3784: 树上的路径

3784: 树上的路径

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 459  Solved: 152
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Description

给定一个N个结点的树，结点用正整数1..N编号。每条边有一个正整数权值。用d(a,b）表示从结点a到结点b路边上经过边的权值。其中要求a<b.将这n*(n-1)/2个距离从大到小排序，输出前M个距离值。

Input

第一行两个正整数N,M

下面N-1行，每行三个正整数a,b,c(a,b<=N，C<=10000)。表示结点a到结点b有一条权值为c的边。

Output

共M行，如题所述.

Sample Input

5 10
1 2 1
1 3 2
2 4 3
2 5 4

Sample Output

7
7
6
5
4
4
3
3
2
1

HINT

N<=50000,M<=Min(300000,n*(n-1) /2 )

Source

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前k大路径？？借助超级钢琴的思路如果我们还能弄出一堆区间，那么还是能用RMQ + 堆解决的路径很多很多？？ -> 点分治事实上，子树上的路径无非分过根or不过根对每棵子树建立一个dfs序，每个点能走到的点都是一个区间事实上，每个点被用了logn次(因为我们点分治呀)因此，每个点在不同重心下，可以由一个区间的信息转移出解然后就能用超级钢琴的思路写了
O(nlog^n)

#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<bitset>#include<algorithm>#include<cstring>#include<map>#include<stack>#include<set>#include<cmath>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 5E4 + 50;const int INF = ~0U>>1;struct E{int to,w;E(int _to = 0,int _w = 0) {to = _to; w = _w;}};struct data1{int Max[16],pos[16];data1() {memset(Max,0,sizeof(Max)); memset(pos,0,sizeof(pos));}};struct data2{int sum,l,r,Root,va,po;data2(int _sum = 0,int _l = 0,int _r = 0,int _Root = 0,int _va = 0,int _po = 0) {sum = _sum; l = _l; r = _r; Root = _Root; va = _va; po = _po;}bool operator < (const data2 &b) const {return va < b.va;}};int n,m,dfs_clock,O,ma,dfn[maxn],out[maxn],siz[maxn],d[maxn];bool Mark[maxn];vector <E> v[maxn];vector <data1> v2[maxn];priority_queue <data2> Q;void dfs2(int x,int from,int Siz){int Max = 0; siz[x] = 1;for (int i = 0; i < v[x].size(); i++) {int to = v[x][i].to;if (to == from || Mark[to]) continue;dfs2(to,x,Siz); siz[x] += siz[to];Max = max(Max,siz[to]);}Max = max(Max,Siz - siz[x]);if (Max < ma) ma = Max,O = x;}void dfs3(int x,int o,int from,int sum){dfn[x] = dfs_clock++;data1 New; New.Max[0] = sum;  New.pos[0] = dfn[x];v2[o].push_back(New);for (int i = 0; i < v[x].size(); i++) {int to = v[x][i].to;if (to == from || Mark[to]) continue;dfs3(to,o,x,sum + v[x][i].w);}out[x] = dfs_clock - 1;}void dfs4(int x,int o,int from,int l,int r,int va,int po){Q.push(data2(v2[o][dfn[x]].Max[0],l,r,o,va + v2[o][dfn[x]].Max[0],po));for (int i = 0; i < v[x].size(); i++) {int to = v[x][i].to;if (to == from || Mark[to]) continue;dfs4(to,o,x,l,r,va,po);}}void dfs1(int x,int Siz){if (Siz == 1) return;dfs_clock = 0; ma = INF;dfs2(x,0,Siz); int o = O;Mark[o] = 1; dfs3(o,o,0,0);for (int i = 1; i < 16; i++)for (int j = 0; j < dfs_clock; j++) {int k = j + (1<<(i-1));if (k >= dfs_clock) break;if (v2[o][j].Max[i-1] > v2[o][k].Max[i-1])v2[o][j].Max[i] = v2[o][j].Max[i-1],v2[o][j].pos[i] = v2[o][j].pos[i-1];else v2[o][j].Max[i] = v2[o][k].Max[i-1],v2[o][j].pos[i] = v2[o][k].pos[i-1];}for (int i = 0; i < v[o].size(); i++) {int to = v[o][i].to;if (Mark[to]) continue;int l = 0,r = dfn[to] - 1;int len = r - l + 1,va,po;int k = r - (1<<d[len]) + 1;if (v2[o][l].Max[d[len]] > v2[o][k].Max[d[len]])va = v2[o][l].Max[d[len]],po = v2[o][l].pos[d[len]];else va = v2[o][k].Max[d[len]],po = v2[o][k].pos[d[len]];dfs4(to,o,0,l,r,va,po);}for (int i = 0; i < v[o].size(); i++) {int to = v[o][i].to;if (Mark[to]) continue;dfs1(to,siz[to]);}}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);#endifcin >> n >> m;int now = 0;for (int i = 1; i <= n; i++) {if (i >= (1<<(now+1))) ++now;d[i] = now;}for (int i = 1; i < n; i++) {int x,y,z; scanf("%d%d%d",&x,&y,&z);v[x].push_back(E(y,z));v[y].push_back(E(x,z));}dfs1(n/2,n);while (m--) {data2 k = Q.top(); Q.pop();printf("%d\n",k.va);if (k.l < k.po) {int l = k.l,r = k.po - 1;int len = r - l + 1,va,po;int K = r - (1<<d[len]) + 1;if (v2[k.Root][l].Max[d[len]] > v2[k.Root][K].Max[d[len]])va = v2[k.Root][l].Max[d[len]],po = v2[k.Root][l].pos[d[len]];else va = v2[k.Root][K].Max[d[len]],po = v2[k.Root][K].pos[d[len]];Q.push(data2(k.sum,l,r,k.Root,va + k.sum,po));}if (k.po < k.r) {int l = k.po + 1,r = k.r;int len = r - l + 1,va,po;int K = r - (1<<d[len]) + 1;if (v2[k.Root][l].Max[d[len]] > v2[k.Root][K].Max[d[len]])va = v2[k.Root][l].Max[d[len]],po = v2[k.Root][l].pos[d[len]];else va = v2[k.Root][K].Max[d[len]],po = v2[k.Root][K].pos[d[len]];Q.push(data2(k.sum,l,r,k.Root,va + k.sum,po));}}return 0;}