【datastruct】codeforces R158 E. Dividing Kingdom
http://codeforces.com/contest/260/problem/E
木有想到好办法，只会暴力处理=。=。。
先将所有坐标离散化，之后将9个块全排列枚举，知道每个位置最终含有节点数有可以通过2分求出x坐标已经y坐标上各自的两条分隔线，但是这样分隔并不一定是一个答案，剩下的判断即转化成求某个[xi，xj]内[yi,yj]的节点数量，这个问题可以用树套树logn^2来解决，最后200行代码ac,还好不怎么debug就通过了，但看着tourist超短的代码就知道我又太暴力了。。
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1 //By Lin 2 #include<cstdio> 3 #include<cstring> 4 #include<map> 5 #include<algorithm> 6 #define maxn 100010 7 #define X first 8 #define Y second 9 using namespace std; 10 11 struct SplayNode{ 12 SplayNode *ch[2],*fa; 13 int key,size; 14 SplayNode(int _key){ 15 key = _key; 16 size = 1; 17 ch[0] = ch[1] = fa = NULL; 18 } 19 void updata(){ 20 size = 1; 21 if ( ch[0] ) size += ch[0]->size; 22 if ( ch[1] ) size += ch[1]->size; 23 } 24 void clear(){ 25 if ( ch[0] ) ch[0]->clear(); 26 if ( ch[1] ) ch[1]->clear(); 27 delete this; 28 } 29 }; 30 31 struct Splaytree{ 32 SplayNode *root; 33 void clear(){ 34 if ( root ) root->clear(); 35 root= NULL; 36 } 37 void Rotate(SplayNode *x ){ 38 SplayNode *y = x->fa; 39 int d = y->ch[1] == x; 40 if ( y->ch[d] = x->ch[d^1] ) y->ch[d]->fa = y; 41 if ( x->fa = y->fa ) x->fa->ch[ x->fa->ch[1]==y ] = x; 42 x->ch[d^1] = y; 43 y->fa = x; 44 y->updata(); 45 x->updata(); 46 } 47 void Splay( SplayNode *x , SplayNode *y ){ 48 while ( x->fa != y ) 49 if ( x->fa->fa == y ) Rotate(x); 50 else{ 51 int d1 = x->fa->fa->ch[0] == x->fa , 52 d2 = x->fa->ch[0] == x; 53 if ( d1 == d2 ) Rotate(x->fa),Rotate(x); 54 else Rotate(x),Rotate(x); 55 } 56 } 57 void insert(SplayNode* &x , SplayNode *tmp){ 58 if ( !x ) x = tmp; 59 else { 60 if ( x->key<=tmp->key ) { 61 insert(x->ch[1],tmp); 62 x->ch[1]->fa = x; 63 } 64 else { 65 insert(x->ch[0],tmp); 66 x->ch[0]->fa = x; 67 } 68 x->updata(); 69 } 70 } 71 72 void insert(int key){ 73 SplayNode *tmp = new SplayNode(key); 74 insert( root , tmp ); 75 Splay( tmp , NULL ); 76 root = tmp; 77 } 78 SplayNode *Select( SplayNode *x ,int key ){ 79 if ( !x ) return NULL; 80 if ( x->key > key ) return Select( x->ch[0] , key ); 81 SplayNode *ret = Select( x->ch[1] , key ); 82 return ret?ret:x; 83 } 84 int ask(int key){ 85 SplayNode *tmp = Select(root,key); 86 if ( !tmp ) return 0; 87 Splay( root = tmp , NULL ); 88 return tmp->ch[0]?tmp->ch[0]->size+1:1; 89 } 90 }A; 91 struct Segtree{ 92 int left[maxn*4],right[maxn*4]; 93 Splaytree splaytree[maxn*4]; 94 void build(int l ,int r,int step ){ 95 left[step] = l , right[step] = r; 96 splaytree[step].clear(); 97 if ( l == r ) return; 98 int mid = ( l+r)/2; 99 build( l , mid , step*2 ); 100 build(mid+1, r , step*2+1 ); 101 } 102 void insert(int