题目链接
题意
一棵树每个点有点权，m次询问，每次询问一条路径上点权在一个区间内的点有多少。
思路
树链剖分+可持久化线段树
在树上建可持久化线段树，维护树链剖分的id序列的权值数量。
每次查询时候在树链上查主席树的结果。
很不幸这种做法T掉了1
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using namespace std;
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
const int N = 1e5 + 5;
ll seg[N << 5];
int lson[N << 5], rson[N << 5];
int root[N], val[N];
int size[N], son[N], dep[N], id[N], top[N], fa[N];
int cur;
struct node
{
int v, nxt;
}edge[2 * N];
int head[N], tot, t;
vector<int> des;
void dfs1(int u, int f, int d)
{
size[u] = 1;
dep[u] = d;
son[u] = -1;
fa[u] = f;
for (int it = head[u]; it != -1; it = edge[it].nxt)
{
int v = edge[it].v;
if (v == f)
continue;
dfs1(v, u, d + 1);
size[u] += size[v];
if (son[u] == -1 || size[v] > size[son[u]])
son[u] = v;
}
}
/*
void dfs2(int u, int f, int t)
{
top[u] = t;
id[u] = ++cur;
if (son[u] != -1)
dfs2(son[u], u, t);
for (int it = head[u]; it != -1; it = edge[it].nxt)
{
int v = edge[it].v;
if (v == f || v == son[u])
continue;
dfs2(v, u, v);
}
}
*/
void push_up(int rt)
{
seg[rt] = seg[lson[rt]] + seg[rson[rt]];
}
void update(int p, int old, int L, int R, int &rt)
{
rt = ++t;
seg[rt] = seg[old];
lson[rt] = lson[old];
rson[rt] = rson[old];
if (L == R)
{
seg[rt] += des[p - 1];
// dbg(p, seg[rt]);
return;
}
int mid = (L + R) >> 1;
if (p <= mid)
update(p, lson[old], L, mid, lson[rt]);
else
update(p, rson[old], mid + 1, R, rson[rt]);
push_up(rt);
}
int query(int l, int r, int L, int R, int rt)
{
// dbg(l, r, L, R, rt);
if (rt == 0)
return 0;
if (l <= L && r >= R)
return seg[rt];
int mid = (L + R) >> 1;
int ans = 0;
if (l <= mid)
ans += query(l, r, L, mid, lson[rt]);
if (r > mid)
ans += query(l, r, mid + 1, R, rson[rt]);
// dbg(ans);
return ans;
}
void dfs2(int u, int f, int t)
{
top[u] = t;
id[u] = ++cur;
int p = lower_bound(des.begin(), des.end(), val[u]) - des.begin() + 1;
update(p, root[cur - 1], 1, des.size(), root[cur]);
// dbg(p, cur);
if (son[u] != -1)
dfs2(son[u], u, t);
for (int it = head[u]; it != -1; it = edge[it].nxt)
{
int v = edge[it].v;
if (v == f || v == son[u])
continue;
dfs2(v, u, v);
}
}
template<class T>
void read(T& ret)
{
ret = 0;
char c;
while ((c = getchar()) > '9' || c < '0');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + c - '0';
c = getchar();
}
}
void init(int n)
{
for (int i = 1; i <= n; i++)
{
son[i] = -1;
head[i] = -1;
root[i] = 0;
}
t = 0;
tot = 0;
lson[0] = rson[0] = 0;
seg[0] = 0;
root[0] = 0;
cur = 0;
des.clear();
}
void add_edge(int u, int v)
{
edge[++tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot;
}
int get_ans(int u, int v, int a, int b)
{
// dbg(u, v, a, b);
int ans = 0;
a = lower_bound(des.begin(), des.end(), a) - des.begin() + 1;
b = upper_bound(des.begin(), des.end(), b) - des.begin();
while (top[u] != top[v])
{
if (dep[top[u]] < dep[top[v]])
swap(u, v);
ans += query(a, b, 1, des.size(), root[id[u]]) - query(a, b, 1, des.size(), root[id[top[u]] - 1]);
// dbg(u, v, top[u], top[v], ans);
u = fa[top[u]];
}
if (dep[u] > dep[v])
swap(u, v);
// dbg(u, v, a, b, id[u], id[v]);
ans += query(a, b, 1, des.size(), root[id[v]]) - query(a, b, 1, des.size(), root[id[u] - 1]);
