bzoj3224 普通平衡树
这道题我先写了值域线段树，一直WA，和网上的标程对拍，也拍不出错误，然后改写SPALY，又WA，也拍不出错误，最后只能用vector水过了。
我把我写的值域线段树、Splay、vector、数据生成器放在下面，若有哪位好心人愿意帮我看看，感激不尽。
值域线段树：
1 /************************************************************** 2 Problem: 3224 3 User: idy002 4 Language: C++ 5 Result: Wrong_Answer 6 ****************************************************************/ 7 8 #include <cstdio> 9 #define fprintf(...) 10 #define maxn 7000000 11 #define minv 0 12 #define inc 10000000 13 #define maxv 20000000 14 15 int son[maxn][2], siz[maxn], ntot; 16 17 int newnode() { 18 int nd = ++ntot; 19 son[nd][0] = son[nd][1] = siz[nd] = 0; 20 return nd; 21 } 22 void pushup( int nd ) { 23 siz[nd] = siz[son[nd][0]]+siz[son[nd][1]]; 24 } 25 void insert( int v, int nd, int lf, int rg ) { 26 fprintf( stderr, "insert( %d %d %d %d )\n", v, nd, lf, rg ); 27 if( lf==rg ) { 28 siz[nd]++; 29 return; 30 } 31 int mid = (lf+rg)>>1; 32 if( !son[nd][ v>mid ] ) son[nd][ v>mid ] = newnode(); 33 if( v<=mid ) insert( v, son[nd][0], lf, mid ); 34 else insert( v, son[nd][1], mid+1, rg ); 35 pushup(nd); 36 } 37 void erase( int v, int nd, int lf, int rg ) { 38 fprintf( stderr, "erase( %d %d %d %d )\n", v, nd, lf, rg ); 39 if( lf==rg ) { 40 siz[nd]--; 41 return; 42 } 43 int mid = (lf+rg)>>1; 44 if( v<=mid ) erase( v, son[nd][0], lf, mid ); 45 else erase( v, son[nd][1], mid+1, rg ); 46 pushup(nd); 47 } 48 int rank( int v, int nd, int lf, int rg ) { 49 fprintf( stderr, "rank( %d %d %d %d )\n", v, nd, lf, rg ); 50 if( lf==rg ) return 1; 51 int mid = (lf+rg)>>1; 52 if( v<=mid ) return rank(v,son[nd][0],lf,mid); 53 else return siz[son[nd][0]]+rank(v,son[nd][1],mid+1,rg); 54 } 55 int nth( int n, int nd, int lf, int rg ) { 56 fprintf( stderr, "nth( %d %d %d %d )\n", n, nd, lf, rg ); 57 if( lf==rg ) return lf; 58 int ls = siz[son[nd][0]]; 59 int mid = (lf+rg)>>1; 60 if( n<=ls ) return nth(n,son[nd][0],lf,mid); 61 else return nth(n-ls,son[nd][1],mid+1,rg); 62 } 63 int gnext( int v, int nd, int lf, int rg ) { 64 fprintf( stderr, "gnext( %d %d %d %d )\n", v, nd, lf, rg ); 65 if( !nd ) return v; 66 if( lf==rg ) return lf>v ? lf : v; 67 int mid = (lf+rg)>>1; 68 if( v<=mid ) { 69 int rt = gnext( v, son[nd][0], lf, mid ); 70 if( rt==v ) return gnext( v, son[nd][1], mid+1, rg ); 71 else return rt; 72 } 73 return gnext( v, son[nd][1], mid+1, rg ); 74 } 75 int gprev( int v, int nd, int lf, int rg ) { 76 fprintf( stderr, "gprev( %d %d %d %d )\n", v, nd, lf, rg ); 77 if( !nd ) return v; 78 if( lf==rg ) return lf<v ? lf : v; 79 int mid = (lf+rg)>>1; 80 if( v>mid ) { 81 int rt = gprev( v, son[nd][1], mid+1, rg ); 82 if( rt==v ) return gprev( v, son[nd][0], lf, mid ); 83 else return rt; 84 } 85 return gprev( v, son[nd][0], lf, mid ); 86 } 87 88 int n; 89 90 int main() { 91 scanf( "%d", &n ); 92 newnode(); 93 for( int i=1,opt,x; i<=n; i++ ) { 94 scanf( "%d%d", &opt, &x ); 95 x += inc; 96 switch(opt) { 97 case 1: 98 insert( x, 1, minv, maxv ); 99 break; 100 case 2: 101 erase( x, 1, minv, maxv ); 102 break; 103 case 3: 104 printf( "%d\n", rank(x,1,minv,maxv) ); 105 break; 106 case 4: 107 printf( "%d\n", nth(x-inc,1,minv,maxv)-inc ); 108 break; 109 case 5: 110 printf( "%d\n", gprev(x,1,minv,maxv)-inc ); 111 break; 112 case 6: 113 printf( "%d\n", gnext(x,1,minv,maxv)-inc ); 114 break; 115 } 116 } 117 }
splay:
1 #include <cstdio> 2 #include <iostream> 3 #define maxn 100020 4 using namespace std; 5 6 7 struct Splay { 8 int key[maxn], pre[maxn], son[maxn][2], siz[maxn], cnt[maxn], root, ntot; 9 10 int newnode( int k, int p ) { 11 int nd = ++ntot; 12 key[nd] = k; 13 pre[nd] = p; 14 son[nd][0] = son[nd][1] = 0; 15 siz[nd] = cnt[nd] = 1; 16 return nd; 17 } 18 void update( int nd ) { 19 siz[nd] = siz[son[nd][0]] + siz[son[nd][1]] + cnt[nd]; 20 } 21 void rotate( int nd, int d ) { 22 int p = pre[nd]; 23 int s = son[nd][!d]; 24 int ss = son[s][d]; 25 26 son[nd][!d] = ss; 27 son[s][d] = nd; 28 if( p ) son[p][ nd==son[p][1] ] = s; 29 else root = s; 30 31 pre[nd] = s; 32 pre[s] = p; 33 if( ss ) pre[ss] = nd; 34 35 update(nd); 36 update(s); 37 } 38 void splay( int nd, int top ) { 39 while( pre[nd]!=top ) { 40 int p = pre[nd]; 41 int nl = nd==son[p][0]; 42 if( pre[p]==top ) { 43 rotate( p, nl ); 44 } else { 45 int pp = pre[p]; 46 int pl = p==son[pp][0]; 47 if( nl==pl ) { 48 rotate( pp, pl ); 49 rotate( p, nl ); 50 } else { 51 rotate( p, nl ); 52 rotate( pp, pl ); 53 } 54 } 55 } 56 } 57 void insert( int k ) { 58 if( !root ) { 59 root = newnode( k, 0 ); 60 return; 61 } 62 int nd = root; 63 while( 1 ) { 64 if( k==key[nd] ) { 65 cnt[nd]++; 66 break; 67 } else { 68 if( !son[nd][ k>key[nd] ] ) { 69 son[nd][ k>key[nd] ] = newnode( k, nd ); 70 break; 71 } 72 nd = son[nd][ k>key[nd] ]; 73 } 74 } 75 update( nd ); 76 splay( nd, 0 ); 77 } 78 int find( int k ) { 79 int nd = root; 80 while( 1 ) { 81 if( k!