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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 }
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原文地址：http://www.cnblogs.com/idy002/p/4282200.html
