• [题解]luogu P2486 [SDOI2011]染色 树剖


    对于树剖里的线段树,我们多维护一个当前区间左端点的颜色和当前区间右端点的颜色
    所以(pushup)时就变成了这样:

    t[p].lc = t[p << 1].lc, t[p].rc = t[p << 1 | 1].rc;
    t[p].s = t[p << 1].s + t[p << 1 | 1].s;
    if (t[p << 1].rc == t[p << 1 | 1].lc) --t[p].s;
    

    还要注意注意在我们跳链时,假如当前id[top[u]]的颜色与id[fa[top[u]]]的颜色相同,我们要记得给答案减1,因为它们是同种颜色属于同一段而我们把这段切成两段加了两次

    #include <cstdio>
    #include <algorithm>
    
    const int N = 1e5 + 30;
    
    int n, m, len, cnt, x, a, b, c;
    int w[N], f[N], dep[N], fa[N], top[N], siz[N], son[N], dfn[N], id[N];
    char ch[5];
    
    struct Edge
    {
        int to, nx;
    }e[N << 1];
    struct SegmentTree
    {
        int l, r, s, lc, rc, lazy;
    }t[N << 2];
    
    int read()
    {
        int s = 0, ff = 1;
        char cc = getchar();
        while (cc < '0' || cc > '9')
        {
            if (cc == '-') ff = -1;
            cc = getchar();
        }
        while (cc >= '0' && cc <= '9')
        {
            s = s * 10 + cc - '0';
            cc = getchar();
        }
        return s * ff;
    }
    
    inline void add_edge(int from, int to)
    {
        e[++len] = (Edge){to, f[from]};
        f[from] = len;
    }
    
    void dfs1(int p, int from)
    {
        fa[p] = from, dep[p] = dep[fa[p]] + 1, siz[p] = 1;
        for (int i = f[p]; i; i = e[i].nx)
        {   
            if (fa[p] == e[i].to) continue;
            dfs1(e[i].to, p), siz[p] += siz[e[i].to];
            if (siz[e[i].to] > siz[son[p]]) son[p] = e[i].to;
        }
    }
    
    void dfs2(int p, int tp)
    {
        dfn[++cnt] = p, id[p] = cnt, top[p] = tp;
        if (son[p]) dfs2(son[p], tp);
        for (int i = f[p]; i; i = e[i].nx)
            if (e[i].to != son[p] && e[i].to != fa[p]) dfs2(e[i].to, e[i].to);
    }
    
    void build(int p, int l, int r)
    {
        t[p].l = l, t[p].r = r;
        if (l == r) { t[p].s = 1; t[p].lc = t[p].rc = w[dfn[l]]; return ; }
        int mid = l + r >> 1;
        build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r);
        t[p].lc = t[p << 1].lc, t[p].rc = t[p << 1 | 1].rc, t[p].s = t[p << 1].s + t[p << 1 | 1].s;
        if (t[p << 1].rc == t[p << 1 | 1].lc) --t[p].s;
    }
    
    inline void spread(int p)
    {
        if (!t[p].lazy) return ;
        t[p << 1].s = t[p << 1 | 1].s = 1;
        t[p << 1].lazy = t[p << 1].lc = t[p << 1].rc = t[p].lazy;
        t[p << 1 | 1].lazy = t[p << 1 | 1].lc = t[p << 1 | 1].rc = t[p].lazy;
        t[p].lazy = 0;
    }
    
    void change(int p, int l, int r, int col)
    {
        if (t[p].l >= l && t[p].r <= r) 
            { t[p].lc = t[p].rc = t[p].lazy = col, t[p].s = 1; return ; }
        spread(p);
        int mid = t[p].l + t[p].r >> 1;
        if (l <= mid) change(p << 1, l, r, col);
        if (mid < r) change(p << 1 | 1, l, r, col);
        t[p].lc = t[p << 1].lc, t[p].rc = t[p << 1 | 1].rc, t[p].s = t[p << 1].s + t[p << 1 | 1].s;
        if (t[p << 1].rc == t[p << 1 | 1].lc) --t[p].s;
    }
    
    int ask(int p, int l, int r)
    {
        if (t[p].l >= l && t[p].r <= r) return t[p].s;
        spread(p);
        int mid = t[p].l + t[p].r >> 1, tot = 0;
        if (l <= mid && mid < r && t[p << 1].rc == t[p << 1 | 1].lc) --tot;
        if (l <= mid) tot += ask(p << 1, l, r);
        if (mid < r) tot += ask(p << 1 | 1, l, r);
        return tot;
    }
    
    int get(int p, int pos)
    {
        if (t[p].l == t[p].r) return t[p].lc;
        spread(p);
        int mid = t[p].l + t[p].r >> 1;
        if (pos <= mid) return get(p << 1, pos);
        else return get(p << 1 | 1, pos);
    }
    
    inline void treechange(int u, int v, int col)
    {
        while (top[u] != top[v])
        {
            if (dep[top[u]] < dep[top[v]]) std::swap(u, v);
            change(1, id[top[u]], id[u], col);
            u = fa[top[u]];
        }
        if (dep[u] < dep[v]) std::swap(u, v);
        change(1, id[v], id[u], col);
    }
    
    inline int treeask(int u, int v)
    {
        int ans = 0, cu, cv;
        while (top[u] != top[v])
        {
            if (dep[top[u]] < dep[top[v]]) std::swap(u, v);
            ans += ask(1, id[top[u]], id[u]);
            cu = get(1, id[top[u]]), cv = get(1, id[fa[top[u]]]);
            if (cu == cv) --ans;
            u = fa[top[u]];
        }
        if (dep[u] < dep[v]) std::swap(u, v);
        ans += ask(1, id[v], id[u]);
        if (ans == 0) return 1;
        return ans;
    }
    
    int main()
    {
        n = read(), m = read();
        for (int i = 1; i <= n; ++i) w[i] = read();
        for (int i = 1; i < n; ++i)
        {
            a = read(), b = read();
            add_edge(a, b), add_edge(b, a);
        }
        dfs1(1, 0), dfs2(1, 1), build(1, 1, n);
        for (int i = 1; i <= m; ++i)
        {
            scanf ("%s", ch);
            if (ch[0] == 'C')
            {
                a = read(), b = read(), c = read();
                treechange(a, b, c);
            }
            else
            {
                a = read(), b = read();
                printf ("%d
    ", treeask(a, b));
            }    
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/martixx/p/13883351.html
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