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Bzoj4817:[SDOI2017]树点涂色

时间:2018-02-10 12:47:29      阅读:161      评论:0      收藏:0      [点我收藏+]

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题面

Bzoj

Sol

做个转化
最开始都是虚边
操作\(1\)就是\(LCT\)里的\(Access\)操作
求的就是路径上虚边的个数+1

然后就好办了
用树链剖分+线段树来维护每个点到根虚边的个数的最大值
操作\(1\)\(Access\)时虚实边的转换,要把原来连的点的\(Splay\)的最左边的点在原树中的子树所有点+1,再把现在连的点做同样的操作-1
操作\(2\):单点查询,记\(deep[x]\)\(x\)到根的虚边个数,那么答案就是\(deep[x]+deep[y]-2*deep[lca]+1\)
操作\(3\):子树最大值

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 5);

IL ll Input(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int size[_], Fa[_], deep[_], top[_], son[_], dfn[_], Index, ed[_], id[_];
int fst[_], nxt[_], to[_], cnt, n, m;
int ch[2][_], fa[_];
int mx[_ << 2], tag[_ << 2];

IL void Add(RG int u, RG int v){
    nxt[cnt] = fst[u]; to[cnt] = v; fst[u] = cnt++;
}

# define lson x << 1, l, mid
# define rson x << 1 | 1, mid + 1, r

IL void Build(RG int x, RG int l, RG int r){
    if(l == r){
        mx[x] = deep[id[l]];
        return;
    }
    RG int mid = (l + r) >> 1;
    Build(lson); Build(rson);
    mx[x] = max(mx[x << 1], mx[x << 1 | 1]);
}

IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){
    if(L <= l && R >= r){
        tag[x] += v, mx[x] += v;
        return;
    }
    RG int mid = (l + r) >> 1;
    if(L <= mid) Modify(lson, L, R, v);
    if(R > mid) Modify(rson, L, R, v);
    mx[x] = max(mx[x << 1], mx[x << 1 | 1]) + tag[x];
}

IL int Query(RG int x, RG int l, RG int r, RG int L, RG int R){
    if(L <= l && R >= r) return mx[x];
    RG int mid = (l + r) >> 1, ans = 0;
    if(L <= mid) ans = Query(lson, L, R);
    if(R > mid) ans = max(ans, Query(rson, L, R));
    return ans + tag[x];
}

# undef lson
# undef rson

IL void Dfs1(RG int u){
    size[u] = 1;
    for(RG int e = fst[u]; e != -1; e = nxt[e]){
        if(size[to[e]]) continue;
        fa[to[e]] = Fa[to[e]] = u, deep[to[e]] = deep[u] + 1;
        Dfs1(to[e]);
        size[u] += size[to[e]];
        if(size[to[e]] > size[son[u]]) son[u] = to[e];
    }
}

IL void Dfs2(RG int u, RG int Top){
    dfn[u] = ++Index, top[u] = Top, id[Index] = u;
    if(son[u]) Dfs2(son[u], Top);
    for(RG int e = fst[u]; e != -1; e = nxt[e])
        if(!dfn[to[e]]) Dfs2(to[e], to[e]);
    ed[u] = Index;
}

IL int LCA(RG int u, RG int v){
    while(top[u] ^ top[v]){
        if(deep[top[u]] > deep[top[v]]) swap(u, v);
        v = Fa[top[v]];
    }
    return deep[u] > deep[v] ? v : u;
}

IL int Son(RG int x){  return ch[1][fa[x]] == x;  }

IL int Isroot(RG int x){  return ch[0][fa[x]] != x && ch[1][fa[x]] != x;  }

IL int Find(RG int x){  while(ch[0][x]) x = ch[0][x]; return x;  } 

IL void Rotate(RG int x){
    RG int y = fa[x], z = fa[y], c = Son(x);
    if(!Isroot(y)) ch[Son(y)][z] = x; fa[x] = z;
    ch[c][y] = ch[!c][x]; fa[ch[c][y]] = y;
    ch[!c][x] = y; fa[y] = x;
}

IL void Splay(RG int x){
    for(RG int y = fa[x]; !Isroot(x); Rotate(x), y = fa[x])
        if(!Isroot(y)) Son(x) ^ Son(y) ? Rotate(x) : Rotate(y);
}

IL void Access(RG int x){
    for(RG int y = 0, ff; x; y = x, x = fa[x]){
        Splay(x);
        if(ch[1][x]) ff = Find(ch[1][x]), Modify(1, 1, n, dfn[ff], ed[ff], 1);
        if(y) ff = Find(y), Modify(1, 1, n, dfn[ff], ed[ff], -1);
        ch[1][x] = y;
    }
}

int main(RG int argc, RG char* argv[]){
    n = Input(); m = Input(); Fill(fst, -1);
    for(RG int i = 1, x, y; i < n; ++i)
        x = Input(), y = Input(), Add(x, y), Add(y, x);
    Dfs1(1), Dfs2(1, 1), Build(1, 1, n);
    for(RG int i = 1; i <= m; ++i){
        RG int opt = Input(), x = Input(), y, ans = 0, lca;
        if(opt == 1) Access(x);
        else if(opt == 2){
            y = Input(), lca = LCA(x, y);
            ans = Query(1, 1, n, dfn[x], dfn[x]) + Query(1, 1, n, dfn[y], dfn[y]);
            ans -= 2 * Query(1, 1, n, dfn[lca], dfn[lca]);
            printf("%d\n", ans + 1);
        }
        else printf("%d\n", Query(1, 1, n, dfn[x], ed[x]) + 1);
    }
    return 0;
}

Bzoj4817:[SDOI2017]树点涂色

标签:online   ret   c++   wap   mes   continue   com   getch   splay   

原文地址:https://www.cnblogs.com/cjoieryl/p/8438227.html

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