查询区间有多少个不同的数
/* * 给出一个序列,查询区间内有多少个不相同的数 */
const int MAXN = 30010;
const int M = MAXN * 100;
int n, q, tot;
int a[MAXN];
int T[MAXN], lson[M], rson[M], c[M];
int build(int l, int r)
{
int root = tot++;
c[root] = 0;
if (l != r)
{
int mid = (l + r) >> 1;
lson[root] = build(l, mid);
rson[root] = build(mid + 1, r);
}
return root;
}
int update(int root, int pos, int val)
{
int newroot = tot++, tmp = newroot;
c[newroot] = c[root] + val;
int l = 1, r = n;
while (l < r)
{
int mid = (l + r) >> 1;
if (pos <= mid)
{
lson[newroot] = tot++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
rson[newroot] = tot++;
lson[newroot] = lson[root];
newroot = rson[newroot];
root = rson[root];
l = mid + 1;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int query(int root, int pos)
{
int ret = 0;
int l = 1, r = n;
while (pos < r)
{
int mid = (l + r) >> 1;
if (pos <= mid)
{
r = mid;
root = lson[root];
}
else
{
ret += c[lson[root]];
root = rson[root];
l = mid + 1;
}
}
return ret + c[root];
}
int main()
{
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
while (scanf("%d", &n) == 1)
{
tot = 0;
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
}
T[n + 1] = build(1, n);
map<int,int> mp;
for (int i = n; i >= 1; i--)
{
if (mp.find(a[i]) == mp.end())
{
T[i] = update(T[i + 1], i, 1);
}
else
{
int tmp = update(T[i + 1], mp[a[i]], -1);
T[i] = update(tmp, i, 1);
}
mp[a[i]] = i;
}
scanf("%d", &q);
while (q--)
{
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", query(T[l], r));
}
}
return 0;
}静态区间第k小
参考题目链接：
POJ 2104 K-th Number
const int MAXN = 100010;
const int M = MAXN * 30;
int n, q, m, tot;
int a[MAXN], t[MAXN];
int T[MAXN], lson[M], rson[M], c[M];
void Init_hash()
{
for (int i = 1; i <= n; i++)
{
t[i] = a[i];
}
sort(t + 1, t + 1 + n);
m = (int)(unique(t + 1, t + 1 + n) - t - 1);
}
int build(int l, int r)
{
int root = tot++; c[root] = 0;
if (l != r)
{
int mid = (l + r) >> 1;
lson[root] = build(l, mid);
rson[root] = build(mid + 1, r);
}
return root;
}
int hash_(int x)
{
return (int)(lower_bound(t + 1, t + 1 + m, x) - t);
}
int update(int root, int pos, int val)
{
int newroot = tot++, tmp = newroot;
c[newroot] = c[root] + val;
int l = 1, r = m;
while (l < r)
{
int mid = (l + r) >> 1;
if (pos <= mid)
{
lson[newroot] = tot++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
rson[newroot] = tot++;
lson[newroot] = lson[root];
newroot = rson[newroot];
root = rson[root];
l = mid + 1;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int query(int left_root, int right_root, int k)
{
int l = 1, r = m;
while ( l < r)
{
int mid = (l + r) >> 1;
if (c[lson[left_root]] - c[lson[right_root]] >= k )
{
r = mid;
left_root = lson[left_root];
right_root = lson[right_root];
}
else
{
l = mid + 1;
k -= c[lson[left_root]] - c[lson[right_root]];
left_root = rson[left_root];
right_root = rson[right_root];
}
}
return l;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
while (scanf("%d%d", &n, &q) == 2)
{
tot = 0;
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
}
Init_hash();
T[n + 1] = build(1, m);
for (int i = n; i; i--)
{
int pos = hash_(a[i]);
T[i] = update(T[i + 1], pos, 1);
}
while (q--)
{
int l, r, k;
scanf("%d%d%d", &l, &r, &k);
printf("%d\n", t[query(T[l], T[r + 1], k)]);
}
}
return 0;
}树上路径点权第k大
/*
* LCA + 主席树
*/
// 主席树部分
const int MAXN = 200010;
const int M = MAXN * 40;
