BZOJ 2243 染色 | 树链剖分模板题进阶版
这道题呢~就是个带区间修改的树链剖分~
如何区间修改？跟树链剖分的区间询问一个道理，再加上线段树的区间修改就好了。
这道题要注意的是，无论是线段树上还是原树上，把两个区间的信息合并的时候，要注意中间相邻两个颜色是否相同。
这代码好长啊啊啊啊
幸好一次过了不然我估计永远也De不出来
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('
')
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;
}