【bzoj3223】Tyvj 1729 文艺平衡树

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题目描述：
您需要写一种数据结构（可参考题目标题），来维护一个有序数列，其中需要提供以下操作：翻转一个区间，例如原有序序列是5 4 3 2 1，翻转区间是[2,4]的话，结果是5 2 3 4 1

输入：

第一行为n,m n表示初始序列有n个数，这个序列依次是(1,2……n-1,n) m表示翻转操作次数
接下来m行每行两个数[l,r] 数据保证 1<=l<=r<=n

输出：

输出一行n个数字，表示原始序列经过m次变换后的结果

样例输入：
5 3
1 3
1 3
1 4

样例输出：

4 3 2 1 5

数据范围：

N,M<=100000

题解：

直接splay，打个rev标记就好了。

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#ifdef WIN32#define LL "%I64d"#else#define LL "%lld"#endif#ifdef CT#define debug(...) printf(__VA_ARGS__)#else#define debug(...)#endif#define R register#define getc() (S==T&&(T=(S=B)+fread(B,1,1<<15,stdin),S==T)?EOF:*S++)#define gmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))#define gmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)char B[1<<15],*S=B,*T=B;inline int FastIn(){R char ch;R int cnt=0;R bool minus=0;while (ch=getc(),(ch < '0' || ch > '9') && ch != '-') ;ch == '-' ?minus=1:cnt=ch-'0';while (ch=getc(),ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';return minus?-cnt:cnt;}#define maxn 100010bool rev[maxn];int fa[maxn] , ch[maxn][2] , size[maxn] , root , n;inline void update (R int x){R int ls = ch[x][0] , rs = ch[x][1];size[x] = size[ls] + size[rs] + 1;}void Build (R int l , R int r , R int rt){if (l > r) return ;R int mid = (l + r) >> 1;fa[mid] = rt;if (mid < rt) ch[rt][0] = mid;else ch[rt][1] = mid;Build(l , mid-1 , mid);Build(mid+1 , r , mid);update(mid);}inline void pushdown (R int x){R int ls = ch[x][0] , rs = ch[x][1];if (rev[x]){if (ls) rev[ls] ^= 1 ;if (rs) rev[rs] ^= 1 ;ch[x][0] = rs ; ch[x][1] = ls;rev[x] = 0;}}inline void rotate(R int x){R int f = fa[x] , gf = fa[f] , d = (ch[f][1] == x);if (f==root) root = x , ch [0][0] = x;(ch[f][d] = ch[x][d ^ 1]) >0 ? fa[ch[f][d]] = f : 0;(fa[x] = gf )> 0 ? ch[gf][ch[gf][1]==f] = x : 0;fa[ch[x][d^1] = f] = x;update (f);}inline void splay(R int x , R int rt){while (fa[x]!=rt){R int f = fa[x] , gf = fa[f];if (gf != rt) rotate ( (ch[gf][1] == f) ^ (ch[f][1] == x) ? x : f );rotate (x);}update(x);}int find(R int x , R int rank){if (rev[x]) pushdown(x);R int ls = ch[x][0] , rs = ch[x][1] ,lsize = size[ls];if (lsize+1 == rank) return x;if (lsize >= rank ) return find(ls , rank);else return find(rs , rank -lsize -1);}inline int prepare (R int l,R int r){R int x = find (root ,l-1 );splay(x , 0);x = find(root , r+1) ;splay(x , root);return ch[x][0];}inline void rever(R int l,R int r){R int x = prepare (l,r);rev[x] ^= 1;pushdown(x);}inline void print(R int x){if (!x) return;if (rev[x]) pushdown(x);R int ls = ch[x][0] , rs = ch[x][1];print(ls);if (x != 1 && x!=n )printf("%d ",x -1 );print(rs);}int main(){n = FastIn()+2 ;R int m = FastIn();Build(1 , n , 0);root = ( 1 + n ) >> 1;for (; m ; m--){R int l = FastIn() + 1 , r = FastIn() + 1;rever( l , r );}splay(1,0);print(root);return 0;}

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