int n, q, m, TOT;
int a[MAXN], t[MAXN];
int T[MAXN], lson[M], rson[M], c[M];
void Init_hash()
{
for (int i = 1; i <= n; i++)
{
t[i] = a[i];
}
sort(t + 1, t + 1 + n);
m = (int)(unique(t + 1, t + n + 1) - t - 1);
return ;
}
int build(int l, int r)
{
int root = TOT++;
c[root] = 0;
if (l != r)
{
int mid = (l + r) >> 1;
lson[root] = build(l, mid);
rson[root] = build(mid + 1, r);
}
return root;
}
int hash_(int x)
{
return (int)(lower_bound(t + 1, t + 1 + m, x) - t);
}
int update(int root, int pos, int val)
{
int newroot = TOT++, tmp = newroot;
c[newroot] = c[root] + val;
int l = 1, r = m;
while (l < r)
{
int mid = (l + r) >> 1;
if (pos <= mid)
{
lson[newroot] = TOT++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
rson[newroot] = TOT++;
lson[newroot] = lson[root];
newroot = rson[newroot];
root = rson[root];
l = mid + 1;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int query(int left_root, int right_root, int LCA, int k)
{
int lca_root = T[LCA];
int pos = hash_(a[LCA]);
int l = 1, r = m;
while (l < r)
{
int mid = (l + r) >> 1;
int tmp = c[lson[left_root]] + c[lson[right_root]] - 2 * c[lson[lca_root]] + (pos >= l && pos <= mid);
if (tmp >= k)
{
left_root = lson[left_root];
right_root = lson[right_root];
lca_root = lson[lca_root];
r = mid;
}
else
{
k -= tmp;
left_root = rson[left_root];
right_root = rson[right_root];
lca_root = rson[lca_root];
l = mid + 1;
}
}
return l;
}
// LCA部分
int rmq[2 * MAXN]; // rmq数组,就是欧拉序列对应的深度序列
struct ST
{
int mm[2 * MAXN];
int dp[2 * MAXN][20]; // 最小值对应的下标
void init(int n)
{
mm[0] = -1;
for (int i = 1; i <= n; i++)
{
mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
dp[i][0] = i;
}
for (int j = 1; j <= mm[n]; j++)
{
for (int i = 1; i + (1 << j) - 1 <= n; i++)
{
dp[i][j] = rmq[dp[i][j - 1]] < rmq[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
}
}
return ;
}
int query(int a, int b) // 查询[a,b]之间最小值的下标
{
if (a > b)
{
swap(a, b);
}
int k = mm[b - a + 1];
return rmq[dp[a][k]] <= rmq[dp[b - (1 << k) + 1][k]] ? dp[a][k] : dp[b - (1 << k) + 1][k];
}
};
// 边的结构体定义
struct Edge
{
int to, next;
};
Edge edge[MAXN * 2];
int tot, head[MAXN];
int F[MAXN * 2]; // 欧拉序列,就是dfs遍历的顺序,长度为2*n-1,下标从1开始
int P[MAXN]; // P[i]表示点i在F中第一次出现的位置
int cnt;
ST st;
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
return ;
}
void addedge(int u, int v) // 加边,无向边需要加两次
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
return ;
}
void dfs(int u, int pre, int dep)
{
F[++cnt] = u;
rmq[cnt] = dep;
P[u] = cnt;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (v == pre)
{
continue;
}
dfs(v, u, dep + 1);
F[++cnt] = u;
rmq[cnt] = dep;
}
return ;
}
void LCA_init(int root, int node_num) // 查询LCA前的初始化
{
cnt = 0;
dfs(root, root, 0);
st.init(2 * node_num - 1);
return ;
}
int query_lca(int u, int v) // 查询u,v的lca编号
{
return F[st.query(P[u], P[v])];
}
void dfs_build(int u, int pre)
{
int pos = hash_(a[u]);
T[u] = update(T[pre], pos, 1);
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (v == pre)
{
continue;
}
dfs_build(v, u);
}
return ;
}
int main()
{
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
while (scanf("%d%d", &n, &q) == 2)
{
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
}
Init_hash();
init();
TOT = 0;
int u, v;
for (int i = 1; i < n; i++)
{
scanf("%d%d", &u, &v);
addedge(u, v);
addedge(v, u);
}
LCA_init(1, n);
T[n + 1] = build(1, m);
