51nod 1766 线段树维护树的直径

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题意：中文

思路：用线段树维护树的直径，具体的可以参考这篇文章，首先要知道如何合并一个区间，那么要知道左区间的最长路径的两个端点以及右区间的两个端点，合并后的最长路径就是4个端点的6种组合中的一个，知道这个就相对简单点了，路径长度的计算可以用LCA-RMQ来完成

#include <vector>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;typedef unsigned long long ull;const int inf=0x3f3f3f3f;const ll INF=0x3f3f3f3f3f3f3f3fll;const int maxn=100010;struct edge{    int to,cost;    edge(int a,int b){to=a;cost=b;}};vector<edge>G[maxn];bool vis[maxn];int L[maxn*2],E[maxn*2],H[maxn],dis[maxn],dp[2*maxn][20];int k,n;void dfs(int t,int deep){    k++;E[k]=t;L[k]=deep;H[t]=k;    for(unsigned int i=0;i<G[t].size();i++){        edge tt=G[t][i];        if(!vis[tt.to]){            vis[tt.to]=1;            dis[tt.to]=dis[t]+tt.cost;            dfs(tt.to,deep+1);            k++;E[k]=t;L[k]=deep;        }    }}void RMQ_init(){    for(int i=1;i<=2*n-1;i++) dp[i][0]=i;    for(int i=1;(1<<i)<=2*n-1;i++){        for(int j=1;j+(1<<i)-1<=2*n-1;j++){            if(L[dp[j][i-1]]<L[dp[j+(1<<(i-1))][i-1]]) dp[j][i]=dp[j][i-1];            else dp[j][i]=dp[j+(1<<(i-1))][i-1];        }    }}int RMQ(int le,int ri){    le=H[le];ri=H[ri];    if(le>ri) swap(le,ri);    int kk=0;    while((1<<(kk+1))<=ri-le+1) kk++;    if(L[dp[le][kk]]<L[dp[ri-(1<<kk)+1][kk]]) return E[dp[le][kk]];    else return E[dp[ri-(1<<kk)+1][kk]];}int Ts[maxn<<2],Tt[maxn<<2],Tlen[maxn<<2];int u,v,u1,v1,c;int calc(int u,int v){return dis[u]+dis[v]-2*dis[RMQ(u,v)];}void pushup(int node){    int ans=-1;    int len1=calc(Ts[node<<1],Tt[node<<1]);    if(len1>ans) {ans=len1;Ts[node]=Ts[node<<1];Tt[node]=Tt[node<<1];}    int len2=calc(Ts[node<<1],Ts[node<<1|1]);    if(len2>ans) {ans=len2;Ts[node]=Ts[node<<1];Tt[node]=Ts[node<<1|1];}    int len3=calc(Ts[node<<1],Tt[node<<1|1]);    if(len3>ans) {ans=len3;Ts[node]=Ts[node<<1];Tt[node]=Tt[node<<1|1];}    int len4=calc(Tt[node<<1],Ts[node<<1|1]);    if(len4>ans) {ans=len4;Ts[node]=Tt[node<<1];Tt[node]=Ts[node<<1|1];}    int len5=calc(Tt[node<<1],Tt[node<<1|1]);    if(len5>ans) {ans=len5;Ts[node]=Tt[node<<1];Tt[node]=Tt[node<<1|1];}    int len6=calc(Ts[node<<1|1],Tt[node<<1|1]);    if(len6>ans) {ans=len6;Ts[node]=Ts[node<<1|1];Tt[node]=Tt[node<<1|1];}    Tlen[node]=ans;}void buildtree(int le,int ri,int node){    if(le==ri){        Ts[node]=Tt[node]=le;        Tlen[node]=0;        return ;    }    int t=(le+ri)>>1;    buildtree(le,t,node<<1);    buildtree(t+1,ri,node<<1|1);    pushup(node);}void query(int l,int r,int le,int ri,int node,int &tmp1,int &tmp2){    if(l<=le&&ri<=r){        tmp1=Ts[node];tmp2=Tt[node];        return ;    }    int t=(le+ri)>>1,ls,lt,rs,rt;    if(r<=t) query(l,r,le,t,node<<1,tmp1,tmp2);    else if(l>t) query(l,r,t+1,ri,node<<1|1,tmp1,tmp2);    else{        query(l,r,le,t,node<<1,ls,lt);        query(l,r,t+1,ri,node<<1|1,rs,rt);        int ans=-1;        if(calc(ls,lt)>ans){ans=calc(ls,lt);tmp1=ls;tmp2=lt;}        if(calc(ls,rs)>ans){ans=calc(ls,rs);tmp1=ls;tmp2=rs;}        if(calc(ls,rt)>ans){ans=calc(ls,rt);tmp1=ls;tmp2=rt;}        if(calc(lt,rs)>ans){ans=calc(lt,rs);tmp1=lt;tmp2=rs;}        if(calc(lt,rt)>ans){ans=calc(lt,rt);tmp1=lt;tmp2=rt;}        if(calc(rs,rt)>ans){ans=calc(rs,rt);tmp1=rs;tmp2=rt;}    }}inline int getint(){    int res=0;    char c=getchar();    bool mi=false;    while(c<'0' || c>'9') mi=(c=='-'),c=getchar();    while('0'<=c && c<='9') res=res*10+c-'0',c=getchar();    return mi ? -res : res;}int main(){    int m;    while(scanf("%d",&n)!=-1){        for(int i=0;i<=n;i++) G[i].clear();        memset(dis,0,sizeof(dis));        memset(vis,0,sizeof(vis));        for(int i=0;i<n-1;i++){            u=getint();v=getint();c=getint();            G[u].push_back(edge(v,c));            G[v].push_back(edge(u,c));        }        k=0;vis[1]=1;dfs(1,1);RMQ_init();        buildtree(1,n,1);        scanf("%d",&m);        while(m--){            scanf("%d%d%d%d",&u,&v,&u1,&v1);            int ans=0,ans1,ans2,ans3,ans4;            query(u,v,1,n,1,ans1,ans2);            query(u1,v1,1,n,1,ans3,ans4);            ans=max(ans,calc(ans1,ans3));            ans=max(ans,calc(ans1,ans4));            ans=max(ans,calc(ans2,ans3));            ans=max(ans,calc(ans2,ans4));            printf("%d\n",ans);        }    }    return 0;}