Codeforces 587C Duff in the Army

来源：互联网发布：arnold mac 破解方法编辑：程序博客网时间：2024/05/12 16:10

题意：给出一棵树，每个节点有一个权值集合(一个点有多个值)，现在有q个询问，query（u,v,a）表示询问u到v之间的前a小的数。

与 http://blog.csdn.net/mr_xujh/article/details/47271843 差不多

就是这一题的每个节点上有多个值，不用离散化，查询的时候要输出前a小

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <algorithm>using namespace std;const int N = 200010;const int M = N * 40;int n,m,q,tot;int a[N],t[N];int T[N],ls[M],rs[M],c[M],d[N],fa[N][27];vector<int> vec[N],peo[N],ans;void init(){    memset(fa, -1, sizeof(fa));    tot = 0;}void build(int l, int r, int &rt){rt = tot ++;c[rt] = 0;if(l != r){int mid = (l+r)>>1;build(l, mid, ls[rt] );build(mid+1, r, rs[rt]);}}int update(int root, int pos, int val){int newroot = tot++, tmp = newroot;c[newroot] = c[root] + val;int l = 1, r = m;while(l < r){int mid = (l+r)>>1;if(pos <= mid){ls[newroot] = tot++;rs[newroot] = rs[root];newroot = ls[newroot];root = ls[root];r = mid;}else {rs[newroot] = tot++;ls[newroot] = ls[root];newroot = rs[newroot];root = rs[root];l = mid+1;}c[newroot] = c[root] + val;}return tmp;}int query(int lrt, int rrt, int lca, int k){int lcart = T[lca], fart = T[fa[lca][0]];int l = 1, r = m;int cot = c[lrt] + c[rrt] - c[lcart] - c[fart];if(cot<k)return -1;while(l < r){int mid = (l+r)>>1;cot = c[ls[lrt]] + c[ls[rrt]] - c[ls[lcart]] - c[ls[fart]];if(cot >= k){r = mid;lrt = ls[lrt];rrt = ls[rrt];lcart = ls[lcart];fart = ls[fart];}else {l = mid+1;k -= cot;lrt = rs[lrt];rrt = rs[rrt];lcart = rs[lcart];fart = rs[fart];}}return l;}void dfs_build(int u, int pre){fa[u][0] = pre;d[u] = d[pre] + 1;int sz = peo[u].size();T[u] = T[pre];for(int i = 0; i < sz; i++){        T[u] = update(T[u], peo[u][i], 1);}//cout<<"ok"<<endl;    sz = vec[u].size();    for(int i = 0; i < sz; i++){        int v = vec[u][i];        if(v != pre){            dfs_build(v, u);        }    }}void init_fa(){    for(int j = 1; (1<<j) <= n; j++){        for(int i = 1; i <= n; i++){            fa[i][j] = fa[fa[i][j-1]][j-1];        }    }}int LCA(int a, int b){    int i, j;    if(d[a] < d[b])swap(a, b);    for(i = 0; (1<<i) <= d[a]; i++);i--;    for( j = i; j >= 0; j--){        if(d[a] - (1<<j) >= d[b])            a = fa[a][j];    }    if(a == b)return a;    for(j = i; j >= 0; j--){        if(fa[a][j] != -1 && fa[a][j] != fa[b][j]){            a = fa[a][j], b = fa[b][j];        }    }    return fa[a][0];}int main(){int u,v,k;    //freopen("in.txt","r",stdin);while(~scanf("%d%d%d", &n, &m, &q)){init();for(int i = 1; i < n; i++){scanf("%d%d", &u, &v);vec[u].push_back(v);vec[v].push_back(u);}for(int i = 1; i <= m ; i++ ){            scanf("%d", &u);            peo[u].push_back(i);        }build(1, m, T[0]);dfs_build(1, 0);init_fa();while(q--){            ans.clear();scanf("%d%d%d", &u, &v, &k);int lca = LCA(u, v);//cout<<lca<<endl;for(int i = 1; i <= k; i++){                int tmp = query(T[u],T[v],lca,i);                if(tmp>0)ans.push_back(tmp);                else break;}int sz = ans.size();            printf("%d", sz);            for(int j = 0; j < sz; j++) printf(" %d", ans[j]);            puts("");}for(int i = 1; i <= n; i++){vec[i].clear(),peo[i].clear();}}return 0;}

0 0