dfs_build(1, n + 1);
int k;
while (q--)
{
scanf("%d%d%d", &u, &v, &k);
printf("%d\n", t[query(T[u], T[v], query_lca(u, v), k)]);
}
}
return 0;
}动态第区间第k大
参考题目链接：
ZOJ 2112 ???这个有些尴尬，现在正处于系统维护阶段……进不去ZOJ
/*
* 树状数组套主席树
*/
const int MAXN = 60010;
const int M = 2500010;
int n, q, m, tot;
int a[MAXN], t[MAXN];
int T[MAXN], lson[M], rson[M],c[M];
int S[MAXN];
struct Query
{
int kind;
int l, r, k;
} query[10010];
void Init_hash(int k)
{
sort(t, t + k);
m = (int)(unique(t, t + k) - t);
return ;
}
int hash_(int x)
{
return (int)(lower_bound(t, t + m, x) - t);
}
int build(int l, int r)
{
int root = tot++;
c[root] = 0;
if (l != r)
{
int mid = (l + r) / 2;
lson[root] = build(l, mid);
rson[root] = build(mid + 1, r);
}
return root;
}
int Insert(int root, int pos, int val)
{
int newroot = tot++, tmp = newroot;
int l = 0, r = m - 1;
c[newroot] = c[root] + val;
while (l < r)
{
int mid = (l + r) >> 1;
if (pos <= mid)
{
lson[newroot] = tot++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
rson[newroot] = tot++;
lson[newroot] = lson[root];
newroot = rson[newroot];
root = rson[root];
l = mid + 1;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int lowbit(int x)
{
return x & (-x);
}
int use[MAXN];
void add(int x, int pos, int val)
{
while (x <= n)
{
S[x] = Insert(S[x], pos, val);
x += lowbit(x);
}
return ;
}
int sum(int x)
{
int ret = 0;
while (x > 0)
{
ret += c[lson[use[x]]];
x -= lowbit(x);
}
return ret;
}
int Query(int left, int right, int k)
{
int left_root = T[left - 1];
int right_root = T[right];
int l = 0, r = m - 1;
for (int i = left - 1; i; i -= lowbit(i))
{
use[i] = S[i];
}
for (int i = right; i; i -= lowbit(i))
{
use[i] = S[i];
}
while (l < r)
{
int mid = (l + r) / 2;
int tmp = sum(right) - sum(left - 1) + c[lson[right_root]] - c[lson[left_root]];
if (tmp >= k)
{
r = mid;
for (int i = left - 1; i; i -= lowbit(i))
{
use[i] = lson[use[i]];
}
for (int i = right; i; i -= lowbit(i))
{
use[i] = lson[use[i]];
}
left_root = lson[left_root];
right_root = lson[right_root];
}
else
{
l = mid + 1;
k -= tmp;
for (int i = left - 1; i; i -= lowbit(i))
{
use[i] = rson[use[i]];
}
for (int i = right; i; i -= lowbit(i))
{
use[i] = rson[use[i]];
}
left_root = rson[left_root];
right_root = rson[right_root];
}
}
return l;
}
void Modify(int x, int p, int d)
{
while (x <= n)
{
S[x] = Insert(S[x], p, d);
x += lowbit(x);
}
return ;
}
int main()
{
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
int Tcase;
scanf("%d", &Tcase);
while (Tcase--)
{
scanf("%d%d", &n, &q);
tot = 0;
m = 0;
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
t[m++] = a[i];
}
char op[10];
for (int i = 0; i < q; i++)
{
scanf("%s", op);
if (op[0] == 'Q')
{
query[i].kind = 0;
scanf("%d%d%d", &query[i].l, &query[i].r, &query[i].k);
}
else
{
query[i].kind = 1;
scanf("%d%d", &query[i].l, &query[i].r);
t[m++] = query[i].r;
}
}
Init_hash(m);
T[0] = build(0, m - 1);
for (int i = 1; i <= n; i++)
{
T[i] = Insert(T[i - 1], hash_(a[i]), 1);
}
for (int i = 1; i <= n; i++)
{
S[i] = T[0];
}
for (int i = 0; i < q; i++)
{
if (query[i].kind == 0)
{
printf("%d\n", t[Query(query[i].l, query[i].r, query[i].k)]);
}
else
{
Modify(query[i].l, hash_(a[query[i].l]), -1);
Modify(query[i].l, hash_(query[i].r), 1);
a[query[i].l] = query[i].r;
}
}
}
return 0;
}2017.8.6 修改 静态区间第K大 为 静态区间第K小
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