POJ 3580SuperMemo

Description

Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A₁, A₂, ... A_n}. Then the host performs a series of operations and queries on the sequence which consists:

1. ADD x y D: Add D to each number in sub-sequence {A_x ... A_y}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
2. REVERSE x y: reverse the sub-sequence {A_x ... A_y}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
3. REVOLVE x y T: rotate sub-sequence {A_x ... A_y} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
4. INSERT x P: insert P after A_x. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
5. DELETE x: delete A_x. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
6. MIN x y: query the participant what is the minimum number in sub-sequence {A_x ... A_y}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2

To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.

Input

The first line contains n (n ≤ 100000).

The following n lines describe the sequence.

Then follows M (M ≤ 100000), the numbers of operations and queries.

The following M lines describe the operations and queries.

Output

For each "MIN" query, output the correct answer.

Sample Input

51 2 3 4 52ADD 2 4 1MIN 4 5

Sample Output

5

解决完splay终极boss以后这种题写起来就十分轻松了。

#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<functional>using namespace std;typedef unsigned long long ull;typedef long long LL;const int maxn = 1e5 + 10;int n, m, root, a[maxn], l, r, c;char s[20];struct Splays{const static int maxn = 2e5 + 10;//节点个数const static int INF = 0x7FFFFFFF;//int最大值int ch[maxn][2], F[maxn], sz;//左右儿子，父亲节点和节点总个数int A[maxn], S[maxn], C[maxn], R[maxn], I[maxn];int Node(int f, int a) { S[sz] = 1; C[sz] = A[sz] = a; R[sz] = I[sz] = ch[sz][0] = ch[sz][1] = 0; F[sz] = f; return sz++; }//申请一个新节点void clear(){ sz = 1; S[0] = ch[0][0] = ch[0][1] = F[0] = 0; C[0] = INF; }//清空操作void Pushdown(int x){if (R[x]){R[ch[x][0]] ^= 1;R[ch[x][1]] ^= 1;swap(ch[x][0], ch[x][1]);R[x] = 0;}if (I[x]){if (ch[x][0]) A[ch[x][0]] += I[x], C[ch[x][0]] += I[x];if (ch[x][1]) A[ch[x][1]] += I[x], C[ch[x][1]] += I[x];I[ch[x][0]] += I[x]; I[ch[x][1]] += I[x];  I[x] = 0;}}void Count(int x){C[x] = min(A[x], min(C[ch[x][0]], C[ch[x][1]])); S[x] = S[ch[x][0]] + S[ch[x][1]] + 1;}void rotate(int x, int k){int y = F[x]; ch[y][!k] = ch[x][k]; F[ch[x][k]] = y;if (F[y]) ch[F[y]][y == ch[F[y]][1]] = x;F[x] = F[y];    F[y] = x;ch[x][k] = y;C[x] = C[y];S[x] = S[y];    Count(y);}void Splay(int x, int r){for (int fa = F[r]; F[x] != fa;){if (F[F[x]] == fa) { rotate(x, x == ch[F[x]][0]); return; }int y = x == ch[F[x]][0], z = F[x] == ch[F[F[x]]][0];y^z ? (rotate(x, y), rotate(x, z)) : (rotate(F[x], z), rotate(x, y));}}void build(int fa, int &x, int l, int r){if (l > r) return;int mid = l + r >> 1;x = Node(fa, a[mid]);build(x, ch[x][0], l, mid - 1);build(x, ch[x][1], mid + 1, r);Count(x);}void find(int &x, int k){for (int i = x; i;){Pushdown(i);if (S[ch[i][0]] > k) { i = ch[i][0]; continue; }if (S[ch[i][0]] == k){ Splay(i, x); x = i; return; }k -= S[ch[i][0]] + 1;i = ch[i][1];}}void Add(int&x){scanf("%d%d%d", &l, &r, &c);find(x, l - 1);find(ch[x][1], r - l + 1);int g = ch[ch[x][1]][0];I[g] += c;A[g] += c;C[g] += c;Count(ch[x][1]);Count(x);}void Reverse(int &x){scanf("%d%d", &l, &r);find(x, l - 1);find(ch[x][1], r - l + 1);R[ch[ch[x][1]][0]] ^= 1;}void Revolve(int &x){scanf("%d%d%d", &l, &r, &c);find(x, l - 1);find(ch[x][1], r - l + 1);int g = ch[ch[x][1]][0]; c %= (r - l + 1);if (!c) return;int s = r - l + 1 - c;find(g, s);int t = ch[g][0];ch[g][0] = 0;Count(g);find(g, S[g] - 1);ch[g][1] = t;F[t] = g;Count(g);}void Insert(int &x){scanf("%d%d", &l, &r);find(x, l);find(ch[x][1], 0);ch[ch[x][1]][0] = Node(ch[x][1], r);Count(ch[x][1]);Count(x);}void Delete(int &x){scanf("%d%d", &l);find(x, l - 1);find(ch[x][1], 1);ch[ch[x][1]][0] = 0;Count(ch[x][1]);Count(x);}void Min(int &x){scanf("%d%d", &l, &r);find(x, l - 1);find(ch[x][1], r - l + 1);printf("%d\n", C[ch[ch[x][1]][0]]);}}solve;int main(){while (scanf("%d", &n) != EOF){a[0] = a[n + 1] = root = 0;solve.clear();for (int i = 1; i <= n; i++) scanf("%d", &a[i]);solve.build(0, root, 0, n + 1);scanf("%d", &m);while (m--){scanf("%s", s);if (!strcmp(s, "ADD")) solve.Add(root);if (!strcmp(s, "REVERSE")) solve.Reverse(root);if (!strcmp(s, "REVOLVE")) solve.Revolve(root);if (!strcmp(s, "INSERT")) solve.Insert(root);if (!strcmp(s, "DELETE")) solve.Delete(root);if (!strcmp(s, "MIN")) solve.Min(root);}}return 0;}