BZOJ 3631 [JLOI2014]松鼠的新家 | 树上差分

链接

BZOJ 3631

题解

看起来是树剖？实际上树上差分就可以解决……

当要给一条路径(u, v) +1的时候，给d[u] += 1, d[v] += 1, d[lca(u, v)] -= 1, d[fa[lca(u, v)]] -= 1。

注意这道题中路径的终点是不 +１的。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define enter putchar('\n')
#define space putchar(' ')
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c > '9' || c < '0')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 300005;
int n, a
, ans
, dep
, stk
, top, pos
, seq[2 * N][20], tot, lg[2 * N], fa
;
int ecnt, adj
, nxt[2 * N], go[2 * N];

void add(int u, int v){
go[++ecnt] = v;
nxt[ecnt] = adj[u];
adj[u] = ecnt;
}
void dfs(int u){
seq[++tot][0] = u, pos[u] = tot, stk[++top] = u;
for(int e = adj[u], v; e; e = nxt[e])
if(v = go[e], v != fa[u]){
fa[v] = u, dep[v] = dep[u] + 1;
dfs(v);
seq[++tot][0] = u;
}
}
int Min(int u, int v){
return dep[u] < dep[v] ? u : v;
}
void init(){
for(int i = 1, j = 0; i <= tot; i++)
lg[i] = i == (1 << (j + 1)) ? ++j : j;
for(int j = 1; (1 << j) <= tot; j++)
for(int i = 1; i + (1 << j) - 1 <= tot; i++)
seq[i][j] = Min(seq[i][j - 1], seq[i + (1 << (j - 1))][j - 1]);
}
int lca(int u, int v){
int l = pos[u], r = pos[v];
if(l > r) swap(l, r);
int j = lg[r - l + 1];
return Min(seq[l][j], seq[r - (1 << j) + 1][j]);
}
void path_modify(int u, int v, int x){
int anc = lca(u, v);
ans[u] += x, ans[v] += x, ans[anc] -= x, ans[fa[anc]] -= x;
}

int main(){

read(n);
for(int i = 1; i <= n; i++) read(a[i]);
for(int i = 1, u, v; i < n; i++)
read(u), read(v), add(u, v), add(v, u);
dfs(1);
init();
for(int i = 1; i < n; i++)
path_modify(a[i], a[i + 1], 1), path_modify(a[i + 1], a[i + 1], -1);
for(int i = top; i; i--)
ans[fa[stk[i]]] += ans[stk[i]];
for(int i = 1; i <= n; i++)
write(ans[i]), enter;

return 0;
}