);
19 s = fa[s];
20 }
21}
22bool cmp(int a ,int b){return a>b;}
23void solve(int k,int u,int v)
24{
25 int LCA = E[rmq_min(R[u],R[v])];
26 vector<int>path;
27 get_path(path,u,LCA);
28 get_path(path,v,LCA);
29 path.PB(val[LCA]);
30 sort(path.begin(),path.end(),cmp);
31 int siz = path.size();
32 // for ( int i = 0 ; i < siz ; i++) printf("%d ",path[i]);
33 // printf("\n");
34 if (k>siz)
35 {
36 puts("invalid request!");
37 return;
38 }
39 printf("%d\n",path[k-1]);
1}
2int main()
3{
4 #ifndef ONLINE_JUDGE
5 freopen("./in.txt","r",stdin);
6 #endif
7 while (~scanf("%d%d",&n,&q))
8 {
9 ms(val,0);
10 for ( int i = 1 ;i <= n ; i++) scanf("%d",&val[i]);
11 for ( int i = 1 ; i <= n-1 ; i++)
12 {
13 int u,v;
14 scanf("%d%d",&u,&v);
15 edge[u].PB(MP(v,1));
16 edge[v].PB(MP(u,1));
17 }
18 p = 0 ;
19 dfs(1,0,0,-1);
20 rmq_init();
21 while (q--)
22 {
23 int k,u,v;
24 scanf("%d%d%d",&k,&u,&v);
25 if (k==0)
26 {
27 val[u] = v;
28 }
29 else
30 {
31 solve(k,u,v);
32 }
33 }
34 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}