#树链剖分，线段树#洛谷 1505 [国家集训队]旅游
分析
边权那就将边权变为深度更大的点的点权，
相反数会影响最值和求和，剩下就是模板了
注意两次相反数的标记要抵消掉
代码
#include <cstdio>
#include <cctype>
#define rr register
using namespace std;
const int N=200011; struct node{int y,w,next;}e[N<<1];
int top[N],fat[N],wsum[N<<2],wmin[N<<2],wmax[N<<2],lazy[N<<2];
int tot,n,K=1,as[N],dfn[N],dep[N],big[N],son[N],A[N],a[N];
inline signed iut(){
rr int ans=0,f=1; rr char c=getchar();
while (!isdigit(c)) f=(c=='-')?-f:f,c=getchar();
while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
return ans*f;
}
inline void print(int ans){
if (ans<0) putchar('-'),ans=-ans;
if (ans>9) print(ans/10);
putchar(ans%10+48);
}
inline void Swap(int &x,int &y){rr int t=x; x=y; y=t;}
inline signed min(int a,int b){return a<b?a:b;}
inline signed max(int a,int b){return a>b?a:b;}
inline void pup(int k){
wsum[k]=wsum[k<<1]+wsum[k<<1|1],
wmin[k]=min(wmin[k<<1],wmin[k<<1|1]),
wmax[k]=max(wmax[k<<1],wmax[k<<1|1]);
}
inline void ptag(int k){
Swap(wmin[k],wmax[k]),wsum[k]=-wsum[k],
wmin[k]=-wmin[k],wmax[k]=-wmax[k],lazy[k]^=1;
}
inline void pdown(int k){ptag(k<<1),ptag(k<<1|1),lazy[k]=0;}
inline void build(int k,int l,int r){
if (l==r){
wmin[k]=wmax[k]=wsum[k]=a[l];
return;
}
rr int mid=(l+r)>>1;
build(k<<1,l,mid);
build(k<<1|1,mid+1,r);
pup(k);
}
inline void single(int k,int l,int r,int x,int y){
if (l==r) {wmin[k]=wmax[k]=wsum[k]=y,lazy[k]=0; return;}
rr int mid=(l+r)>>1; if (lazy[k]) pdown(k);
if (x<=mid) single(k<<1,l,mid,x,y);
else single(k<<1|1,mid+1,r,x,y);
pup(k);
}
inline void update(int k,int l,int r,int x,int y){
if (l==x&&r==y) {ptag(k); return;}
rr int mid=(l+r)>>1; if (lazy[k]) pdown(k);
if (y<=mid) update(k<<1,l,mid,x,y);
else if (x>mid) update(k<<1|1,mid+1,r,x,y);
else update(k<<1,l,mid,x,mid),update(k<<1|1,mid+1,r,mid+1,y);
pup(k);
}
inline signed query(int k,int l,int r,int x,int y,int z){
if (l==x&&r==y){
if (!z) return wmin[k];
else if (z==1) return wmax[k];
else return wsum[k];
}
rr int mid=(l+r)>>1; if (lazy[k]) pdown(k);
if (y<=mid) return query(k<<1,l,mid,x,y,z);
else if (x>mid) return query(k<<1|1,mid+1,r,x,y,z);
else{
rr int ans1=query(k<<1,l,mid,x,mid,z);
rr int ans2=query(k<<1|1,mid+1,r,mid+1,y,z);
if (z==1) return max(ans1,ans2);
else if (z==2) return ans1+ans2;
else return min(ans1,ans2);
}
}
inline void Update(int x,int y){
for (;top[x]!=top[y];x=fat[top[x]]){
if (dep[top[x]]<dep[top[y]]) x^=y,y^=x,x^=y;
update(1,1,n,dfn[top[x]],dfn[x]);
}
if (dep[x]>dep[y]) x^=y,y^=x,x^=y;
if (dfn[x]<dfn[y]) update(1,1,n,dfn[x]+1,dfn[y]);
}
inline signed Query(int x,int y,int z){
rr int ans=(z==2)?0:((z==1)?-1001:1001);
for (;top[x]!=top[y];x=fat[top[x]]){
if (dep[top[x]]<dep[top[y]]) x^=y,y^=x,x^=y;
rr int tans=query(1,1,n,dfn[top[x]],dfn[x],z);
if (z==1) ans=max(ans,tans);
else if (z==2) ans+=tans;
else ans=min(ans,tans);
}
if (dep[x]>dep[y]) x^=y,y^=x,x^=y;
if (dfn[x]<dfn[y]){
rr int tans=query(1,1,n,dfn[x]+1,dfn[y],z);
if (z==1) ans=max(ans,tans);
else if (z==2) ans+=tans;
else ans=min(ans,tans);
}
return ans;
}
inline void dfs1(int x,int fa){
dep[x]=dep[fa]+1,fat[x]=fa,son[x]=1;
for (rr int i=as[x],mson=-1;i;i=e[i].next)
if (e[i].y!=fa){
dfs1(e[i].y,x); A[e[i].y]=e[i].w;
son[x]+=son[e[i].y];
if (son[e[i].y]>mson) big[x]=e[i].y,mson=son[e[i].y];
}
}
inline void dfs2(int x,int linp){
dfn[x]=++tot,top[x]=linp,a[dfn[x]]=A[x];
if (!big[x]) return; dfs2(big[x],linp);
for (rr int i=as[x];i;i=e[i].next)
if (e[i].y!=fat[x]&&e[i].y!=big[x])
dfs2(e[i].y,e[i].y);
}
signed main(){
n=iut();
for (rr int i=1;i<n;++i){
rr int x=iut()+1,y=iut()+1,w=iut();
e[++K]=(node){y,w,as[x]},as[x]=K;
e[++K]=(node){x,w,as[y]},as[y]=K;
}
dfs1(1,0),dfs2(1,1),build(1,1,n);
rr char ss[5];
for (rr int T=iut();T;--T){
scanf("%s",ss);
rr int x=iut()+1,y=iut()+1;
switch (ss[0]){
case 'C':{
--x; --y;
rr int X=e[x<<1].y,Y=e[x<<1|1].y;
if (dep[X]<dep[Y]) X^=Y,Y^=X,X^=Y;
single(1,1,n,dfn[X],y);
break;
}
case 'N':{
Update(x,y);
break;
}
case 'S':{
print(Query(x,y,2)),putchar(10);
break;
}
case 'M':{
if (ss[1]=='I') print(Query(x,y,0));
else print(Query(x,y,1));
putchar(10);
break;
}
}
}
return 0;
}
