SDOI2011染色

Posted 2022-04-12 shl-blog

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