P3038 [USACO11DEC] Grass Planting G - 重链剖分

本题可以说是板题 P3384 的弱化版，只不过要改的变成了边权

边权很好处理，只需要将每个边的边权下放到两端点深度比较深的上面就好了（因为每个点连比它浅的节点必定只有一条边）。

那么就将模板改一下就好了

代码如下：

#include <cstdio>
using namespace std;
const int N = 1e5 + 5;
int n, m;
int head[N], e_cnt;
int siz[N], son[N], fa[N], dep[N], id[N], cnt, top[N];
struct edge {
    int to, nxt;
}e[N << 1];
void swap(int &a, int &b) {
    a ^= b;
    b ^= a;
    a ^= b;
    return ;
}
void add(int u, int v) {
    e[++ e_cnt] = (edge){v, head[u]};
    head[u] = e_cnt;
}
void dfs1(int sx, int ffa) {
    fa[sx] = ffa;
    dep[sx] = dep[ffa] + 1;
    siz[sx] = 1;
    int maxn = -1;
    for (int i = head[sx];i;i = e[i].nxt) {
        if (e[i].to == ffa) continue;
        dfs1 (e[i].to, sx);
        siz[sx] += siz[e[i].to];
        if (siz[e[i].to] > maxn) {
            maxn = e[i].to;
            son[sx] = e[i].to;
        }
    }
    return ;
}
void dfs2(int sx, int topf) {
    id[sx] = ++ cnt;
    top[sx] = topf;
    if (son[sx] != 0)
        dfs2 (son[sx], topf);
    for (int i = head[sx];i;i = e[i].nxt) {
        if (e[i].to == fa[sx] || e[i].to == son[sx]) continue;
        dfs2 (e[i].to, e[i].to);
    }
    return ;
}
struct sgt {
    int sum[N << 2], lz[N << 2];
    void PushUp(int u) {
        sum[u] = sum[u << 1] + sum[u << 1 | 1];
        return ;
    }
    void PushDown(int u, int s) {
        if (lz[u]) {
            sum[u << 1] += lz[u] * (s - (s >> 1));
            sum[u << 1 | 1] += lz[u] * (s >> 1);
            lz[u << 1] += lz[u];
            lz[u << 1 | 1] += lz[u];
            lz[u] = 0;
        }
        return ;
    }
    void build(int u, int l, int r) {
        if (l == r) {
            sum[u] = 0;
            return ;
        }
        int mid = l + r >> 1;
        build (u << 1, l, mid);
        build (u << 1 | 1, mid + 1, r);
        PushUp (u);
    }
    void modify(int u, int l, int r, int L, int R, int k) {
        if (L <= l && r <= R) {
            sum[u] += (r - l + 1) * k;
            lz[u] += k;
            return ;
        }
        PushDown (u, r - l + 1);
        int mid = l + r >> 1;
        if (L <= mid)
            modify (u << 1, l, mid, L, R, k);
        if (mid < R)
            modify (u << 1 | 1, mid + 1, r, L, R, k);
        PushUp (u);
        return ;
    }
    void change(int x, int y) {
        while (top[x] != top[y]) {
            if (dep[top[x]] > dep[top[y]])
                swap (x, y);
            modify (1, 1, n, id[top[y]], id[y], 1);
            y = fa[top[y]];
        }
        if (dep[x] > dep[y])
            swap (x, y);
        modify (1, 1, n, id[x] + 1, id[y], 1);
        return ;
    }
    int query(int u, int l, int r, int L, int R) {
        if (L <= l && r <= R)
            return sum[u];
        int ans = 0, mid = l + r >> 1;
        PushDown (u, r - l + 1);
        if (L <= mid)
            ans += query (u << 1, l, mid, L, R);
        if (mid < R)
            ans += query (u << 1 | 1, mid + 1, r, L, R);
        return ans;
    }
}tre;
int main() {
    scanf ("%d%d", &n, &m);
    for (int i = 1;i < n; ++ i) {
        int u, v;
        scanf ("%d%d", &u, &v);
        add(u, v), add(v, u);
    }
    dfs1(1, 1);
    dfs2(1, 1);
    tre.build(1, 1, n);
    while (m --) {
        char opt;
        scanf (" %c", &opt);
        if (opt == 'P') {
            int x, y;
            scanf ("%d%d", &x, &y);
            tre.change (x, y);
        }
        if (opt == 'Q') {
            int x, y;
            scanf ("%d%d", &x, &y);
            if (dep[x] > dep[y])
                printf ("%d\n", tre.query(1, 1, n, id[x], id[x]));
            else printf ("%d\n", tre.query(1, 1, n, id[y], id[y]));
        }
    }
    return 0;
}