return ans;
}
int main()
{
int n, m;
while (scanf("%d%d", &n, &m) != EOF)
{
init(n);
for (int i = 1; i <= n; i++)
{
read(val[i]);
des.push_back(val[i]);
}
des.push_back(0);
des.push_back(0x3f3f3f3f);
for (int i = 1; i < n; i++)
{
int u, v;
read(u);
read(v);
add_edge(u, v);
add_edge(v, u);
}
sort(des.begin(), des.end());
des.erase(unique(des.begin(), des.end()), des.end());
dfs1(1, 0, 1);
dfs2(1, 0, 1);
while (m--)
{
int u, v, a, b;
read(u);
read(v);
read(a);
read(b);
printf("%d ", get_ans(u, v, a, b));
}
putchar('\n');
}
return 0;
}
赛后我自己去测试，这种做法跑最大的数据比正解慢很多。1
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203.86user 0.04system 0:03.91elapsed 99%CPU (0avgtext+0avgdata 21884maxresident)k
0inputs+9824outputs (0major+4724minor)pagefaults 0swaps
13.40user 0.10system 0:13.52elapsed 99%CPU (0avgtext+0avgdata 36392maxresident)k
0inputs+9824outputs (0major+8545minor)pagefaults 0swaps
AC
3.86user 0.03system 0:03.90elapsed 99%CPU (0avgtext+0avgdata 21784maxresident)k
0inputs+9824outputs (0major+4725minor)pagefaults 0swaps
13.45user 0.09system 0:13.55elapsed 99%CPU (0avgtext+0avgdata 36288maxresident)k
0inputs+9824outputs (0major+8546minor)pagefaults 0swaps
AC
3.86user 0.04system 0:03.91elapsed 99%CPU (0avgtext+0avgdata 21884maxresident)k
0inputs+9808outputs (0major+4724minor)pagefaults 0swaps
13.65user 0.09system 0:13.77elapsed 99%CPU (0avgtext+0avgdata 36292maxresident)k
0inputs+9808outputs (0major+8546minor)pagefaults 0swaps
AC
3.95user 0.03system 0:03.98elapsed 99%CPU (0avgtext+0avgdata 21784maxresident)k
0inputs+9816outputs (0major+4726minor)pagefaults 0swaps
13.90user 0.07system 0:13.99elapsed 99%CPU (0avgtext+0avgdata 36288maxresident)k
0inputs+9816outputs (0major+8545minor)pagefaults 0swaps
AC
离线树状数组（或线段树）+树链剖分
这个做法也很好理解，但是开始的时候深陷主席树的泥潭不能自拔。
离线求出小于等于某个数字的点有多少，查询结果的时候用之前离线处理的点相减就行了。1
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using namespace std;
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
const int N = 1e5 + 5;
ll val[N];
int size[N], son[N], dep[N], id[N], top[N], fa[N];
int cur;
struct node
{
int v, nxt;
}edge[2 * N];
int head[N], tot, t;
// vector<int> des;
ll sum[N];
int lowbit(int x)
{
return x & (-x);
}
void add(int x, ll v)
{
while (x <= cur)
{
// dbg(x);
sum[x] += v;
x += lowbit(x);
}
}
ll query(int x)
{
ll ans = 0;
while (x)
{
ans += sum[x];
x -= lowbit(x);
}
return ans;
}
void dfs1(int u, int f, int d)
{
size[u] = 1;
dep[u] = d;
son[u] = -1;
fa[u] = f;
for (int it = head[u]; it != -1; it = edge[it].nxt)
{
int v = edge[it].v;
if (v == f)
continue;
dfs1(v, u, d + 1);
size[u] += size[v];
if (son[u] == -1 || size[v] > size[son[u]])
son[u] = v;
}
}
/*
void dfs2(int u, int f, int t)
{
top[u] = t;
id[u] = ++cur;
if (son[u] != -1)
dfs2(son[u], u, t);
for (int it = head[u]; it != -1; it = edge[it].nxt)
{
int v = edge[it].v;
if (v == f || v == son[u])
continue;
dfs2(v, u, v);
}
}
void push_up(int rt)
{
seg[rt] = seg[lson[rt]] + seg[rson[rt]];
}
void update(int p, int old, int L, int R, int &rt)
{
rt = ++t;
seg[rt] = seg[old];
lson[rt] = lson[old];
rson[rt] = rson[old];
if (L == R)
{
seg[rt] += des[p - 1];
