对应HDU题目:点击打开链接
Play with Chain
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4571 Accepted Submission(s): 1859
Problem Description
YaoYao is fond of playing his chains. He has a chain containing n diamonds on it. Diamonds are numbered from 1 to n.
At first, the diamonds on the chain is a sequence: 1, 2, 3, …, n.
He will perform two types of operations:
CUT a b c: He will first cut down the chain from the ath diamond to the bth diamond. And then insert it after the cth diamond on the remaining chain.
For example, if n=8, the chain is: 1 2 3 4 5 6 7 8; We perform “CUT 3 5 4”, Then we first cut down 3 4 5, and the remaining chain would be: 1 2 6 7 8. Then we insert “3 4 5” into the chain before 5th diamond, the chain turns out to be: 1 2 6 7 3 4 5 8.
FLIP a b: We first cut down the chain from the ath diamond to the bth diamond. Then reverse the chain and put them back to the original position.
For example, if we perform “FLIP 2 6” on the chain: 1 2 6 7 3 4 5 8. The chain will turn out to be: 1 4 3 7 6 2 5 8
He wants to know what the chain looks like after perform m operations. Could you help him?
Input
There will be multiple test cases in a test data.
For each test case, the first line contains two numbers: n and m (1≤n, m≤3*100000), indicating the total number of diamonds on the chain and the number of operations respectively.
Then m lines follow, each line contains one operation. The command is like this:
CUT a b c // Means a CUT operation, 1 ≤ a ≤ b ≤ n, 0≤ c ≤ n-(b-a+1).
FLIP a b // Means a FLIP operation, 1 ≤ a < b ≤ n.
The input ends up with two negative numbers, which should not be processed as a case.
Output
For each test case, you should print a line with n numbers. The ith number is the number of the ith diamond on the chain.
Sample Input
8 2 CUT 3 5 4 FLIP 2 6 -1 -1
Sample Output
1 4 3 7 6 2 5 8
题意:
n个数一开始是1~n顺序排列,有两个操作。
CUT(a, b, c)操作表示把第a到第b个数取下放到新组成的第c个数后面;
FLIP(a, b)操作表示把第a到第b个数翻转;
比如样例: n = 8
1 2 3 4 5 6 7 8
CUT(3, 5, 4)后数列变成 1 2 6 7 3 4 5 8
FLIP(2, 6)后数列变成1 4 3 7 6 2 5 8
问m次操作后的数列为?
思路:
伸展树的基础操作,区间截断,区间翻转。
区间截断:对于CUT(a, b, c)
1)提取区间[a, b]
具体方法:Splay(a - 1, T),Splay(b +1, T->right);即把第a - 1个数旋转到根,把第b + 1个数旋转到根的右儿子;那以根的右儿子的左儿子为根的子树就是所有区间[a, b]内的值;把它剪下。
2)把第c个数旋转到根,把第c + 1个数旋转到根的右儿子;那根的右儿子的左儿子肯定是空的。
3)把剪下的子树接在根的右儿子的左儿子。
区间翻转:对于FLIP(a, b)
1)提取区间[a, b]
2)在左儿子做翻转标志(注意不是简单的赋值为1,而是要做异或操作,即原来是1的标记为0,是0的标记为1)
3)在适当的地方加Push_down向下更新(跟线段树区间更新一样)
首先要明白的一点是二叉树结点的信息是最终中序遍历(一定是按下标先后顺序)输出的值,而不是下标的值;那怎样取下标为a的结点呢?