轻松 A 掉了这道题目，但是线段树这么长，码着也怪累的，而且还不好调，再看题目我们惊奇地发现每次 Q 操作只找一条边，下放后就相当于只查询单点，那么区间修改，单点查询就可以用树状数组来替代线段树，所以就有了二代代码

#include <cstdio>
#define lowbit(x) (x) & (-x)
using namespace std;
const int N = 1e5 + 5;
int head[N], e_cnt;
int n, m;
int fa[N], siz[N], son[N], dep[N];
int id[N], cnt, top[N];
struct edge {
    int nxt, to;
}e[N << 1];
void add(int u, int v) {
    e[++e_cnt] = (edge){head[u], v};
    head[u] = e_cnt;
}
void swap(int &a, int &b) {
    a ^= b;
    b ^= a;
    a ^= b;
}
void dfs1(int sx, int ffa) {
    fa[sx] = ffa;
    siz[sx] = 1;
    dep[sx] = dep[ffa] + 1;
    int maxn = -1;
    for (int i = head[sx];i;i = e[i].nxt) {
        if (e[i].to == ffa) continue;
        dfs1 (e[i].to, sx);
        if (siz[e[i].to] > maxn) {
            maxn = siz[e[i].to];
            son[sx] = e[i].to;
        }
        siz[sx] += siz[e[i].to];
    }
    return ;
}
void dfs2(int sx, int topf) {
    top[sx] = topf;
    id[sx] = ++ cnt;
    if (son[sx]) dfs2 (son[sx], topf);
    for (int i = head[sx];i;i = e[i].nxt) {
        if (e[i].to == fa[sx] || e[i].to == son[sx]) continue;
        dfs2 (e[i].to, e[i].to);
    }
    return ;
}
int tre[N];
void modify(int x, int k) {
    for (;x <= n;x += lowbit(x))
        tre[x] += k;
    return ;
}
int query(int x) {
    int ans = 0;
    for (;x > 0;x -= lowbit(x))
        ans += tre[x];
    return ans;
}
void change(int u, int v) {
    while (top[u] != top[v]) {
        if (dep[top[u]] > dep[top[v]]) swap (u, v);
        modify (id[top[v]], 1);
        modify (id[v] + 1, -1);
        v = fa[top[v]];
    }
    if (dep[u] > dep[v]) swap (u, v);
    modify (id[u] + 1, 1);
    modify (id[v] + 1, -1);
    return ;
}
int main() {
    scanf ("%d%d", &n, &m);
    for (int i = 1;i < n; ++ i) {
        int u, v;
        scanf ("%d%d", &u, &v);
        add(u, v), add(v, u);
    }
    dfs1 (1, 1);
    dfs2 (1, 1);
    while (m --) {
        char opt;
        scanf (" %c", &opt);
        if (opt == 'P') {
            int u, v;
            scanf ("%d%d", &u, &v);
            change (u, v);
        }
        else if (opt == 'Q') {
            int u, v;
            scanf ("%d%d", &u, &v);
            if (dep[u] > dep[v])
                printf ("%d\n", query (id[u]));
            else printf ("%d\n", query (id[v]));
        }
    }
    return 0;
}