标签：int 题解 make fa sec 留坑 dp define
这题卡常……而且目前还没有卡过去
首先以原树重心为根，向所有子树重心连边，可以建立一棵点分树
点分树有两个性质：
一个是树高只有log层
另一个是两点在点分树上的lca一定在原树上两点间的树上路径上
所以在原树上不断删点，并统计当前子树中的信息就好
至于如何统计，令 \(dp[i][j][k][l]\) 表示分治中心为 \(i\) ，到点 \(j\) ，第一条边颜色为 \(k\) ，最后一条边颜色为 \(j\) 的最大得分
转移挺好写的，询问时暴力枚举相关连边的颜色
留个坑，纯点分治还不会写呢
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define reg register int
#define fir first
#define sec second
#define make make_pair
#define pb push_back
#define min2(a, b) ((a)<(b)?(a):(b))
#define max2(a, b) ((a)>(b)?(a):(b))
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, m, q;
int head[N], size;
struct edge{int to, next, val;}e[N*3];
inline void add(int s, int t, int w) {e[++size].to=t; e[size].val=w; e[size].next=head[s]; head[s]=size;}
namespace force{
bool none[N];
int dfs(int u, int to, int fa, int now, int sum) {
//cout<<"dfs "<<u<<' '<<to<<' '<<fa<<' '<<now<<' '<<sum<<endl;
if (u==to) return sum;
int ans=0;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa || none[v]) continue;
ans=max(ans, dfs(v, to, u, e[i].val, sum+(now!=e[i].val)));
}
if (!ans) none[u]=1;
return ans;
}
void solve() {
for (int i=1,u,v,w; i<=m; ++i) {
u=read(); v=read(); w=read();
add(u, v, w); add(v, u, w);
}
q=read();
if (!q) exit(0);
for (int i=1,x,y; i<=q; ++i) {
x=read(); y=read();
memset(none, 0, sizeof(bool)*(n+5));
printf("%d\n", dfs(x, y, 0, 0, 0));
}
exit(0);
}
}
namespace task1{
bool none[N];
int dfs(int u, int to, int fa, int now, int sum, int dis) {
//cout<<"dfs "<<u<<' '<<to<<' '<<fa<<' '<<now<<' '<<sum<<endl;
if (dis>60) {none[u]=1; return 0;}
if (u==to) return sum;
int ans=0;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa || none[v]) continue;
ans=max(ans, dfs(v, to, u, e[i].val, sum+(now!=e[i].val), dis+1));
}
if (!ans) none[u]=1;
return ans;
}
void solve() {
for (int i=1,u,v,w; i<=m; ++i) {
u=read(); v=read(); w=read();
add(u, v, w); add(v, u, w);
}
q=read();
if (!q) exit(0);
for (int i=1,x,y; i<=q; ++i) {
x=read(); y=read();
memset(none, 0, sizeof(bool)*(n+5));
printf("%d\n", dfs(x, y, 0, 0, 0, 0));
}
exit(0);
}
}
namespace task{
int fa[N][25], dep[N], lg[N], rot, siz[N], msiz[N], sumsiz;
bool rm[N];
struct st{int i, j, k; st() {} st(int a, int b, int c):i(a),j(b),k(c){}};
inline bool operator == (st a, st b) {return a.i==b.i&&a.j==b.j&&a.k==b.k;}
inline bool operator < (st a, st b) {return a.k<b.k;}
struct pair_hash{inline size_t operator () (pair<int, int> p) const {return hash<int>()(p.fir*p.sec+p.fir);}};
unordered_map<pair<int, int>, vector<int>*, pair_hash> mp{5000, pair_hash()};
unordered_map<pair<int, int>, vector<st>*, pair_hash> dp{20000, pair_hash()};
unordered_map<pair<int, int>, int, pair_hash> tem{50, pair_hash()};
struct edge{int to, next;}e2[N<<1];
inline void add(int s, int t) {e2[++size].to=t; e2[size].next=head[s]; head[s]=size;}
vector<int> e[N];
void getrt(int u, int fa) {
//cout<<"getrt"<<u<<' '<<fa<<endl;
siz[u]=1; msiz[u]=0;
for (register auto v:e[u]) {
if (!rm[v] && v!=fa) {
getrt(v, u);
siz[u]+=siz[v];
msiz[u]=max(msiz[u], siz[v]);
}
}
//cout<<"sumsiz: "<<sumsiz<<' '<<sumsiz-siz[u]<<' '<<msiz[u]<<' '<<msiz[rot]<<' '<<rot<<endl;