=key[nd] ) nd = son[nd][ k>key[nd] ]; 82 else break; 83 } 84 return nd; 85 } 86 void erase( int k ) { 87 int nd = find(k); 88 cnt[nd]--; 89 update(nd); 90 splay(nd,0); 91 } 92 int gnext( int k ) { 93 insert( k ); 94 int nd = find( k ); 95 cnt[nd]--; 96 update(nd); 97 splay(nd,0); 98 int rnd = son[nd][1]; 99 if( !rnd ) fprintf( stderr, "gnext k=%d\n", k ); 100 int rt = key[rnd]; 101 while( son[rnd][0] ) { 102 rnd = son[rnd][0]; 103 rt = min( rt, key[rnd] ); 104 } 105 splay( rnd,0 ); 106 return rt; 107 } 108 int gprev( int k ) { 109 insert( k ); 110 int nd = find( k ); 111 cnt[nd]--; 112 update(nd); 113 splay(nd,0); 114 int lnd = son[nd][0]; 115 if( !lnd ) fprintf( stderr, "gprev k=%d\n", k ); 116 int rt = key[lnd]; 117 while( son[lnd][1] ) { 118 lnd = son[lnd][1]; 119 rt = max( rt, key[lnd] ); 120 } 121 splay(lnd,0); 122 return rt; 123 } 124 int rank( int k ) { 125 int nd = root; 126 int rt = 0; 127 while(1) { 128 int ls = siz[son[nd][0]]; 129 int cs = cnt[nd]; 130 if( k==key[nd] ) { 131 rt += ls+1; 132 splay( nd,0 ); 133 return rt; 134 } 135 if( k<key[nd] ) { 136 nd = son[nd][0]; 137 } else { 138 nd = son[nd][1]; 139 rt += ls+cs; 140 } 141 } 142 } 143 int nth( int n ) { 144 int nd = root; 145 while(1) { 146 int ls = siz[son[nd][0]]; 147 int cs = cnt[nd]; 148 if( n<=ls ) { 149 nd = son[nd][0]; 150 } else if( n<=ls+cs ) { 151 splay( nd, 0 ); 152 return key[nd]; 153 } else { 154 n -= ls+cs; 155 nd = son[nd][1]; 156 } 157 } 158 } 159 void print( int nd ) { 160 if( !nd ) return; 161 print( son[nd][0] ); 162 fprintf( stderr, "%d(%d) ", key[nd], cnt[nd] ); 163 print( son[nd][1] ); 164 } 165 }; 166 167 int n; 168 Splay T; 169 int main() { 170 scanf( "%d", &n ); 171 while(n--) { 172 int opt, x; 173 scanf( "%d%d", &opt, &x ); 174 switch( opt ) { 175 case 1: 176 // fprintf( stderr, "insert(%d)\n", x ); 177 T.insert(x); 178 break; 179 case 2: 180 // fprintf( stderr, "erase(%d)\n", x ); 181 T.erase(x); 182 break; 183 case 3: 184 printf( "%d\n", T.rank(x) ); 185 break; 186 case 4: 187 printf( "%d\n", T.nth(x) ); 188 break; 189 case 5: 190 printf( "%d\n", T.gprev(x) ); 191 break; 192 case 6: 193 printf( "%d\n", T.gnext(x) ); 194 break; 195 } 196 //T.print( T.root ); 197 //fprintf( stderr, "\n" ); 198 } 199 }
vector:
1 #include <cstdio> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5 6 7 vector<int> vc; 8 void insert( int x ) { 9 vc.insert( lower_bound(vc.begin(),vc.end(),x), x ); 10 } 11 void erase( int x ) { 12 vc.erase( lower_bound(vc.begin(),vc.end(),x) ); 13 } 14 int rank( int x ) { 15 return lower_bound(vc.begin(),vc.end(),x)-vc.begin()+1; 16 } 17 int nth( int n ) { 18 return vc[n-1]; 19 } 20 int prev( int x ) { 21 return *--lower_bound(vc.begin(),vc.end(),x); 22 } 23 int next( int x ) { 24 return *upper_bound(vc.begin(),vc.end(),x); 25 } 26 27 int main() { 28 int n; 29 scanf( "%d", &n ); 30 while(n--) { 31 int opt, x; 32 scanf( "%d%d", &opt, &x ); 33 switch(opt) { 34 case 1: 35 insert( x ); 36 break; 37 case 2: 38 erase(x); 39 break; 40 case 3: 41 printf( "%d\n", rank(x) ); 42 break; 43 case 4: 44 printf( "%d\n", nth(x) ); 45 break; 46 case 5: 47 printf( "%d\n", prev(x) ); 48 break; 49 case 6: 50 printf( "%d\n", next(x) ); 51 break; 52 } 53 } 54 }
