题意
对一个树上维护两种操作，一种是把x到y间的点都染成c色，另一种是求x到y间的点有多少个颜色块，比如112221由“11”，“222”，“1”三块组成。
思路
这题的关键是要如何维护这个颜色块，我们可以利用线段树，记录每个区间的块数，区间左端点，区间右端点的颜色。合并中如果两个区间相邻点颜色相同，个数要减去1.
查询也是一样的，链与链间的相邻点也要考虑清楚。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ ＼＼ Λ＿Λ 来了老弟 ＼('ㅅ') > ⌒ヽ / へ＼ / / ＼＼ ﾚ ノ ヽ_つ / / / /| ( (ヽ | |、＼ | 丿 ＼ ⌒) | | ) / 'ノ ) Lﾉ */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const ll mod = 2147483648; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 1e5+9; int a[maxn],b[maxn]; int dp[maxn],sz[maxn],fa[maxn],son[maxn]; vector<int>mp[maxn]; void dfs1(int u,int f,int deep){ dp[u] = deep; fa[u] = f; sz[u] = 1; int mx = 0; for(int i=0; i<mp[u].size(); i++){ int v = mp[u][i]; if(v == f) continue; dfs1(v, u, deep+1); sz[u] += sz[v]; if(sz[v] > mx) {mx = sz[v], son[u] = v;} } } int top[maxn],id[maxn],cnt = 0; void dfs2(int u,int f,int topf){ top[u] = topf; id[u] = ++cnt; b[cnt] = a[u]; if(son[u])dfs2(son[u], u, topf); for(int i=0; i<mp[u].size(); i++){ int v = mp[u][i]; if(v == f || v == son[u]) continue; dfs2(v, u, v); } } int tag[maxn<<2],lazy[maxn<<2],ly[maxn<<2],rz[maxn<<2]; void pushup(int rt){ tag[rt] = tag[rt<<1] + tag[rt<<1|1]; rz[rt] = rz[rt<<1]; ly[rt] = ly[rt<<1|1]; if(ly[rt<<1] == rz[rt<<1|1])tag[rt] --; } void build(int l,int r,int rt){ if(l == r){ tag[rt] = 1; ly[rt] = rz[rt] = b[l]; return ; } int mid = (l + r) >> 1; build(l, mid, rt<<1); build(mid+1,r,rt<<1|1); pushup(rt); // cout<<l<<" "<<r<<" "<<rz[rt] << " " << ly[rt]<<endl; } void pushdown(int l,int r,int rt){ tag[rt<<1] = tag[rt<<1|1] = 1; ly[rt<<1] = rz[rt<<1] = lazy[rt]; ly[rt<<1|1] = rz[rt<<1|1] = lazy[rt]; lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt]; lazy[rt] = 0; } void update(int L, int R, int c, int l, int r,int rt){ if(l >= L && r <= R){ tag[rt] = 1; ly[rt] = rz[rt] = c; lazy[rt] = c; return; } int mid = (l + r) >> 1; if(lazy[rt]) pushdown(l, r, rt); if(mid >= L) update(L, R, c, l, mid, rt<<1); if(mid < R) update(L,R,c,mid+1,r,rt<<1|1); pushup(rt); } int n,m; void solve(int x,int y,int c){ while(top[x] != top[y]){ if(dp[top[x]] < dp[top[y]]) swap(x, y); update(id[top[x]], id[x], c, 1, n, 1); x = fa[top[x]]; } if(dp[x] > dp[y]) swap(x, y); update(id[x], id[y], c, 1, n, 1); } int query(int L,int R, int l,int r,int rt,int &tmpl,int &tmpr){ if(l >= L && r<= R){ if(l == L) tmpl = rz[rt]; if(r == R) tmpr = ly[rt]; return tag[rt]; } int mid = (l + r) >> 1; if(lazy[rt])pushdown(l, r, rt); int res = 0; int ql = -1, qr = -1; if(mid >= L) { res += query(L, R, l, mid, rt<<1,tmpl,tmpr); ql = ly[rt<<1]; } if(mid < R){ res += query(L, R, mid +1, r, rt<<1|1,tmpl,tmpr); qr = rz[rt<<1|1]; } if(ql == qr && ql != -1) res --; pushup(rt); return res; } int cal(int x,int y){ int res = 0,lax = -1,lay=-1; while(top[x] != top[y]){ int tmpl = -1,tmpr=-1; if(dp[top[x]] > dp[top[y]]) { res += query(id[top[x]], id[x], 1, n, 1, tmpl, tmpr);//tmp,flag 1,r if(lax == tmpr) res--; lax = tmpl; x = fa[top[x]]; } else { res += query(id[top[y]], id[y], 1, n, 1,tmpl,tmpr); if(lay == tmpr) res--; lay = tmpl; y = fa[top[y]]; } // cout<<tmpl<<" !! "<<tmpr<<endl; } if(dp[x] > dp[y]){ int tmpl = -1,tmpr=-1; res += query(id[y], id[x], 1, n, 1,tmpl,tmpr); if(tmpr == lax) res--; if(tmpl == lay) res--; } else { int tmpl = -1,tmpr=-1; res += query(id[x], id[y], 1, n, 1,tmpl,tmpr); if(tmpl == lax) res--; if(tmpr == lay) res--; } return res; } int main(){ scanf("%d%d", &n, &m); rep(i, 1, n) scanf("%d", &a[i]); rep(i, 1, n-1) { int u,v; scanf("%d%d", &u, &v); mp[u].pb(v); mp[v].pb(u); } dfs1(1, 1, 1); dfs2(1, 1, 1); build(1, n, 1); while(m--){ char str[5]; scanf("%s", str); if(str[0] == 'C') { int x,y,z; scanf("%d%d%d", &x, &y, &z); solve(x, y, z); } else { int x,y; scanf("%d%d", &x, &y); printf("%d\n", cal (x,y)); } } return 0; }