[Bzoj2243][SDOI2011]染色(线段树&&树剖||LCT)

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=2243

线段树+树链剖分，在线段树需要每次用lt和rt两个数组记录当前区间的左右边界的颜色，向上更新时需要判断左区间的右边界是否和右区间的左边界相等。在剖分求LCA的过程中需要在求值之后查询与下一次求值的边界是否相等。

 #include<bits/stdc++.h>
 #define lson l,mid,i<<1
 #define rson mid+1,r,i<<1|1
 using namespace std;
 typedef long long ll;
 const int maxn = ;
 const int INF = 2e9;
 struct node {
     int s, e, next;
 }edge[maxn * ];
 int n, m;
 int son[maxn], top[maxn], tid[maxn], fat[maxn], siz[maxn], dep[maxn], rak[maxn];
 int head[maxn], len, dfx;
 //siz保存以i为根的子树节点个数，top保存i节点所在链的顶端节点,son保存i节点的重儿子,fat保存i节点的父亲节点
 //dep保存i节点的深度(根为1),,tid保存i节点dfs后的新编号，rak保存新编号i对应的节点(rak[i]=j,tid[j]=i)。
 void init() {
     memset(head, -, sizeof(head));
     len = , dfx = ;
 }
 void add(int s, int e) {//邻接表存值
     edge[len].s = s;
     edge[len].e = e;
     edge[len].next = head[s];
     head[s] = len++;
 }
 //搜出每个节点的siz,son,fat,dep
 void dfs1(int x, int fa, int d) {
     siz[x] = , son[x] = -, fat[x] = fa, dep[x] = d;
     for (int i = head[x]; i != -; i = edge[i].next) {
         int y = edge[i].e;
         if (y == fa)
             continue;
         dfs1(y, x, d + );
         siz[x] += siz[y];
         if (son[x] == - || siz[y] > siz[son[x]])
             son[x] = y;
     }
 }
 //搜出每个节点的top,tid,rak
 void dfs2(int x, int c) {
     top[x] = c;
     tid[x] = ++dfx;
     rak[dfx] = x;
     if (son[x] == -)
         return;
     dfs2(son[x], c);
     for (int i = head[x]; i != -; i = edge[i].next) {
         int y = edge[i].e;
         if (y == fat[x] || y == son[x])
             continue;
         dfs2(y, y);
     }
 }
 int a[maxn];
 int cr[maxn * ];
 int rt[maxn * ];
 int lt[maxn * ];
 int lazy[maxn * ];
 void up(int i) {
     lt[i] = lt[i << ], rt[i] = rt[i <<  | ];
     cr[i] = cr[i << ] + cr[i <<  | ];
     if (lt[i <<  | ] == rt[i << ])
         cr[i]--;
 }
 void down(int i) {
     if (lazy[i] != -) {
         lt[i << ] = lt[i <<  | ] = rt[i << ] = rt[i <<  | ] = lazy[i];
         cr[i << ] = cr[i <<  | ] = cr[i];
         lazy[i << ] = lazy[i <<  | ] = lazy[i];
         lazy[i] = -;
     }
 }
 void build(int l, int r, int i) {
     lazy[i] = -;
     if (l == r) {
         cr[i] = ;
         rt[i] = a[rak[l]];
         lt[i] = a[rak[l]];
         return;
     }
     int mid = (l + r) >> ;
     build(lson);
     build(rson);
     up(i);
 }
 void update(int L, int R, int k, int l, int r, int i) {
     if (L <= l && r <= R) {
         cr[i] = ;
         lazy[i] = k;
         lt[i] = rt[i] = k;
         return;
     }
     down(i);
     int mid = (l + r) >> ;
     if (L <= mid)
         update(L, R, k, lson);
     if (R > mid)
         update(L, R, k, rson);
     up(i);
 }
 int qquery(int L, int R, int l, int r, int i) {
     if (L <= l && r <= R) return cr[i];
     down(i);
     int mid = (l + r) >> ;
     if (R <= mid) return qquery(L, R, lson);
     if (L > mid) return qquery(L, R, rson);
     int ans1 = qquery(L, R, lson);
     int ans2 = qquery(L, R, rson);
     int ans = ans1 + ans2;
     if (rt[i << ] == lt[i <<  | ]) ans--;
     return ans;
 }

 int dquery(int k, int l, int r, int i) {
     if (l == r) {
         return lt[i];
     }
     int mid = (l + r) / ;
     down(i);
     if (k <= mid)
         return dquery(k, lson);
     else
         return dquery(k, rson);
 }
 int solve(int x, int y, int w, int flg) {
     int ans = ;
     while (top[x] != top[y]) {
         if (dep[top[x]] < dep[top[y]])
             swap(x, y);
         if (!flg)
             update(tid[top[x]], tid[x], w, , n, );
         else {
             ans += qquery(tid[top[x]], tid[x], , n, );
             if (dquery(tid[top[x]], , n, ) == dquery(tid[fat[top[x]]], , n, ))
                 ans--;
         }
         x = fat[top[x]];
     }
     if (dep[x] < dep[y])
         swap(x, y);
     if (!flg)
         update(tid[y], tid[x], w, , n, );
     else {
         ans += qquery(tid[y], tid[x], , n, );
         return ans;
     }
 }
 int main() {
     while (scanf("%d%d", &n, &m) != EOF) {
         for (int i = ; i <= n; i++)
             scanf("%d", &a[i]);
         init();
         int x, y, z;
         for (int i = ; i < n - ; i++) {
             scanf("%d%d", &x, &y);
             add(x, y);
             add(y, x);
         }
         dfs1(, , );
         dfs2(, );
         build(, n, );
         while (m--) {
             char s[];
             scanf("%s", s);
             if (s[] == 'C') {
                 scanf("%d%d%d", &x, &y, &z);
                 solve(x, y, z, );
             }
             else {
                 scanf("%d%d", &x, &y);
                 printf("%d\n", solve(x, y, , ));
             }
         }
     }
 }