msiz[u]=max(msiz[u], sumsiz-siz[u]);
if (msiz[u]<msiz[rot]) rot=u;
}
void info(int u, int fa) {
//cout<<"info "<<u<<' '<<fa<<endl;
for (register auto v:e[u]) {
if (v==fa || rm[v]) continue;
if (rot==u) {
if (dp.find(make(u, v))==dp.end()) dp[make(u, v)]=new vector<st>;
auto t1=mp[make(u, v)]; auto t2=dp[make(u, v)];
for (register auto it:*t1)
t2->pb(st(it, it, 1));
}
else {
if (dp.find(make(rot, v))==dp.end()) dp[make(rot, v)]=new vector<st>;
for (register auto it:*dp[make(rot, u)])
for (register auto t:*mp[make(u, v)])
tem[make(it.i, t)] = max(tem[make(it.i, t)], it.k+(it.j!=t));
register auto t1=dp[make(rot, v)];
for (register auto it:tem) t1->pb(st(it.fir.fir, it.fir.sec, it.sec));
//assert(tem.size()<=9);
tem.clear();
}
info(v, u);
//cout<<"size: "<<dp.size()<<endl;
}
}
void build(int u) {
//cout<<"build "<<u<<endl;
rm[u]=1;
info(u, 0);
for (register auto v:e[u]) {
rot=0;
if (!rm[v]) {
sumsiz=siz[v];
getrt(v, u);
add(u, rot), add(rot, u);
//info(rot, u, t);
build(rot);
}
}
}
void dfs(int u, int pa) {
//cout<<"dfs "<<u<<' '<<pa<<endl;
for (reg i=1; i<25; ++i)
if (dep[u]>=(1<<i)) fa[u][i] = fa[fa[u][i-1]][i-1];
else break;
for (reg i=head[u],v; ~i; i=e2[i].next) {
v = e2[i].to;
if (v!=pa) dep[v]=dep[u]+1, fa[v][0]=u, dfs(v, u);
}
}
int lca(int a, int b) {
if (dep[a]<dep[b]) swap(a, b);
while (dep[a]>dep[b]) a=fa[a][lg[dep[a]-dep[b]]-1];
if (a==b) return a;
for (reg i=lg[dep[a]]-1; ~i; --i)
if (fa[a][i]!=fa[b][i])
a=fa[a][i], b=fa[b][i];
return fa[a][0];
}
void solve() {
for (reg i=1; i<=n; ++i) lg[i]=lg[i-1]+(1<<lg[i-1]==i);
pair<int, int> t1, t2;
for (reg i=1,u,v,w; i<=m; ++i) {
u=read(); v=read(); w=read();
e[u].pb(v); e[v].pb(u);
t1=make(u, v), t2=make(v, u);
if (mp.find(t1)==mp.end()) mp[t1]=new vector<int>;
mp[t1]->pb(w);
if (mp.find(t2)==mp.end()) mp[t2]=new vector<int>;
mp[t2]->pb(w);
}
for (reg i=1,siz; i<=n; ++i) {
sort(e[i].begin(), e[i].end());
siz=unique(e[i].begin(), e[i].end())-e[i].begin();
e[i].resize(siz);
}
for (register auto it:mp) {
sort(it.sec->begin(), it.sec->end());
int siz=unique(it.sec->begin(), it.sec->end())-it.sec->begin();
it.sec->resize(min2(siz, 3));
}
msiz[0]=sumsiz=n;
getrt(1, 0);
int root=rot;
//cout<<"root: "<<root<<endl;
build(rot);
dep[root]=1;
dfs(root, 0);
//for (auto it:dp) cout<<it.fir.i<<' '<<it.fir.j<<' '<<it.fir.k<<' '<<it.fir.h<<' '<<it.sec<<endl;
q=read(); if (!q) exit(0);
for (reg i=1,x,y,t,ans; i<=q; ++i) {
x=read(); y=read();
if (x==y) {puts("0"); continue;}
t=lca(x, y); ans=0;
//cout<<"lca: "<<x<<' '<<y<<' '<<t<<endl;
if (t==x || t==y) {
//puts("pos1");
if (t!=x) swap(x, y);
for (register auto it:*dp[make(t, y)]) ans=max(ans, it.k);
printf("%d\n", ans);
continue;
}
for (register auto i:*dp[make(t, x)])
for (register auto j:*dp[make(t, y)])
ans = max(ans, i.k+j.k-(i.i==j.i));
printf("%d\n", ans);
}
//cout<<"size: "<<dp.size()<<endl;
exit(0);
}
}
signed main()
{
memset(head, -1, sizeof(head));
n=read(); m=read();
//if (n<100000) force::solve();
//else task1::solve();
task::solve();
return 0;
}
标签：int,题解,make,fa,sec,留坑,dp,define 来源： https://www.cnblogs.com/narration/p/15139768.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。