利用每个结点的sz,它表示以该结点为子树的的所有元素个数(T->sz = T->left->sz + T->right->sz + 1)
从根结点p开始
1)如果(左子树的元素个数加上1(1表示p结点) )sum = p->left->sz + 1;if (sum == a) ,那就找到了,返回p结点
2)sum比a小,说明要往右走(p = p->right);同时a -= sum;跳到步骤1
3)sum比a大,说明要往左走(p = p->right);a无需变动;跳到步骤1
怎样书写Push_down函数
就把左右子树对调,然后取消该结点标记,把两棵子树添加标记。
注意:修改时要注意Push_down的地方,注意父亲指针;注意更新sz;用动态的方法时要注意儿子是否为空。
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> #define MAX(x, y) ((x) > (y) ? (x) : (y)) typedef struct TREE { int data; TREE *fa, *l, *r; int sz; //以该结点为根的树的总结点数 bool flag; //翻转标记 }Tree; bool space; void Push_down(Tree *T) { if(NULL == T) return; //左右子树对调 if(T->flag){ Tree *tmp; tmp = T->r; T->r = T->l; T->l = tmp; T->flag = 0; if(T->l) T->l->flag ^= 1; if(T->r) T->r->flag ^= 1; } } void Init(Tree *&T, int n) { int i; bool k = 0; Tree *cur, *pre; for(i = n + 1; i > -1; i--){ cur = (Tree *)malloc(sizeof(Tree)); cur->data = i; cur->fa = cur->l = cur->r = NULL; cur->sz = 1; cur->flag = 0; if(k){ cur->r = pre; pre->fa = cur; cur->sz = pre->sz + 1; } if(!k) k = 1; pre = cur; } T = cur; } void PreOrder(Tree *T) { if(NULL == T) return; printf("%d ", T->data); PreOrder(T->l); PreOrder(T->r); } void MidOrder(Tree *T, int n) { if(NULL == T) return; Push_down(T); MidOrder(T->l, n); if(T->data > 0 && T->data < n + 1){ if(space) printf(" "); else space = 1; printf("%d", T->data); } MidOrder(T->r, n); } void R_rotate(Tree *x) { Tree *y = x->fa; Tree *z = y->fa; Tree *k = x->r; int sx = x->sz, sy = y->sz, sk = 0; if(k) sk = k->sz; y->l = k; x->r = y; if(z){ if(y == z->l) z->l = x; else z->r = x; } if(k) k->fa = y; y->fa = x; x->fa = z; y->sz = sy - sx + sk; x->sz = sx - sk + y->sz; } void L_rotate(Tree *x) { Tree *y = x->fa; Tree *z = y->fa; Tree *k = x->l; int sx = x->sz, sy = y->sz, sk = 0; if(k) sk = k->sz; y->r = k; x->l = y; if(z){ if(y == z->r) z->r = x; else z->l = x; } if(k) k->fa = y; y->fa = x; x->fa = z; y->sz = sy - sx + sk; x->sz = sx - sk + y->sz; } //寻找第x个数的结点 Tree *FindTag(Tree *T, int x) { if(NULL == T) return NULL; Push_down(T); Tree *p; p = T; int sum = (p->l ? p->l->sz : 0) + 1; while(sum != x && p) { if(sum < x){ p = p->r; x -= sum; } else p = p->l; Push_down(p); sum = (p->l ? p->l->sz : 0) + 1; } Push_down(p); return p; } void Splay(int x, Tree *&T) { Push_down(T); Tree *p, *X, *end, *new_t; end = T->fa; new_t = T; if(end) new_t = T->fa; X = FindTag(new_t, x); while(X->fa != end) { p = X->fa; if(end == p->fa){ //p是根结点 if(X == p->l) R_rotate(X); else L_rotate(X); break; } //p不是根结点 if(X == p->l){ if(p == p->fa->l){ R_rotate(p); //LL R_rotate(X); //LL } else{ R_rotate(X); //RL L_rotate(X); } } else{ if(p == p->fa->r){ //RR L_rotate(p); L_rotate(X); } else{ //LR L_rotate(X); R_rotate(X); } } } T = X; } void CUT(Tree *&T, int a, int b, int c) { //取[a,b] Splay(a - 1, T); Splay(b + 1, T->r); //剪[a,b] Tree *tmp; tmp = T->r->l; tmp->fa = NULL; T->r->l = NULL; T->r->sz -= tmp->sz; T->sz -= tmp->sz; //移动第c个数到根结点,第c+1个数到根结点右儿子 //这样根结点右儿子的左儿子必然为空,就可以把剪掉的放上去 Splay(c, T); Splay(c + 1, T->r); //接[a, b] T->r->l = tmp; tmp->fa = T->r; T->r->sz += tmp->sz; T->sz += tmp->sz; } void FLIP(Tree *&T, int a, int b) { //取[a,b] Splay(a - 1, T); Splay(b + 1, T->r); //标记T->r->l T->r->l->flag ^= 1; } void FreeTree(Tree *T) { if(NULL == T) return; FreeTree(T->l); FreeTree(T->r); free(T); } int main() { //freopen("in.txt", "r", stdin); Tree *T; int n, q, a, b, c; char s[6]; while(scanf("%d%d", &n, &q), n >= 0 && q >= 0) { space = 0; T = NULL; Init(T, n); while(q--) { scanf("%s", s); if('C' == s[0]){ scanf("%d%d%d", &a, &b, &c); CUT(T, a + 1, b + 1, c + 1); } else{ scanf("%d%d", &a, &b); FLIP(T, a + 1, b + 1); } } MidOrder(T, n); printf("\n"); FreeTree(T); } return 0; }