数据生成器(有一个参数，是随机数种子）：
1 #include <cstdio> 2 #include <set> 3 #include <cstdlib> 4 #define R(l,r) ((rand())%((r)-(l)+1)+(l)) 5 using namespace std; 6 7 int all = 1000; //指令数 8 int n = 500; //最开始的插入的数的数量 9 int maxd = 200; //最多删除多少个数 10 11 int a[101100], mn, mx; 12 int d[101100]; 13 14 int main( int argc, char **argv ) { 15 srand(atoi(argv[1])); 16 17 freopen( "input", "w", stdout ); 18 printf( "%d\n", all ); 19 int p = all-n; 20 for( int i=1; i<=n; i++ ) 21 printf( "1 %d\n", a[i]=R(-n,n) ); 22 int remain = maxd; 23 for( int i=1; i<=p; i++ ) { 24 AGAIN: 25 if( i%10000==0 ) fprintf( stderr, "%d\n", i ); 26 int opt = R(2,6); 27 int x; 28 switch( opt ) { 29 case 2: 30 if( remain ) { 31 int ind; 32 do 33 ind = R(1,n); 34 while( !(!d[ind] && a[ind]!=mn && a[ind]!=mx) ); 35 remain--; 36 d[ind] = 1; 37 printf( "2 %d\n", a[ind] ); 38 } else goto AGAIN; 39 break; 40 case 3: 41 { 42 int ind; 43 do 44 ind = R(1,n); 45 while( !(!d[ind]) ); 46 printf( "3 %d\n", a[ind] ); 47 } 48 break; 49 case 4: 50 { 51 set<int> st; 52 for( int i=1; i<=n; i++ ) 53 if( !d[i] ) st.insert(a[i]); 54 int x = R(1,st.size()); 55 printf( "4 %d\n", x ); 56 } 57 break; 58 case 5: 59 if( mx==mn ) goto AGAIN; 60 { 61 printf( "5 %d\n", R(mn+1,mx) ); 62 } 63 break; 64 case 6: 65 if( mx==mn ) goto AGAIN; 66 { 67 printf( "6 %d\n", R(mn,mx-1) ); 68 } 69 break; 70 } 71 } 72 }
20150708
终于填了这个坑。
将splay中rank(k)的返回值定义为小于k的数的个数加一。
然后k的前趋就是nth(rank(k)-1)
k的后继就是nth(rank(k+1))
1 /************************************************************** 2 Problem: 3224 3 User: idy002 4 Language: C++ 5 Result: Accepted 6 Time:776 ms 7 Memory:2772 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 12 const int N = 100000 + 10; 13 14 struct Splay { 15 int son[N][2], fat[N], key[N], siz[N], ntot, root; 16 17 inline void update( int nd ) { 18 siz[nd] = siz[son[nd][0]] + siz[son[nd][1]] + 1; 19 } 20 void rotate( int nd, int d ) { 21 int p = fat[nd]; 22 int s = son[nd][!d]; 23 int ss = son[s][d]; 24 25 if( p ) son[p][ nd==son[p][1] ] = s; 26 else root=s; 27 son[nd][!d] = ss; 28 son[s][d] = nd; 29 30 fat[nd] = s; 31 fat[s] = p; 32 if( ss ) fat[ss]=nd; 33 34 update(nd); 35 update(s); 36 } 37 void splay( int nd, int top=0 ) { 38 while( fat[nd]!