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HDU 3487 Play with Chain

时间:2015-04-09 13:42:08      阅读:132      评论:0      收藏:0      [点我收藏+]

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题意:对序列取出连续的一段接到剩下的第k个值后面,或者把一段序列反转。

解题思路:splay 区间操作。

解题代码:

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  1 // File Name: hdu3487.cpp
  2 // Author: darkdream
  3 // Created Time: 2015年04月09日 星期四 10时16分12秒
  4 
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 #define maxn 310000
 26 using namespace std;
 27 
 28 char str[10];
 29 int ta,tb,tc;
 30 int n , m ; 
 31 int lastans[maxn];
 32 struct SplayTree{
 33     int sz[maxn];
 34     int ch[maxn][2];
 35     int pre[maxn];
 36     int root ,top1,top2;
 37     int ss[maxn],que[maxn];
 38     int rev[maxn];
 39     inline void Rotate(int x, int f){
 40         int y = pre[x];
 41         push_down(y);
 42         push_down(x);
 43         ch[y][!f] = ch[x][f];
 44         pre[ch[x][f]] = y ; 
 45         pre[x] = pre[y];
 46         if(pre[x]) ch[pre[y]][ch[pre[y]][1] == y]  = x; 
 47         ch[x][f] = y; 
 48         pre[y] = x; 
 49         push_up(y);
 50     }
 51     inline void Splay(int x, int goal){
 52         push_down(x);
 53         while(pre[x] != goal){
 54             if(pre[pre[x]] == goal){
 55                 Rotate(x,ch[pre[x]][0] == x);
 56             }else{
 57                 int y = pre[x],z = pre[y];
 58                 int f = (ch[z][0] == y);
 59                 if(ch[y][f] == y ){
 60                     Rotate(x,!f),Rotate(x,f);
 61                 }else{
 62                     Rotate(y,f),Rotate(x,f);
 63                 }
 64             }
 65         }
 66         push_up(x);
 67         if(goal == 0 ) root = x;
 68     }
 69     inline void RotateTo(int k ,int goal){
 70         int x = root ; 
 71         push_down(x);
 72         while(sz[ch[x][0]] != k ){
 73             if(k < sz[ch[x][0]]){
 74                 x = ch[x][0];
 75             }else{
 76                 k -= (sz[ch[x][0]] +1) ; 
 77                 x = ch[x][1];
 78             }
 79             push_down(x);
 80         }
 81         Splay(x,goal);
 82     }
 83     inline void NewNode(int &x ,int c){
 84          if(top2) x = ss[--top2];
 85          else x = ++ top1;
 86          ch[x][0] = ch[x][1] = pre[x] = 0 ; 
 87          sz[x] = 1; 
 88          
 89          val[x] = c ;
 90          rev[x] = 0 ;
 91 
 92     }
 93     inline void push_down(int x){
 94         if(rev[x])
 95         {
 96             swap(ch[x][0],ch[x][1]);
 97             rev[ch[x][0]] ^= 1; 
 98             rev[ch[x][1]] ^= 1; 
 99             rev[x] = 0 ; 
100         }
101     }
102     inline void push_up(int x){
103         sz[x] = 1 + sz[ch[x][0]] + sz[ch[x][1]];
104     }
105     inline void makeTree(int &x, int l ,int r ,int f){
106         if(l > r) return ;
107         int m = (l + r) >> 1; 
108         NewNode(x, m);
109         makeTree(ch[x][0] , l , m-1, x);
110         makeTree(ch[x][1] , m+1 , r, x);
111         pre[x] = f; 
112         push_up(x);
113     }
114     inline void init(int n){
115         rev[0] = ch[0][1] = ch[0][0] = pre[0] = sz[0] = 0 ;
116         root  = top1 =0 ; 
117         NewNode(root,-1);
118         NewNode(ch[root][1],-1);
119         pre[ch[root][1]] = root ;
120         sz[root] = 2; 
121     
122         makeTree(ch[ch[root][1]][0],1,n,ch[root][1]);
123         push_up(ch[root][1]);
124         push_up(root);
125     }
126     inline void cut(int l,int r, int si){
127     
128         RotateTo(l-1,0);
129         RotateTo(r+1,root);
130 
131         int del = ch[ch[root][1]][0];
132         ch[ch[root][1]][0] = 0 ; 
133         push_up(ch[root][1]);
134         push_up(root);
135         RotateTo(si,0);
136         RotateTo(si+1,root);
137         pre[del] = ch[root][1];
138         ch[ch[root][1]][0] = del ;
139         Splay(ch[ch[root][1]][0],0);
140         //push_up(ch[root][1]);
141         //push_up(root);
142     }
143     inline void flip(int l , int r )
144     {
145         RotateTo(l-1,0);
146         RotateTo(r+1,root);
147         rev[ch[ch[root][1]][0]] ^= 1; 
148         Splay(ch[ch[root][1]][0],0);
149     }
150     void print(int x,int k ){
151         if(x == 0 )
152             return;
153         push_down(x);
154         print(ch[x][0],k);
155         lastans[k+sz[ch[x][0]]] = val[x]; 
156         print(ch[x][1],k+1+sz[ch[x][0]]);
157     }
158     int val[maxn];
159 
160 }spt;
161 int main(){ 
162     while(scanf("%d %d",&n,&m) != EOF)
163     {
164        if(n == -1)
165            break;
166        spt.init(n);
167        for(int i = 1;i <= m;i ++)
168        {
169            scanf("%s",str);
170            if(str[0] == C)
171            {
172               scanf("%d %d %d",&ta,&tb,&tc);
173               spt.cut(ta,tb,tc);
174            }else{
175               scanf("%d %d",&ta,&tb);
176               spt.flip(ta,tb);
177            }
178        }
179        spt.print(spt.root,0);
180        for(int i = 1;i <= n; i ++)
181          printf(i == 1?"%d":" %d",lastans[i]);
182        printf("\n");
183     }
184 return 0;
185 }
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HDU 3487 Play with Chain

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原文地址:http://www.cnblogs.com/zyue/p/4409370.html

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