$ \rightarrow $ 戳我进CF原题

E. Tourists

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time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output

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There are $ n $ cities in Cyberland, numbered from $ 1 $ to $ n $ , connected by m bidirectional roads.
The $ j $ -th road connects city $ a_j $ and $ b_j $ .
　
For tourists, souvenirs are sold in every city of Cyberland. In particular, city $ i $ sell it at a price of $ w_i $ .
　
Now there are $ q $ queries for you to handle. There are two types of queries:
　

• " C $ a w $ ": The price in city $ a $ is changed to $ w $.

• " A $ a b $ ": Now a tourist will travel from city $ a $ to $ b $ .
  He will choose a route, he also doesn't want to visit a city twice.
  He will buy souvenirs at the city where the souvenirs are the cheapest (possibly exactly at city $ a $ or $ b $ ).
  You should output the minimum possible price that he can buy the souvenirs during his travel.
  　

More formally, we can define routes as follow:
　

• A route is a sequence of cities $ [x_1, x_2, ..., x_k] $ , where $ k $ is a certain positive integer.

• For any $ 1 ≤ i < j ≤ k, xi ≠ xj $ .

• For any $ 1 ≤ i < k $ , there is a road connecting $ x_i $ and $ x_{i + 1} $ .

• The minimum price of the route is $ min(w_{x_1}, w_{x_2}, ..., w_{x_k}) $ .

• The required answer is the minimum value of the minimum prices of all valid routes from $ a $ to $ b $ .
  　

Input

The first line of input contains three integers $ n, m, q (1 ≤ n, m, q ≤ 10^5) $ , separated by a single space.
　
Next $ n $ lines contain integers $ w_i (1 ≤ w_i ≤ 10^9) $ .
　
Next $ m $ lines contain pairs of space-separated integers $ a_j $ and $ b_j (1 ≤ a_j, b_j ≤ n, a_j ≠ b_j) $ .
　
It is guaranteed that there is at most one road connecting the same pair of cities.
There is always at least one valid route between any two cities.
　
Next $ q $ lines each describe a query. The format is " C $ a w $ " or " A $ a b $ " $ (1 ≤ a, b ≤ n, 1 ≤ w ≤ 10^9) $ .
　

Output

For each query of type "A", output the corresponding answer.
　

Examples

input1

 3 3 31231 22 31 3
A 2 3
C 1 5
A 2 3

output1

 12

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input2

 7 9 412345671 22 51 52 33 42 45 66 75 7
A 2 3
A 6 4
A 6 7
A 3 3

output2

 2153

　

Note

For the second sample, an optimal routes are:
　
From $ 2 $ to $ 3 $ it is $ [2, 3] $ .
　
From $ 6 $ to $ 4 $ it is $ [6, 5, 1, 2, 4] $ .
　
From $ 6 $ to $ 7 $ it is $ [6, 5, 7] $ .
　
From $ 3 $ to $ 3 $ it is $ [3] $ .

　

题目大意

• $ n $ 个点 $ m $ 条边的无向图，每个点的纪念品都有一个价格，执行 $ q $ 次操作，分为两类

• 改变一个点的纪念品价格

• 询问 $ x $ 到 $ y $ 的任意简单路径上最便宜的纪念品

• $ n,m,q \le 100000

　

题解

结论：一个点数大于等于3的点双连通分量中对于任意不同的三点 $ a,b,c $ ，
必定存在一条简单路径从 $ a $ 走到 $ b $ 经过 $ c $ 。
　

• 一个 $ v-DCC $ 中的点肯定能通过简单路径互相到达

• 把点双连通分量缩点，形成一棵树，树上包括“割点”和“缩点后的新点”

　

-对于每个缩成的点，用 $ set $ 维护对应 $ v-DCC $ 中除了“最高割点”之外的纪念品的最小价格

• 换言之，割点纪念品的价格只在树上“割点”本身和它父亲对应的 $ v-DCC $ 中维护

　

• 最后再加上动态树或树链剖分即可解决

　

