传送门
题解:
第一问:求出倍增LCA然后dis加减一下即可。
第二问:按照倍增思想往上跳,预判一下如果要跳过LCA就反着跳。
水题,TLE了两次因为倍增预处理不小心写进for循环了,对过样例不造成影响。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=10004;
int n;
int head[MAXN],edge,dis[MAXN],dep[MAXN];
struct EDGE {
int v,nxt,w;
}e[MAXN<<1];
int f[15][MAXN];
char ss[5];
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
inline void adde(int u,int v,int w) {
e[edge].nxt=head[u],e[edge].v=v,e[edge].w=w,head[u]=edge++;
e[edge].nxt=head[v],e[edge].v=u,e[edge].w=w,head[v]=edge++;
}
void dfs(int p,int fa) {
for (int i=head[p];~i;i=e[i].nxt) {
int v=e[i].v;
if (v^fa) {
f[0][v]=p;
dep[v]=dep[p]+1;
dis[v]=dis[p]+e[i].w;
dfs(v,p);
}
}
}
inline void da() {
for (int j=1;(1<<j)<=n;++j)
for (register int i=1;i<=n;++i)
f[j][i]=f[j-1][f[j-1][i]];
}
inline int LCA(int x,int y) {
if (dep[x]<dep[y]) x^=y^=x^=y;
int t=dep[x]-dep[y];
for (int i=0;i<=14;++i)
if (t&(1<<i)) x=f[i][x];
if (x==y) return x;
for (int i=14;~i;--i)
if (f[i][x]!=f[i][y]) x=f[i][x],y=f[i][y];
return f[0][x];
}
inline int find(int k,int pos) {
for (int i=0;i<=14;++i)
if (k&(1<<i)) pos=f[i][pos];
return pos;
}
int main() {
// freopen("spoj qot2.in","r",stdin);
int kase=read();
while (kase--) {
edge=0;
memset(head,-1,sizeof(head));
n=read();
for (int i=1;i<n;++i) {
int u=read(),v=read(),w=read();
adde(u,v,w);
}
dis[1]=dep[1]=0,dfs(1,0);
da();
while (scanf("%s",ss)&&ss[1]^'O') {
if (ss[0]=='D') {
int x=read(),y=read(),lca=LCA(x,y);
printf("%d\n",dis[x]+dis[y]-(dis[lca]<<1));
}
else {
int x=read(),y=read(),k=read()-1,lca=LCA(x,y);
if (dep[x]-dep[lca]>=k) printf("%d\n",find(k,x));
else printf("%d\n",find(dep[x]+dep[y]-(dep[lca]<<1)-k,y));
}
}
}
return 0;
}