BZOJ 2243 染色 | 樹鏈剖分模板題進階版

這道題呢~就是個帶區間修改的樹鏈剖分~
如何區間修改？跟樹鏈剖分的區間詢問一個道理，再加上線段樹的區間修改就好了。
這道題要註意的是，無論是線段樹上還是原樹上，把兩個區間的信息合並的時候，要註意中間相鄰兩個顏色是否相同。

這代碼好長啊啊啊啊
幸好一次過了不然我估計永遠也De不出來

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
#define space putchar(' ')
 

#define enter putchar('\n')
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9'
 
)
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}
char readchar(){
    char c;
    while(c = getchar(), c < 'A' || c > 'Z'
 
);
    return c;
}
const int N = 100005;
int n, m;
int ecnt, adj[N], nxt[2*N], go[2*N];
int tot, pos[N], idx[N], fa[N], son[N], sze[N], top[N], dep[N], val[N];
int le[4*N], ri[4*N], data[4*N], lazy[4*N];
void add(int u, int v){
    go[++ecnt] = v;
    nxt[ecnt] = adj[u];
    adj[u] = ecnt;
}
void pushup(int k){
    data[k] = data[k << 1] + data[k << 1 | 1];
    if(ri[k << 1] == le[k << 1 | 1]) data[k]--;
    le[k] = le[k << 1], ri[k] = ri[k << 1 | 1];
}
void pushdown(int k){
    if(lazy[k] == -1) return;
    lazy[k << 1] = lazy[k << 1 | 1] = lazy[k];
    le[k << 1] = le[k << 1 | 1] = lazy[k];
    ri[k << 1] = ri[k << 1 | 1] = lazy[k];
    data[k << 1] = data[k << 1 | 1] = 1;
    lazy[k] = -1;
}
void build(int k, int l, int r){
    lazy[k] = -1;
    if(l == r) return (void)(data[k] = 1, le[k] = ri[k] = val[idx[l]]);
    int mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    pushup(k);
}
void change(int k, int l, int r, int ql, int qr, int x){
    if(ql <= l && qr >= r) return (void)(data[k] = 1, le[k] = ri[k] = lazy[k] = x);
    pushdown(k);
    int mid = (l + r) >> 1;
    if(ql <= mid) change(k << 1, l, mid, ql, qr, x);
    if(qr > mid) change(k << 1 | 1, mid + 1, r, ql, qr, x);
    pushup(k);
}
int query(int k, int l, int r, int ql, int qr){
    if(ql <= l && qr >= r) return data[k];
    pushdown(k);
    int mid = (l + r) >> 1;
    if(qr <= mid) return query(k << 1, l, mid, ql, qr);
    if(ql > mid) return query(k << 1 | 1, mid + 1, r, ql, qr);
    return query(k << 1, l, mid, ql, qr) + query(k << 1 | 1, mid + 1, r, ql, qr) - (ri[k << 1] == le[k << 1 | 1]);
}
int getcol(int k, int l, int r, int p){
    if(lazy[k] != -1) return lazy[k];
    if(l == r) return le[k];
    int mid = (l + r) >> 1;
    if(pos[p] <= mid) return getcol(k << 1, l, mid, p);
    else return getcol(k << 1 | 1, mid + 1, r, p);
}
void path_change(int u, int v, int x){
    if(top[u] == top[v]){
        if(pos[u] > pos[v]) swap(u, v);
        change(1, 1, n, pos[u], pos[v], x);
        return;
    }
    if(dep[top[u]] > dep[top[v]]) swap(u, v);
    change(1, 1, n, pos[top[v]], pos[v], x);
    path_change(u, fa[top[v]], x);
}
int path_query(int u, int v){
    if(top[u] == top[v]){
        if(pos[u] > pos[v]) swap(u, v);
        return query(1, 1, n, pos[u], pos[v]);
    }
    if(dep[top[u]] > dep[top[v]]) swap(u, v);
    int same = (getcol(1, 1, n, top[v]) == getcol(1, 1, n, fa[top[v]]));
    return path_query(fa[top[v]], u) + query(1, 1, n, pos[top[v]], pos[v]) - same;
}
void init(){
    static int que[N], qr;
    que[qr = 1] = 1;
    for(int ql = 1, u; ql <= qr; ql++){
        u = que[ql], sze[u] = 1;
        for(int e = adj[u], v; e; e = nxt[e])
            if(v = go[e], v != fa[u])
                fa[v] = u, dep[v] = dep[u] + 1, que[++qr] = v;
    }
    for(int ql = qr, u; ql; ql--){
        u = que[ql];
        sze[fa[u]] += sze[u];
        if(sze[u] >= sze[son[fa[u]]]) son[fa[u]] = u;
    }
    for(int ql = 1, u; ql <= qr; ql++)
        if(!top[u = que[ql]])
            for(int v = u; v; v = son[v])
                idx[++tot] = v, pos[v] = tot, top[v] = u;
    build(1, 1, n);
}
int main(){
    read(n), read(m);
    for(int i = 1; i <= n; i++)
        read(val[i]);
    for(int i = 1, u, v; i < n; i++)
        read(u), read(v), add(u, v), add(v, u);
    init();
    char op;
    int a, b, c;
    while(m--){
        op = readchar(), read(a), read(b);
        if(op == 'Q') write(path_query(a, b)), enter;
        else read(c), path_change(a, b, c);
    }
    return 0;
}

BZOJ 2243 染色 | 樹鏈剖分模板題進階版