[JLOI2014]松鼠的新家

嘟嘟嘟

这还是一道树链剖分板子题呀！

从1到n - 1枚举a[i]，每一次使节点a[i]到a[i + 1]的路径加1，但这样的话除a[1]，每一个点都多加了一个1，所以输出答案的时候减1即可。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<cctype>
  8 #include<stack>
  9 #include<queue>
 10 #include<vector>
 11 using namespace std;
 12 #define enter puts("")
 13 #define space putchar(' ')
 14 #define Mem(a) memset(a, 0, sizeof(a))
 15 typedef long long ll;
 16 typedef double db;
 17 const int INF = 0x3f3f3f3f;
 18 const db eps  =1e-8;
 19 const int maxn = 3e5 + 5;
 20 inline ll read()
 21 {
 22     ll ans = 0;
 23     char ch = getchar(), last = ' ';
 24     while(!isdigit(ch)) {last = ch; ch = getchar();}
 25     while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
 26     if(last == '-') ans = -ans;
 27     return ans;
 28 }
 29 inline void write(ll x)
 30 {
 31     if(x < 0) putchar('-'), x = -x;
 32     if(x >= 10) write(x / 10);
 33     putchar(x % 10 + '0');
 34 }
 35 
 36 int n, a[maxn];
 37 vector<int> v[maxn];
 38 
 39 bool vis[maxn];
 40 int fa[maxn], dep[maxn], size[maxn], son[maxn];
 41 void dfs1(int now)
 42 {
 43     vis[now] = 1; size[now] = 1;
 44     for(int i = 0; i < (int)v[now].size(); ++i)
 45     {
 46         if(!vis[v[now][i]])
 47         {
 48             fa[v[now][i]] = now;
 49             dep[v[now][i]] = dep[now] + 1;
 50             dfs1(v[now][i]);
 51             size[now] += size[v[now][i]];
 52             if(!son[now] || size[v[now][i]] > size[son[now]]) son[now] = v[now][i];
 53         }
 54     }
 55 }
 56 int top[maxn], dfsx[maxn], pos[maxn], cnt = 0;
 57 void dfs2(int now)
 58 {
 59     vis[now] = 1;
 60     dfsx[now] = ++cnt; pos[cnt] = now;
 61     if(son[now])
 62     {
 63         top[son[now]] = top[now];
 64         dfs2(son[now]);
 65     }
 66     for(int i = 0; i < (int)v[now].size(); ++i)
 67     {
 68         if(!vis[v[now][i]] && v[now][i] != son[now])
 69         {
 70             top[v[now][i]] = v[now][i];
 71             dfs2(v[now][i]);
 72         }
 73     }
 74 }
 75 
 76 int l[maxn << 2], r[maxn << 2];
 77 ll sum[maxn << 2], lazy[maxn << 2];
 78 void build(int L, int R, int now)
 79 {
 80     l[now] = L; r[now] = R;
 81     if(L == R) return;
 82     int mid = (L + R) >> 1;
 83     build(L, mid, now << 1);
 84     build(mid + 1, R, now << 1 | 1);
 85 }
 86 void pushdown(int now)
 87 {
 88     if(lazy[now])
 89     {
 90         sum[now << 1] += (r[now << 1] - l[now << 1] + 1) * lazy[now];
 91         sum[now << 1 | 1] += (r[now << 1 | 1] - l[now << 1 | 1] + 1) * lazy[now];
 92         lazy[now << 1] += lazy[now];
 93         lazy[now << 1 | 1] += lazy[now];
 94         lazy[now] = 0;
 95     }
 96 }
 97 void update(int L, int R, int now)
 98 {
 99     if(l[now] == L && r[now] == R) 
100     {
101         sum[now] += r[now] - l[now] + 1; 
102         lazy[now]++; return;
103     }
104     pushdown(now);
105     int mid = (l[now] + r[now]) >> 1;
106     if(R <= mid) update(L, R, now << 1);
107     else if(L > mid) update(L ,R, now << 1 | 1);
108     else {update(L, mid, now << 1); update(mid + 1, R, now << 1 | 1);}
109     sum[now] = sum[now << 1] + sum[now << 1 | 1];
110 }
111 ll query(int id, int now)
112 {
113     if(l[now] == r[now]) return sum[now];
114     pushdown(now);
115     int mid = (l[now] + r[now]) >> 1;
116     if(id <= mid) return query(id, now << 1);
117     else return query(id, now << 1 | 1);
118 }
119 
120 void update2(int x, int y)
121 {
122     
123     while(top[x] != top[y])
124     {
125         if(dep[top[x]] < dep[top[y]]) swap(x, y);
126         update(dfsx[top[x]], dfsx[x], 1);
127         x = fa[top[x]];    
128     }
129     if(dfsx[x] > dfsx[y]) swap(x, y);
130     update(dfsx[x], dfsx[y], 1);
131 }
132 
133 ll ans[maxn];
134 
135 int main()
136 {
137     n = read();
138     for(int i = 1; i <= n; ++i) a[i] = read();
139     for(int i = 1; i < n; ++i)
140     {
141         int x = read(), y = read();
142         v[x].push_back(y); v[y].push_back(x);
143     }
144     dfs1(1); Mem(vis); dfs2(1);
145     build(1, cnt, 1);
146     for(int i = 1; i < n; ++i) update2(a[i], a[i + 1]);        
147     for(int i = 1; i <= n; ++i) {write(query(dfsx[i], 1) - (a[1] == i ? 0 : 1)); enter;}
148     return 0;
149 }