SDOI2015

核心思路

题目要求的就是关键点的极小联通子树。可以算半个结论题了。这个结论也不难想

再求出图的DFS序之后,所有关键点{a1,a2,...,an}\{a_1,a_2,...,a_n\}的极小联通子树的边权和的两倍为

i=1n1dist(ai,ai+1+dist(an,a1))\sum_{i=1}^{n-1}\mathrm{dist}(a_i,a_{i+1}+\mathrm{dist}(a_n,a_1))

手玩一下很好理解。

那么每次修改xx的贡献就为

dist(x,pre)+dist(x,nxt)dist(pre,nxt)\mathrm{dist}(x,pre)+\mathrm{dist}(x,nxt)-\mathrm{dist}(pre,nxt)

dist\mathrm{dist}就拿LCA直接维护就好了,找前驱和后继用STLset最方便

完整代码

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#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<set>
using namespace std;
#define int ll
#define REP(i,e,s) for(register int i=e; i<=s; i++)
#define DREP(i,e,s) for(register int i=e; i>=s; i--)
#define ll long long
#define DE(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG(a) DE("DEBUG: %d\n",a)
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)

int read() {
int x=0,f=1,ch=getchar();
while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}

const int MAXN=100000+10;

int head[MAXN],_next[MAXN<<1],to[MAXN<<1],weigh[MAXN<<1],cnt;

void addedge(int u,int v,int w) {
cnt++;
_next[cnt]=head[u];
head[u]=cnt;
to[cnt]=v;
weigh[cnt]=w;
}

int dfn[MAXN],tot,fa[MAXN][21],dept[MAXN],dist[MAXN];

int lca(int u,int v) {
if(dept[u]>dept[v]) swap(u,v);
int len=dept[v]-dept[u];
DREP(i,20,0) if((1<<i)&len) v=fa[v][i];
if(u==v) return u;
DREP(i,20,0) if(fa[u][i]!=fa[v][i]) u=fa[u][i],v=fa[v][i];
return fa[u][0];
}

int dis(int u,int v) {
return dist[u]+dist[v]-2*dist[lca(u,v)];
}
int have[MAXN],pos[MAXN],id[MAXN];

void dfs(int u,int f) {
dfn[u]=++tot;id[tot]=u;
for(int i=head[u]; i; i=_next[i]) {
int v=to[i];
if(dfn[v]) continue;
fa[v][0]=u;
dept[v]=dept[u]+1;
dist[v]=dist[u]+weigh[i];
dfs(v,u);
}
}

set<int> s;

signed main() {
int n=read(),m=read();
REP(i,1,n-1) {
int u=read(),v=read(),w=read();
addedge(u,v,w);
addedge(v,u,w);
}

dfs(1,0);


REP(j,1,20) REP(i,1,n) fa[i][j]=fa[fa[i][j-1]][j-1];
int ans=0;
REP(i,1,m) {
int x=dfn[read()],y,z;
set<int> :: iterator it;

if(!have[id[x]]) {
s.insert(x);

y=id[(it=s.lower_bound(x))==s.begin()?*--s.end():*--it];
z=id[(it=s.upper_bound(x))==s.end()?*s.begin():*it];

x=id[x];
ans+=(dis(x,y)+dis(x,z)-dis(y,z));
have[x]=1;

}
else {
y=id[(it=s.lower_bound(x))==s.begin()?*--s.end():*--it];
z=id[(it=s.upper_bound(x))==s.end()?*s.begin():*it];
s.erase(x);
x=id[x];

ans-=(dis(x,y)+dis(x,z)-dis(y,z));
have[x]=0;
}
printf("%lld\n",ans);
}
return 0;
}

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