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Codeforces 835 F. Roads in the Kingdom

时间:2018-07-12 18:13:59      阅读:51      评论:0      收藏:0      [点我收藏+]

标签:ram   bool   multiset   code   lock   ret   The   dom   struct   

\(>Codeforces\space835 F. Roads in the Kingdom<\)

题目大意 : 给你一棵 \(n\) 个点构成的树基环树,你需要删掉一条环边,使其变成一颗树,并最小化删掉环边后的树的直径。

\(n \leq 2 \times 10^5\) 树的边权 $ \leq 10^9 $

解题思路 :

考虑最终树的直径可能由两部分组成,答案是其中的最大值

第一种就是外向树内的直径的最大值,这个只需要随便\(dp\)一下即可,这里不过多讨论

第二种情况树的直径经过原来的环,必然是环上的一段加上两端对应的外向树的 \(maxdep\)

\(l_i\) 为将环展开后环上第 \(i\) 个点到环左端点的距离,\(dep_i\) 为 环上第 \(i\) 个点对应的外向树的最大深度

设选取的环的左端点为 \(x\) 右端点为 \(y\) ,那么经过环上 \(x \rightarrow y\) 一段的树的直径就是可以表示为

\(l_y - l_x + dep_x + dep_y\)

问题自此转化为对于每一种把环展开的方式,求 \(\max(l_y - l_x + dep_x + dep_y)\)

\(\max(l_y + dep_y) - \min(l_x - dep_x) \ y \neq x\) 用两个\(set\) 分别维护即可

/*program by mangoyang*/
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
    int f = 0, ch = 0; x = 0;
    for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = 1;
    for(; isdigit(ch); ch = getchar()) x = x * 10 + Qch - 48;
    if(f) x = -x;
}
#define int ll
#define N (600005)
int a[N], b[N], head[N], nxt[N], cnt;
int st[N], ct[N], vis[N], success;
int g[N], dep[N], c[N], s[N], tot, top, n;
struct Node{ 
    int val, id;
    bool operator < (const Node &A) const{ return val < A.val; }
};
multiset<Node> s1, s2;
inline void add(int x, int y, int z){
    a[++cnt] = y, b[cnt] = z; nxt[cnt] = head[x], head[x] = cnt;
}
inline void Getcircle(int u, int fa){
    vis[u] = 1;
    for(int p = head[u]; p; p = nxt[p]){
        int v = a[p];
        if(v == fa) continue;
        if(!vis[v]){
            st[++top] = v, ct[top] = b[p];
            Getcircle(v, u);
        }
        else{
            int pos = 1;
            for(int i = 1; i <= top; i++)
                if(st[i] == v){ pos = i; break; }
            for(int i = pos; i <= top; i++)
                c[++tot] = st[i], s[tot] = ct[i];
            s[1] = b[p];
            return (void) (success = 1);
                    
        }
        if(success) return;
    }
    if(success) return; vis[u] = 0, top--;
}
inline int dfs(int u, int fa){
    int mx = u;
    for(int p = head[u]; p; p = nxt[p]){
        int v = a[p];
        if(v != fa && !vis[v]){
            dep[v] = dep[u] + b[p];
            int now = dfs(v, u);
            if(dep[now] > dep[mx]) mx = now;
        } 
    }
    return mx;    
}
inline int dfs2(int u, int fa, int dis, int t){
    int res = dis;
    for(int p = head[u]; p; p = nxt[p]){
        int v = a[p];
        if(v != fa){
            if(vis[v] && t) continue;
            if(!vis[v]){
                int now = dfs2(v, u, dis + b[p], t);
                if(now > res) res = now;
            }
            else{
                int now = dfs2(v, u, dis + b[p], 1);
                if(now > res) res = now;
            }
        }
    }
    return res;
}
inline ll calc1(){
    multiset<Node>::iterator it2 = s2.begin();
    int id = it2 -> id;
    s1.erase(s1.find((Node){s[id] + g[id], id}));
    multiset<Node>::iterator it1 = s1.end(); it1--;
    int ans = it1 -> val - it2 -> val;
    s1.insert((Node){s[id] + g[id], id});
    return ans;
}
inline ll calc2(){
    multiset<Node>::iterator it1 = s1.end(); it1--;
    int id = it1 -> id;
    s2.erase(s2.find((Node){s[id] - g[id], id}));
    multiset<Node>::iterator it2 = s2.begin();
    int ans = it1 -> val - it2 -> val;
    s2.insert((Node){s[id] - g[id], id});
    return ans;
}
inline ll calc(){ return max(calc1(), calc2()); }
main(){
    read(n);
    if(n <= 2) return puts("0"), 0;
    for(int i = 1, x, y, z; i <= n; i++){
        read(x), read(y), read(z);
        add(x, y, z), add(y, x, z);
    }
    ll ans = 0;
    st[++top] = 1, Getcircle(1, 0);
    for(int i = 1; i <= n; i++) vis[i] = 0;
    for(int i = 1; i <= tot; i++) vis[c[i]] = 1;
    for(int i = 1; i <= tot; i++){
        int mx = dfs(c[i], 0); 
        g[i] = dep[mx], ans = max(ans, g[i]);
        ans = max(ans, dfs2(mx, 0, 0, mx == c[i] ? 1 : 0));
    }
    for(int i = 1; i <= tot; i++) g[i+tot] = g[i];
    for(int i = 1; i <= tot; i++) s[i+tot] = s[i];
    tot *= 2;         
    for(int i = 1; i <= tot; i++) s[i] += s[i-1];
    for(int i = 1; i <= tot / 2; i++){
        s1.insert((Node){s[i] + g[i], i});
        s2.insert((Node){s[i] - g[i], i});
    }
    ll tmp = calc();    
    for(int i = 2; i <= tot / 2; i++){
        int l = i, r = i + tot / 2 - 1;
        s1.insert((Node){s[r] + g[r], r});
        s2.insert((Node){s[r] - g[r], r});
        s1.erase(s1.find((Node){s[l-1] + g[l-1], l - 1}));
        s2.erase(s2.find((Node){s[l-1] - g[l-1], l - 1}));
        tmp = min(tmp, calc());
    }
    cout << Max(ans, tmp);
    return 0;
}                                     

Codeforces 835 F. Roads in the Kingdom

标签:ram   bool   multiset   code   lock   ret   The   dom   struct   

原文地址:https://www.cnblogs.com/mangoyang/p/9300365.html

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