[ZJOI 2018]历史

标签：head   ons   ret   mat   tmp   names   char   long   c++   

题意:给定一棵树和点的\(Access\)次数，求切换链的最大值。

考虑修改时实边与虚边的贡献，用\(LCT\)维护此树。

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2000010;
const int INF = 0x7fffffff;
#define int long long
inline int read()
{
    int q=0,f=1;char ch=getchar();
    while(!isdigit(ch)){
        if(ch==‘-‘) f=-1;ch=getchar();
    }
    while(isdigit(ch)){
        q=q*10+ch-‘0‘;ch=getchar();
    }
    return q*f;
}
int head[maxn];
int cnt;
struct edge
{
    int nxt;
    int to;
}e[maxn<<1];
inline void add(int u,int v){
    e[++cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt;
    return;
}

int ans;

struct LCT{
    int ch[maxn][2];
    int fa[maxn];
    int stack[maxn];
    int val[maxn];
    int sum[maxn];
    int isum[maxn];
    inline bool isroot(int x){
        return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
    }
    inline void push_up(int now){
        sum[now] = val[now] + sum[ch[now][0]]+sum[ch[now][1]]+isum[now];
    }
    inline void rotate(int x){
        int f=fa[x];
        int ff = fa[f];
        int y = (ch[f][0] == x);
        bool flag = isroot(f);
        fa[x] = ff;
        fa[f] = x;
        fa[ch[x][y]] = f;
        if(!flag) ch[ff][ch[ff][1]==f] = x;
        ch[f][!y] = ch[x][y];
        ch[x][y] = f;
        push_up(f);
    }
    inline void splay(int x){
        while(!isroot(x)){
            int f = fa[x];
            if(!isroot(f)){
                if((ch[fa[f]][0] == f)^(ch[f][0] == x)) rotate(x);
                else rotate(f);
            }
            rotate(x);
        }
        push_up(x);
    }
    inline void access(int x,int v,int nt){
        for(;x;nt = x,x = fa[x]){
            splay(x);
            int res = sum[ch[x][1]] + val[x] + isum[x];
            if(ch[x][1]){
                ans -=((res-sum[ch[x][1]]) << 1);
            }
            else if(res + 1 <= val[x] * 2){
                ans -= 2*(res - val[x]);
            }
            else ans -= (res - 1);
            sum[x] += v;
            isum[x] += v;
            res += v;
            if(res+1 > sum[ch[x][1]]*2){
                isum[x] += sum[ch[x][1]];
                ch[x][1] = 0;
            }
            if(res + 1<= sum[nt]*2){
                ch[x][1] = nt;
                isum[x] -= sum[ch[x][1]];
            }
            if(ch[x][1]){
                ans += 2*(res - sum[ch[x][1]]);
            }
            else if(res + 1 <= val[x]*2){
                ans += 2*(res - val[x]);
            }
            else ans += (res - 1);
        }
    }
    inline void update(int x,int f){
        splay(x);
        int res = sum[ch[x][1]] + val[x] + isum[x];
        if(ch[x][1]) ans -= 2*(res - sum[ch[x][1]]);
        else if(res + 1 <= val[x] * 2){
            ans -= 2*(res - val[x]);
        }
        else ans -= (res - 1);
        val[x] += f;
        sum[x] += f;
        res += f;
        if(res + 1 > sum[ch[x][1]] * 2){
            isum[x] += sum[ch[x][1]];
            ch[x][1] = 0;
        }
        if(ch[x][1]){
            ans += 2*(res - sum[ch[x][1]]);
        }
        else if(res + 1 <= val[x] * 2){
            ans += (res - val[x]) * 2;
        }
        else ans += (res - 1);
        access(fa[x],f,x);
    }
    inline void dfs(int x,int f){
        fa[x] = f;
        sum[x] = val[x];
        int maxm = val[x];
        int i;
        int tmp = x;
        for(int i = head[x];i;i=e[i].nxt){
            int y = e[i].to;
            if(y == f) continue;
            dfs(y,x);
            sum[x] += sum[y];
            if(sum[y] > maxm){
                tmp = y;
                maxm = sum[y];
            }
        }
        ans += min(sum[x]-1,2*(sum[x]-maxm));
        if(tmp != x && sum[x] + 1 <= maxm * 2) ch[x][1] = tmp;
        isum[x] = sum[x] - val[x] - sum[ch[x][1]];
    }
    inline void ins(){
        dfs(1,0);
    }
}lct;

signed main()
{
    int n = read(),m=read();
    for(int i = 1;i <= n; ++i){
        lct.val[i] = read();
    }
    for(int i = 1;i < n; ++i){
        int u=read(),v=read();
        add(u,v);
        add(v,u);
    }
    lct.ins();
    cout<<ans<<endl;
    for(int i = 1;i <=m; ++i){
        int x = read(),v = read();
        lct.update(x,v);
        cout<<ans<<endl;    
    }
    return 0;
    
}

[ZJOI 2018]历史

标签：head   ons   ret   mat   tmp   names   char   long   c++   

原文地址：https://www.cnblogs.com/akoasm/p/9419221.html