hdu 4603 Color the Tree

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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; }



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