hdu 4603 Color the Tree

这道题细节真的非常多

首先能够想到a和b的最优策略一定是沿着a和b在树上的链走，走到某个点停止，然后再依次占据和这个点邻接的边

所以，解决这道题的过程例如以下：

预处理阶段：

step 1：取随意一个点为根节点。找出父子关系而且对这个树进行dp。求出从某个节点出发往下所包括的全部边的权值总和  复杂度O（n）

step 2：从tree dp 的结果中计算对于某个节点。从某条边出发所包括的边的综合。而且对其从大到小进行排序 复杂度O（n*logn）

step 3：dfs求出这颗树的欧拉回路，以及每一个点的dfn，而且按欧拉回路的顺序计算每一个节点的深度 复杂度O（2*n）

step 4：利用sparse table算法初始化step 3中的深度序列 复杂度 O（n*logn）

step 5：计算出从某个节点往上走2的n次方步所到达的节点  复杂度O（n*logn）

查询阶段：

关键是找到两点的 LCA 以及相遇点，而且找到一条或两条所经过且和相遇点邻接的边

分几种情况讨论

1. 两个点在一起

2.两个点之间的距离为1

3.dep[a] == dep[b]

4.dep[a] > dep[b] + 1

5.dep[a] < dep[b]

6.dep[a] == dep[b]+1

ps：少考虑第六种情况wa了一个下午

#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
#include<ctime>
using namespace std;
#ifdef _WIN32
typedef __int64 i64;
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
typedef long long i64;
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)		memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?

(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const long long inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 100111;
const int maxlevel = 21;

struct Node
{
	int now;
	int to;
	int c;
	int tot;
	int ss;
	const bool operator <  (const Node& cmp) const {
		return tot > cmp.tot;
	}
};

int all;
int n, m;
vector<Node>g[maxn];
int t[maxn];
int dep[maxn];

int df;
int dfn[maxn];
int dfv[maxn * 2];
int st[maxn * 2][maxlevel];
int up[maxn][maxlevel];

int dp[maxn];   //down sum
int xtof[maxn];
int ftox[maxn];

int vis[maxn];

int lg2[maxn*2];

void dfs(int now)
{
	vis[now] = true;
	int to;
	for (int i = 0; i < (int)g[now].size(); i++) {
		to = g[now][i].to;
		if (!vis[to])	{
			t[to] = now;
			dfs(to);
		}
	}
}

int treedp(int now)
{
	int to,id;
	dp[now] = 0;
	for (int i = 0; i < (int)g[now].size(); i++) {
		to = g[now][i].to;
		if (to != t[now]) {
			int temp = treedp(to) + g[now][i].c;
			g[now][i].tot = temp;
			dp[now] += temp;
		}
		else {
			id = i;
		}
	}
	if (t[now] != -1) {
		g[now][id].tot = all - dp[now];
	}
	return dp[now];
}

void euler_circuit(int now ,int step)
{
	dep[now] = step;
	dfn[now] = df;
	dfv[df++] = now;
	int to;
	for (int i = 0; i < (int)g[now].size(); i++) {
		to = g[now][i].to;
		if (to != t[now]) {
			euler_circuit(to,step+1);
			dfv[df++] = now;
		}
	}
}

void get_up_node()
{
	for (int i = 1; i <= n; i++) {
		up[i][0] = t[i];
	}
	int to;
	for (int step = 1; step < maxlevel; step++) {
		for (int now = 1; now <= n; now++) {
			to = up[now][step - 1];
			if (to == -1) {
				up[now][step] = -1;
			}
			else {
				up[now][step] = up[to][step - 1];
			}
		}
	}
}

void sparse_table()
{
	for (int i = 1; i < df; i++){
		st[i][0] = dep[dfv[i]];
	}

	int to;
	for (int step = 1; step <= lg2[n] + 1; step++){
		for (int now = 1; now < df; now++) {
			to = now + (1 << (step - 1));
			if (to < df){
				st[now][step] = min(st[now][step - 1], st[to][step - 1]);
			}
			else{
				st[now][step] = st[now][step - 1];
			}
		}
	}
}

void relation()
{
	int to;
	for (int now = 1; now <= n; now++){
		for (int i = 0; i < (int)g[now].size(); i++){
			to = g[now][i].to;
			if (to == t[now]){
				xtof[now] = i;
			}
			else{
				ftox[to] = i;
			}
		}
	}
}

int rmq(int l,int r)
{
	return min(st[l][lg2[r - l]], st[r - (1 << lg2[r - l])][lg2[r - l]] );
}