// dbg(p, seg[rt]);
return;
}
int mid = (L + R) >> 1;
if (p <= mid)
update(p, lson[old], L, mid, lson[rt]);
else
update(p, rson[old], mid + 1, R, rson[rt]);
push_up(rt);
}
int query(int l, int r, int L, int R, int rt)
{
// dbg(l, r, L, R, rt);
if (rt == 0)
return 0;
if (l <= L && r >= R)
return seg[rt];
int mid = (L + R) >> 1;
int ans = 0;
if (l <= mid)
ans += query(l, r, L, mid, lson[rt]);
if (r > mid)
ans += query(l, r, mid + 1, R, rson[rt]);
// dbg(ans);
return ans;
}
*/
void dfs2(int u, int f, int t)
{
top[u] = t;
id[u] = ++cur;
// update(p, root[cur - 1], 1, des.size(), root[cur]);
// dbg(p, cur);
if (son[u] != -1)
dfs2(son[u], u, t);
for (int it = head[u]; it != -1; it = edge[it].nxt)
{
int v = edge[it].v;
if (v == f || v == son[u])
continue;
dfs2(v, u, v);
}
}
template<class T>
void read(T& ret)
{
ret = 0;
char c;
while ((c = getchar()) > '9' || c < '0');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + c - '0';
c = getchar();
}
}
void init(int n)
{
for (int i = 1; i <= n; i++)
{
son[i] = -1;
head[i] = -1;
}
tot = 0;
cur = 0;
// des.clear();
memset(sum, 0, sizeof(sum));
}
void add_edge(int u, int v)
{
edge[++tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot;
}
ll get_ans(int u, int v)
{
// dbg(u, v, a, b);
ll ans = 0;
while (top[u] != top[v])
{
if (dep[top[u]] < dep[top[v]])
swap(u, v);
//ans += query(a, b, 1, des.size(), root[id[u]]) - query(a, b, 1, des.size(), root[id[top[u]] - 1]);
ans += query(id[u]) - query(id[top[u]] - 1);
// dbg(u, v, top[u], top[v], ans);
u = fa[top[u]];
}
if (dep[u] > dep[v])
swap(u, v);
// dbg(u, v, a, b, id[u], id[v]);
// ans += query(a, b, 1, des.size(), root[id[v]]) - query(a, b, 1, des.size(), root[id[u] - 1]);
ans += query(id[v]) - query(id[u] - 1);
return ans;
}
struct nn
{
int a, b, u, v, id;
/*
int to1, to2;
bool operator < (const nn& w) const
{
return id < w.id;
}
*/
}que[N];
struct wa
{
int a, u, v;
ll ans;
int type;
wa (int u = 0, int v = 0, int a = 0, int type = 0): u(u), v(v), a(a), type(type) {}
}wai[N * 3];
int num[3 * N];
bool cmp(int a, int b)
{
if (wai[a].a == wai[b].a)
return wai[a].type < wai[b].type;
return wai[a].a < wai[b].a;
}
int main()
{
int n, m;
while (scanf("%d%d", &n, &m) != EOF)
{
init(n);
for (int i = 1; i <= n; i++)
{
read(val[i]);
}
for (int i = 1; i < n; i++)
{
int u, v;
read(u);
read(v);
add_edge(u, v);
add_edge(v, u);
}
dfs1(1, 0, 1);
dfs2(1, 0, 1);
int now = 0;
for (int i = 1; i <= m; i++)
{
int u, v, a, b;
read(que[i].u);
read(que[i].v);
read(que[i].a);
read(que[i].b);
wai[++now] = wa(que[i].u, que[i].v, que[i].a - 1, 2);
num[now] = now;
wai[++now] = wa(que[i].u, que[i].v, que[i].b, 2);
num[now] = now;
}
for (int i = 1; i <= n; i++)
wai[++now] = wa(i, 0, val[i], 1), num[now] = now;
sort(num + 1, num + now + 1, cmp);
for (int i = 1; i <= now; i++)
{
// dbg(i, wai[num[i]].u, wai[num[i]].type, num[i]);
if (wai[num[i]].type == 2)
wai[num[i]].ans = get_ans(wai[num[i]].u, wai[num[i]].v);
else
add(id[wai[num[i]].u], wai[num[i]].a);
}
for (int i = 1; i <= m; i++)
printf("%lld%c", wai[2 * i].ans - wai[2 * i - 1].ans, i == m? '\n': ' ');
}
return 0;
}