【SDOI2011】染色
作者：互联网
Description
给定一棵树和若干操作，每次可以选取树上任意两点之间的路径染成一种颜色，或是查询任意两点之间路径上有多少段颜色。
Solution
树链剖分不解释，主要分析线段树的维护，在线段树区间上我们维护区间左端的颜色、右端的颜色、整个区间颜色段的数量。那么当我们合并两个区间时，我们首先将两区间的颜色段数量相加，如果左区间右端的颜色等于右区间左端的颜色，那么答案减一。
剩下的就是线段树和树链剖分的基本操作，在此不再赘述。
Code
1 #include <iostream>
2 #include <cstdio>
3 using namespace std;
4 typedef long long ll;
5 inline int read() {
6 int ret = 0, op = 1;
7 char c = getchar();
8 while (c < '0' || c > '9') {
9 if (c == '-') op = -1;
10 c = getchar();
11 }
12 while (c <= '9' && c >= '0') {
13 ret = ret * 10 + c - '0';
14 c = getchar();
15 }
16 return ret * op;
17 }
18 struct node {
19 int next, to;
20 } a[100010 << 1];
21 struct segment {
22 int lc, rc, cnt;
23 int tag;
24 } s[100010 << 2];
25 int n, m, num, head[100010], in[100010];
26 void add(int from, int to) {
27 a[++num].next = head[from];
28 a[num].to = to;
29 head[from] = num;
30 }
31 int size[100010], top[100010], son[100010], dep[100010], f[100010];
32 int seg[100010 << 2], rev[100010 << 2], tot;
33 void dfs1(int u, int fa) {
34 dep[u] = dep[fa] + 1;
35 f[u] = fa;
36 size[u] = 1;
37 for (int i = head[u]; i; i = a[i].next)
38 if (a[i].to != fa) {
39 dfs1(a[i].to, u);
40 size[u] += size[a[i].to];
41 if (size[son[u]] < size[a[i].to]) son[u] = a[i].to;
42 }
43 return ;
44 }
45 void dfs2(int u, int fa) {
46 if (son[u]) {
47 top[son[u]] = top[u];
48 seg[son[u]] = ++tot;
49 rev[tot] = son[u];
50 dfs2(son[u], u);
51 }
52 for (int i = head[u]; i; i = a[i].next)
53 if (!top[a[i].to]) {
54 top[a[i].to] = a[i].to;
55 seg[a[i].to] = ++tot;
56 rev[tot] = a[i].to;
57 dfs2(a[i].to, a[i].to);
58 }
59 }
60 void pushup(int now) {
61 s[now].lc = s[now << 1].lc;
62 s[now].rc = s[now << 1 | 1].rc;
63 s[now].cnt = s[now << 1].cnt + s[now << 1 | 1].cnt;
64 if (s[now << 1].rc == s[now << 1 | 1].lc) s[now].cnt --;
65 }
66 void pushdown(int now) {
67 if (s[now].tag != -1) {
68 s[now << 1].lc = s[now << 1].rc = s[now << 1 | 1].lc = s[now << 1 | 1].rc = s[now].tag;
69 s[now << 1].cnt = s[now << 1 | 1].cnt = 1;
70 s[now << 1].tag = s[now << 1 | 1].tag = s[now].tag;
71 s[now].tag = -1;
72 }
73 }
74 void build(int now, int l, int r) {
75 s[now].tag = -1;
76 if (l == r) {
77 s[now].lc = s[now].rc = in[rev[l]];
78 s[now].cnt = 1;
79 return ;
80 }
81 int mid = l + r >> 1;
82 build(now << 1, l, mid);
83 build(now << 1 | 1, mid + 1, r);
84 pushup(now);
85 return ;
86 }
87 void update(int now, int l, int r, int x, int y, int val) {
88 if (x == l && r == y) {
89 s[now].lc = s[now].rc = s[now].tag = val;
90 s[now].cnt = 1;
91 return ;
92 }
93 int mid = l + r >> 1;
94 pushdown(now);
95 if (y <= mid) update(now << 1, l, mid, x, y, val);
96 else if (x > mid) update(now << 1 | 1, mid + 1, r, x, y, val);
97 else {
98 update(now << 1, l, mid, x, mid, val);
99 update(now << 1 | 1, mid + 1, r, mid + 1, y, val);
100 }
101 pushup(now);
102 return ;
103 }
104 void find(int x, int y, int val) {
105 while (top[x] != top[y]) {
106 if (dep[top[x]] < dep[top[y]]) swap(x, y);
107 update(1, 1, tot, seg[top[x]], seg[x], val);
108 x = f[top[x]];
109 }
110 if (dep[x] > dep[y]) swap(x, y);
111 update(1, 1, tot, seg[x], seg[y], val);
112 }
113 int query(int now, int l, int r, int x, int y) {
114 if (x == l && r == y) return s[now].cnt;
115 pushdown(now);
116 int mid = l + r >> 1;
117 if (y <= mid) return query(now << 1, l, mid, x, y);
118 else if (x > mid) return query(now << 1 | 1, mid + 1, r, x, y);
119 else {
120 int retl = query(now << 1, l, mid, x, mid);
121 int retr = query(now << 1 | 1, mid + 1, r, mid + 1, y);
122 int ret = retl + retr;
123 if (s[now << 1].rc == s[now << 1 | 1].lc) ret--;
124 return ret;
125 }
126 }
127 int query(int now, int l, int r, int x) {
128 if (l == r) return s[now].lc;
129 int mid = l + r >> 1;
130 pushdown(now);
131 if (x <= mid) return query(now << 1, l, mid, x);
132 else return query(now << 1 | 1, mid + 1, r, x);
133 }
134 int find(int x, int y) {
135 int ans = 0;
136 while (top[x] != top[y]) {
137 if (dep[top[x]] < dep[top[y]]) swap(x, y);
138 ans += query(1, 1, tot, seg[top[x]], seg[x]);
139 if (query(1, 1, tot, seg[top[x]]) == query(1, 1, tot, seg[f[top[x]]])) ans--;
140 x = f[top[x]];
141 // cout << ans << endl;
142 }
143 if (dep[x] > dep[y]) swap(x, y);
144 ans += query(1, 1, tot, seg[x], seg[y]);
145 // cout << ans << endl;
146 return ans;
147 }
148 int main() {
149 n = read(); m = read();
150 for (int i = 1; i <= n; ++i) in[i] = read();
151 for (int i = 1; i < n; ++i) {
152 int x = read(), y = read();
153 add(x, y); add(y, x);
154 }
155 dfs1(1, 0);
156 seg[1] = tot = 1;
157 rev[1] = 1;
158 top[1] = 1;
159 dfs2(1, 1);
160 build(1, 1, tot);
161 while (m--) {
162 char op;
163 int x, y, z;
164 cin >> op;
165 cin >> x >> y;
166 if (op == 'C') {
167 cin >> z;
168 find(x, y, z);
169 }
170 else {
171 printf("%d\n", find(x, y));
172 }
173 }
174 return 0;
175 }
AC Code
标签：颜色,int,染色,seg,SDOI2011,return,now,op 来源： https://www.cnblogs.com/shl-blog/p/11184072.html