int calculate(int x,bool first,int id1,int id2=-1)
{
	if (id2 != -1){
		if (id1 > id2){
			swap(id1, id2);
		}
	}
	int sum = g[x][0].ss;
	sum -= g[x][id1].tot;
	if (id2 != -1){
		sum -= g[x][id2].tot;
	}
	int size = (int)g[x].size() - 1;
	if (size >= 1){
		sum += g[x][1].ss;
	}
	int ans = g[x][0].ss;
	if (id1 % 2 ){
		if (id1 + 1 <= size){
			ans -= g[x][id1 + 1].ss;
			if (id1 + 2 <= size){
				ans += g[x][id1 + 2].ss;
			}
		}
		if (id2 != -1){
			if (id2 % 2){
				ans -= g[x][id2].ss;
				if (id2 + 1 <= size){
					ans += g[x][id2 + 1].ss;
				}
			}
			else{
				if (id2 + 1 <= size){
					ans -= g[x][id2 + 1].ss;
					if (id2 + 2 <= size){
						ans += g[x][id2 + 2].ss;
					}
				}
			}
		}
	}
	else{
		ans -= g[x][id1].ss;
		if (id1 + 1 <= size){
			ans += g[x][id1 + 1].ss;
		}
		if (id2 != -1){
			if (id2 % 2){
				ans -= g[x][id2].ss;
				if (id2 + 1 <= size){
					ans += g[x][id2 + 1].ss;
				}
			}
			else{
				if (id2 + 1 <= size){
					ans -= g[x][id2 + 1].ss;
					if (id2 + 2 <= size){
						ans += g[x][id2 + 2].ss;
					}
				}
			}
		}
	}
	if (first) return ans;
	else return sum - ans;
}

int go_up(int now, int x)
{
	int step = 0;
	while (x) {
		if (x & 1) {
			now = up[now][step];
		}
		step++;
		x >>= 1;
	}
	return now;
}

int find(int a,int b)
{
	int l = dfn[a];
	int r = dfn[b];
	if (l == r){
		return g[a][0].ss;
	}
	if (l > r){
		swap(l, r);
	}
	int lca = rmq(l, r + 1);   //dep
	if (dep[a] - lca + dep[b] - lca == 1){
		if (dep[a] == lca){
			return g[b][xtof[b]].tot + calculate(b, false, xtof[b]);
		}
		else if (dep[b] == lca){
			return g[b][ftox[a]].tot + calculate(b, false, ftox[a]);
		}
	}
	else if (dep[a] > dep[b]+1){
		int temp = dep[a] - dep[b];
		int mid = lca + temp / 2;
		int child = go_up(a, dep[a] - mid - 1);
		if (temp % 2){
			return g[t[child]][ftox[child]].tot + calculate(t[child], false, ftox[child], xtof[t[child]]);
		}
		else{
			return g[t[child]][ftox[child]].tot + calculate(t[child], true, ftox[child], xtof[t[child]]);
		}
	}
	else if (dep[a] == dep[b] + 1) {
		int ca = go_up(a, dep[a] - lca - 1);
		int cb = go_up(b, dep[b] - lca - 1);
		int meet = t[ca];
		return g[meet][ftox[ca]].tot + calculate(meet, false, ftox[ca], ftox[cb]);
	}
	else if (dep[a] < dep[b]){
		int temp = dep[b] - dep[a];
		int mid = lca + (temp + 1)/ 2;
		int child = go_up(b, dep[b] - mid - 1);
		if (temp % 2){
			return g[t[child]][xtof[t[child]]].tot + calculate(t[child], false, xtof[t[child]], ftox[child]);
		}
		else{
			return g[t[child]][xtof[t[child]]].tot + calculate(t[child], true, xtof[t[child]], ftox[child]);
		}
	}
	else if(dep[a] == dep[b]) {
		int ca = go_up(a, dep[a] - lca - 1);
		int cb = go_up(b, dep[b] - lca - 1);
		int meet = t[ca];
		return g[meet][ftox[ca]].tot + calculate(meet, true, ftox[ca], ftox[cb]);
	}
	assert(false);
}

void start()
{
	for (int i = 1; i <= n; i++) {
		vis[i] = false;
	}
	t[0] = t[1] = -1;
	dfs(1);
	treedp(1);

	for (int now = 1; now <= n; now++) {
		sort(g[now].begin(), g[now].end());
		for (int i =(int) g[now].size() - 1; i >= 0; i--) {
			g[now][i].ss = g[now][i].tot;
			if (i + 3 <= (int)g[now].size()) {
				g[now][i].ss += g[now][i + 2].ss;
			}
		}
	}
	df = 1;
	euler_circuit(1, 0);
	get_up_node();
	sparse_table();
	relation();
}

int main()
{
	for (int i = 0; i < maxlevel; i++){
		if ( (1<<i) < maxn*2)
		lg2[1 << i] = i;
	}
	for (int i = 3; i < maxn*2; i++) {
		if (!lg2[i]){
			lg2[i] = lg2[i - 1];
		}
	}

	int T;
	cin >> T;
	while (T--) {
		all = 0;
		cin >> n >> m;
		for (int i = 0; i <= n; i++){
			g[i].clear();
		}
		Node node;
		for (int i = 1; i <= n - 1; i++) {
			//cin >> node.now >> node.to >> node.c;
			SS(node.now);
			SS(node.to);
			SS(node.c);
			g[node.now].push_back(node);
			swap(node.now, node.to);
			g[node.now].push_back(node);
			all += node.c;
		}
		start();
		int a, b;
		for (int i = 1; i <= m; i++){
			//cin >> a >> b;
			SS(a); SS(b);
			//cout << find(a, b) << endl;
			printf("%d\n", find(a, b));
		}
	}
	return 0;
}