简单的两种操作,一种删除某段区间,加在第I个点的后面,另一个是翻转区间。都是splay的简单操作。
悲剧一:pushdown时候忘记让lz=0
悲剧二:删除区间,加在某点之后的时候忘记修改其父亲节点。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 300010 12 #define LL long long 13 #define INF 0xfffffff 14 #define key_value ch[ch[root][1]][0] 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 using namespace std; 19 struct splay_tree 20 { 21 int pre[N],size[N]; 22 int ch[N][2]; 23 int root,tot,num,tot1; 24 int key[N],lz[N]; 25 void dfs(int x) 26 { 27 if(x) 28 { 29 dfs(ch[x][0]); 30 printf("结点%2d:左儿子 %2d 右儿子 %2d 父结点 %2d size=%2d,key=%2d lz = %d\n", 31 x,ch[x][0],ch[x][1],pre[x],size[x],key[x],lz[x]); 32 dfs(ch[x][1]); 33 } 34 } 35 void debug() 36 { 37 printf("root:%d\n",root); 38 dfs(root); 39 } 40 //以上用于debug*/ 41 void newnode(int &x,int v,int fa)//新建一结点 42 { 43 x = ++tot; 44 ch[x][0]=ch[x][1] = 0; 45 pre[x] = fa; 46 lz[x] = 0; 47 size[x] = 1; 48 key[x] = v; 49 } 50 void pushdown(int w) 51 { 52 if(lz[w]) 53 { 54 int l = ch[w][0],r = ch[w][1]; 55 swap(ch[w][0],ch[w][1]); 56 57 lz[l]^=lz[w]; 58 lz[r]^=lz[w]; 59 lz[w] = 0; 60 } 61 } 62 void pushup(int w)//由儿子更新其父亲 63 { 64 size[w] = size[ch[w][0]]+size[ch[w][1]]+1; 65 //cout<<s[w][0]<<" "<<s[w][1]<<endl; 66 } 67 void rotate(int r,int kind)//旋转操作,根据kind进行左旋和右旋 68 { 69 int y = pre[r]; 70 pushdown(y); 71 pushdown(r); 72 ch[y][!kind] = ch[r][kind]; 73 pre[ch[r][kind]] = y; 74 if(pre[y]) 75 { 76 ch[pre[y]][ch[pre[y]][1]==y] = r; 77 } 78 pre[r] = pre[y]; 79 ch[r][kind] = y; 80 pre[y] = r; 81 pushup(y); 82 pushup(r); 83 } 84 void splay(int r,int goal)//将r结点旋至goal下 85 { 86 pushdown(r); 87 while(pre[r]!=goal) 88 { 89 if(pre[pre[r]]==goal) 90 { 91 rotate(r,ch[pre[r]][0]==r); 92 } 93 else 94 { 95 int y = pre[r]; 96 int kind = (ch[pre[y]][0]==y); 97 if(ch[y][kind]==r) 98 { 99 rotate(r,!kind); 100 rotate(r,kind); 101 } 102 else 103 { 104 rotate(y,kind); 105 rotate(r,kind); 106 } 107 } 108 } 109 pushup(r); 110 if(goal==0) root = r; 111 } 112 int get_k(int k)//得到第k个的结点 113 { 114 int r = root; 115 pushdown(r); 116 while(size[ch[r][0]]+1!=k) 117 { 118 if(size[ch[r][0]]>=k) 119 r = ch[r][0]; 120 else 121 { 122 k-=(size[ch[r][0]]+1);//根据左右结点的数量来确定第k个节点在哪里 123 r = ch[r][1]; 124 } 125 pushdown(r); 126 } 127 pushup(r); 128 return r; 129 } 130 void cut(int l,int r,int k) 131 { 132 splay(get_k(l),0); 133 splay(get_k(r+2),root); 134 pushdown(ch[root][1]); 135 int nod = key_value; 136 ch[ch[root][1]][0] = 0; 137 pre[nod] = 0; 138 pushup(ch[root][1]); 139 pushup(root); 140 //debug(); 141 splay(get_k(k),0); 142 splay(get_k(k+1),root); 143 ch[ch[root][1]][0] = nod; 144 pre[nod] = ch[root][1]; 145 pushup(ch[root][1]); 146 pushup(root); 147 // debug(); 148 } 149 void filp(int l,int r) 150 { 151 //cout<<get_k(l)<<" "<<get_k(r+2)<<endl; 152 splay(get_k(l),0); 153 splay(get_k(r+2),root); //cout<<","<<endl;debug(); 154 lz[key_value]^=1; 155 //swap(ch[key_value][0],ch[key_value][1]); 156 pushup(ch[root][1]); 157 pushup(root); 158 } 159 void work(int n) 160 { 161 int i; 162 for(i = 1 ;i < n ;i++) 163 { 164 printf("%d ",key[get_k(i+1)]); 165 } 166 printf("%d\n",key[get_k(n+1)]); 167 //debug(); 168 } 169 void build(int &x,int l,int r,int fa) 170 { 171 int m = (l+r)>>1; 172 if(l>r) return ; 173 newnode(x,m,fa); 174 build(ch[x][0],l,m-1,x); 175 build(ch[x][1],m+1,r,x); 176 pushup(x); 177 } 178 void init(int o) 179 { 180 size[0] = ch[0][0] = ch[0][1] = key[0] = lz[0] = 0; 181 root = tot = 0; 182 newnode(root,0,0); 183 newnode(ch[root][1],0,root); 184 build(ch[ch[root][1]][0],1,o,ch[root][1]); 185 size[root] = 2; 186 pushup(ch[root][1]); 187 pushup(root); 188 } 189 }SP; 190 int main() 191 { 192 int n,q; 193 while(scanf("%d%d",&n,&q)!=EOF) 194 { 195 if(n==-1&&q==-1) break; 196 SP.init(n); 197 while(q--) 198 { 199 char sq[20]; 200 int k,x,y; 201 scanf("%s",sq); 202 if(sq[0]==‘C‘) 203 { 204 scanf("%d%d%d",&x,&y,&k); 205 SP.cut(x,y,k+1); 206 // SP.debug(); 207 // SP.work(n); 208 } 209 else 210 { 211 //SP.debug(); 212 scanf("%d%d",&x,&y); 213 SP.filp(x,y); 214 //SP.debug(); 215 } 216 //SP.work(n); 217 } 218 SP.work(n); 219 } 220 return 0; 221 }
hdu3487Play with Chain(splay),布布扣,bubuko.com
原文:http://www.cnblogs.com/shangyu/p/3796693.html