LCT的做法好像更简明，只不过down的时候记得交换lv和rv.

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = ;
 ll fa[maxn], ch[maxn][], siz[maxn], val[maxn], sum[maxn], lv[maxn], rv[maxn], lazy2[maxn], lazy[maxn], st[maxn];//父亲节点,左右儿子节点,当前子数节点个数,当前值，lazy标记,辅助数组。
 inline bool isroot(int x) {//判断x是否为所在splay的根
     return ch[fa[x]][] != x && ch[fa[x]][] != x;
 }
 inline void pushup(int x) {
     lv[x] = ch[x][] ? lv[ch[x][]] : val[x];
     rv[x] = ch[x][] ? rv[ch[x][]] : val[x];
     if (ch[x][] && ch[x][])sum[x] = sum[ch[x][]] + sum[ch[x][]] +  - (rv[ch[x][]] == val[x]) - (lv[ch[x][]] == val[x]);
     else if (ch[x][])sum[x] = sum[ch[x][]] + (rv[ch[x][]] != val[x]);
     else if (ch[x][])sum[x] = sum[ch[x][]] + (lv[ch[x][]] != val[x]);
     else sum[x] = ;
     siz[x] = siz[ch[x][]] + siz[ch[x][]] + ;
 }
 inline void pushr(int x) {
     swap(ch[x][], ch[x][]);
     swap(lv[x], rv[x]);
     lazy[x] ^= ;
 }
 inline void pushC(int x, int c) {
     val[x] = lv[x] = rv[x] = c, sum[x] = ;
     lazy2[x] = c;
 }
 inline void pushdown(int x) {
     if (lazy[x]) {
         if (ch[x][])lazy[ch[x][]] ^= ;
         if (ch[x][])lazy[ch[x][]] ^= ;
         swap(ch[x][], ch[x][]);
         lazy[x] = ;
     }
     if (lazy2[x]) {
         if (ch[x][])pushC(ch[x][], lazy2[x]);
         if (ch[x][])pushC(ch[x][], lazy2[x]);
         lazy2[x] = ;
     }
 }
 inline void rotate(int x) {
     int y = fa[x], z = fa[y];
     int k = ch[y][] == x;
     if (!isroot(y))
         ch[z][ch[z][] == y] = x;
     fa[x] = z; ch[y][k] = ch[x][k ^ ]; fa[ch[x][k ^ ]] = y;
     ch[x][k ^ ] = y; fa[y] = x;
     pushup(y);
     pushup(x);
 }
 inline void splay(int x) {
     int f = x, len = ;
     st[++len] = f;
     while (!isroot(f))st[++len] = f = fa[f];
     while (len)pushdown(st[len--]);
     while (!isroot(x)) {
         int y = fa[x];
         int z = fa[y];
         if (!isroot(y))
             rotate((ch[y][] == x) ^ (ch[z][] == y) ? x : y);
         rotate(x);
     }
     pushup(x);
 }
 inline void access(int x) {//打通根节点到x的实链
     for (int y = ; x; x = fa[y = x])
         splay(x), ch[x][] = y, pushup(x);
 }
 inline void makeroot(int x) {//将x变为原树的根
     access(x); splay(x); pushr(x);
 }
 int Findroot(int x) {//找根节点
     access(x), splay(x);
     while (ch[x][])
         pushdown(x), x = ch[x][];
     splay(x);
     return x;
 }
 inline void split(int x, int y) {//将x到y路径变为play
     makeroot(x); access(y); splay(y);
 }
 inline void Link(int x, int y) {//合法连边
     makeroot(x); fa[x] = y;
 }
 inline void cut(int x, int y) {//合法断边
     split(x, y); fa[x] = ch[y][] = ; pushup(y);
 }
 int main() {
     int n, q;
     scanf("%d%d", &n, &q);
     for (int i = ; i <= n; i++)
         scanf("%d", &val[i]), lv[i] = rv[i] = val[i], sum[i] = ;
     int x, y, z;
     for (int i = ; i < n; i++) {
         scanf("%d%d", &x, &y);
         Link(x, y);
     }
     while (q--) {
         char s[];
         scanf("%s", s);
         if (s[] == 'Q') {
             scanf("%d%d", &x, &y);
             split(x, y);
             printf("%d\n", sum[y]);
         }
         else {
             scanf("%d%d%d", &x, &y, &z);
             split(x, y);
             pushC(y, z);
         }
     }
 }