=top ) { 39 int p=fat[nd]; 40 int nl=nd==son[p][0]; 41 if( fat[p]==top ) { 42 rotate( p, nl ); 43 } else { 44 int pp=fat[p]; 45 int pl=p==son[pp][0]; 46 if( nl==pl ) { 47 rotate( pp, pl ); 48 rotate( p, nl ); 49 } else { 50 rotate( p, nl ); 51 rotate( pp, pl ); 52 } 53 } 54 } 55 } 56 int newnode( int p, int k ) { 57 int nd = ++ntot; 58 key[nd] = k; 59 siz[nd] = 1; 60 fat[nd] = p; 61 son[nd][0] = son[nd][1] = 0; 62 return nd; 63 } 64 void insert( int v ) { 65 if( !root ) { 66 root = newnode( 0, v ); 67 return; 68 } 69 int nd = root; 70 while( son[nd][v>key[nd]] ) nd=son[nd][v>key[nd]]; 71 son[nd][v>key[nd]] = newnode( nd, v ); 72 update(nd); 73 splay(nd); 74 } 75 int find( int v ) { 76 int nd = root; 77 while( key[nd]!=v ) nd=son[nd][ v>key[nd] ]; 78 return nd; 79 } 80 void erase( int v ) { 81 int nd = find(v); 82 splay(nd); 83 int lnd = son[nd][0]; 84 int rnd = son[nd][1]; 85 if( !lnd && !rnd ) { 86 root = 0; 87 } else if( !lnd ) { 88 root = rnd; 89 fat[rnd] = 0; 90 } else if( !rnd ) { 91 root = lnd; 92 fat[lnd] = 0; 93 } else { 94 while( son[lnd][1] ) lnd=son[lnd][1]; 95 while( son[rnd][0] ) rnd=son[rnd][0]; 96 splay( lnd, 0 ); 97 splay( rnd, lnd ); 98 son[rnd][0] = 0; 99 update(rnd); 100 update(lnd); 101 } 102 } 103 int nth( int k ) { 104 int nd=root; 105 while(1) { 106 int lz = siz[son[nd][0]]; 107 if( k<=lz ) { 108 nd = son[nd][0]; 109 } else if( k>=lz+2 ) { 110 k -= lz+1; 111 nd = son[nd][1]; 112 } else { 113 splay(nd); 114 return key[nd]; 115 } 116 } 117 } 118 int rank( int v ) { 119 int nd=root; 120 int rt = 1; 121 int last_nd; 122 while(nd) { 123 last_nd = nd; 124 if( key[nd]<v ) { 125 rt += siz[son[nd][0]] + 1; 126 nd = son[nd][1]; 127 } else { 128 nd = son[nd][0]; 129 } 130 } 131 splay(last_nd); 132 return rt; 133 } 134 int prev( int v ) { 135 int k = rank(v); 136 return nth(k-1); 137 } 138 int succ( int v ) { 139 int k = rank(v+1); 140 return nth(k); 141 } 142 }T; 143 144 int n; 145 146 int main() { 147 scanf( "%d", &n ); 148 for( int i=1,opt,x; i<=n; i++ ) { 149 scanf( "%d%d", &opt, &x ); 150 if( opt==1 ) { 151 T.insert(x); 152 } else if( opt==2 ) { 153 T.erase(x); 154 } else if( opt==3 ) { 155 printf( "%d\n", T.rank(x) ); 156 } else if( opt==4 ) { 157 printf( "%d\n", T.nth(x) ); 158 } else if( opt==5 ) { 159 printf( "%d\n", T.prev(x) ); 160 } else { 161 printf( "%d\n", T.succ(x) ); 162 } 163 } 164 }
Update：2018—01—18
来自好心人：我们发现这个Splay的删除是惰性删除的，并且很多地方的更改是无效的。这也就导致了一个问题：已经存在于这棵Splay里面的一些节点其实在原来的序列中是不存在的。那么这就导致了如果直接在树上找前驱后继时遍历节点会遍历到不在原序列中的数。于是就GG了。
In：
5
1 1
1 2
1 3
2 2
5 3
Out：
2
但应该是1对吧......