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<set>
using namespace std;
#define N 200005
multiset<int>s[N];
multiset<int>::iterator it;
vector<int>e[N],G[N];
stack<int>st;
int n,m,q,dfn[N],low[N],tim,cnt,w[N],bel[N];
bool vis[N];
void tarjan(int u,int fa){dfn[u]=low[u]=++tim; st.push(u); vis[u]=1;for(int i=0;i<e[u].size();++i){int v=e[u][i];if(v==fa) continue;if(!dfn[v]){tarjan(v,u);low[u]=min(low[u],low[v]);if(low[v]>=dfn[u]){++cnt; int tmp;G[u].push_back(cnt);do{tmp=st.top(); st.pop();G[cnt].push_back(tmp);s[cnt].insert(w[tmp]);bel[tmp]=cnt;}while(tmp!=v);w[cnt]=*(s[cnt].begin());}} else if(vis[v])low[u]=min(low[u],dfn[v]);}
}
int siz[N],f[N],dep[N],son[N],top[N],id[N],wt[N];
void dfs1(int u){siz[u]=1; for(int i=0;i<G[u].size();++i){int v=G[u][i];dep[v]=dep[u]+1; f[v]=u;dfs1(v);siz[u]+=siz[v];if(siz[v]>siz[son[u]]) son[u]=v; }
}
void dfs2(int u,int topf){top[u]=topf;wt[id[u]=++tim]=u;if(son[u]) dfs2(son[u],topf);for(int i=0;i<G[u].size();++i){int v=G[u][i];if(v==son[u]) continue;dfs2(v,v);}
}
int sum[N<<2];
void build(int o,int l,int r){if(l==r){sum[o]=w[wt[l]];return;}int mid=l+r>>1;build(o<<1,l,mid); build(o<<1|1,mid+1,r);sum[o]=min(sum[o<<1],sum[o<<1|1]);
}
void updata(int o,int l,int r,int u,int val){if(l==r){sum[o]=val;return;}int mid=l+r>>1;if(u<=mid) updata(o<<1,l,mid,u,val);else updata(o<<1|1,mid+1,r,u,val);sum[o]=min(sum[o<<1],sum[o<<1|1]);
}
int check(int o,int l,int r,int u,int v){if(u<=l&&r<=v){return sum[o];}int res=1e9+7,mid=l+r>>1;if(u<=mid) res=min(res,check(o<<1,l,mid,u,v));if(v>mid) res=min(res,check(o<<1|1,mid+1,r,u,v));return res;
}
void modify(int u,int val){if(bel[u]){it=s[bel[u]].find(w[u]);s[bel[u]].erase(it);}w[u]=val;updata(1,1,cnt,id[u],val);if(bel[u]){s[bel[u]].insert(w[u]);w[bel[u]]=*(s[bel[u]].begin());updata(1,1,cnt,id[bel[u]],w[bel[u]]);}
}
int query(int u,int v){int res=1e9+7;while(top[u]!=top[v]){if(dep[top[u]]<dep[top[v]]) swap(u,v);res=min(res,check(1,1,cnt,id[top[u]],id[u]));u=f[top[u]];}if(dep[u]>dep[v]) swap(u,v);res=min(res,check(1,1,cnt,id[u],id[v]));if(u>n&&f[u]) res=min(res,w[f[u]]);return res;
}
int main(){scanf("%d %d %d",&n,&m,&q);cnt=n;for(int i=1;i<=n;++i) scanf("%d",&w[i]);for(int i=1;i<=m;++i){int u,v;scanf("%d %d",&u,&v);e[u].push_back(v);e[v].push_back(u);}tarjan(1,0);dfs1(1);tim=0;dfs2(1,0);build(1,1,cnt);while(q--){char opt[1]; int x,y;scanf("%s %d %d",opt,&x,&y);if(opt[0]=='C') modify(x,y);else printf("%d\n",query(x,y));}return 0;
}
/*
#         42550351
When      2018-09-06 14:42:23
Who       PotremZ
Problem   E - Tourists
Lang      GNU C++11
Verdict   Accepted
Time      421 ms
Memory    41800 KB
*/