x,int y,int step ){ 103 splaytree[step].insert(y); 104 if ( left[step] == right[step] ) return; 105 int mid = ( left[step] + right[step] )/2; 106 if ( x <= mid ) insert( x , y , step*2 ); 107 else insert( x , y , step*2+1 ); 108 } 109 int ask(int l,int r,int y,int step){ 110 if ( l > r ) return 0; 111 if ( left[step] == l && right[step] == r ) 112 return splaytree[step].ask(y); 113 int mid = ( left[step] + right[step] )/2; 114 if ( r <= mid ) return ask(l,r,y,step*2); 115 else if ( l > mid ) return ask(l,r,y,step*2+1); 116 else 117 return ask(l,mid,y,step*2)+ask(mid+1,r,y,step*2+1); 118 } 119 }tree; 120 121 int n; 122 int xx[maxn],yy[maxn],xcnt,ycnt,xsum[maxn],ysum[maxn]; 123 int num[9]; 124 pair<int,int> data[maxn]; 125 map<int,int> mm; 126 127 bool xget(int key, int &ret ){ 128 int g = 0 , h = xcnt; 129 while ( g<=h ) { 130 int mid = (g+h)/2; 131 if ( xsum[mid] == key ) { ret = mid; return true; } 132 if ( xsum[mid] < key ) g = mid+1; 133 else h = mid-1; 134 } 135 return false; 136 } 137 bool yget(int key, int &ret ){ 138 int g = 0 , h = ycnt; 139 while ( g<=h ) { 140 int mid = (g+h)/2; 141 if ( ysum[mid] == key ) { ret = mid; return true; } 142 if ( ysum[mid] < key ) g = mid+1; 143 else h = mid-1; 144 } 145 return false; 146 } 147 int main(){ 148 // A.clear(); 149 // int x,y; 150 // while ( ~scanf("%d%d", &x, &y ) ) { 151 // if ( x == 1 ) A.insert(y); 152 // else printf("%d\n" , A.ask(y) ); 153 // } 154 // return 0; 155 scanf("%d", &n ); 156 for (int i = 0; i<n; i++) { 157 scanf("%d%d", &data[i].X , &data[i].Y ); 158 xx[i] = data[i].X; 159 yy[i] = data[i].Y; 160 } 161 sort( xx , xx+n ); 162 xcnt = unique(xx,xx+n)-xx; 163 xx[xcnt] = xx[xcnt-1]+1; 164 for (int i = 0; i<xcnt; i++) mm[xx[i]] = i+1; 165 for (int i = 0; i<n; i++) xsum[data[i].X = mm[data[i].X]]++; 166 for (int i = 1; i<=xcnt; i++) xsum[i] += xsum[i-1]; 167 mm.clear(); 168 sort( yy , yy+n ); 169 ycnt = unique(yy,yy+n)-yy; 170 yy[ycnt] = yy[ycnt-1]+1; 171 for (int i = 0; i<ycnt; i++) mm[yy[i]] = i+1; 172 for (int i = 0; i<n; i++) ysum[data[i].Y = mm[data[i].Y]]++; 173 for (int i = 1; i<=ycnt; i++) ysum[i] += ysum[i-1]; 174 175 tree.build(1,xcnt,1); 176 for (int i = 0; i<n; i++) tree.insert( data[i].X , data[i].Y , 1 ); 177 for (int i = 0; i<9; i++) scanf("%d", &num[i] ); 178 sort( num, num+9 ); 179 bool flag = false; 180 do{ 181 int x1 = num[0]+num[1]+num[2] , 182 x2 = num[3]+num[4]+num[5] , 183 x3 = num[6]+num[7]+num[8] , 184 y1 = num[0]+num[3]+num[6] , 185 y2 = num[1]+num[4]+num[7] , 186 y3 = num[2]+num[5]+num[8] ; 187 int g , h , l , r; 188 if ( !xget( x1 , g ) ) continue; 189 if ( !xget( x1+x2 , h ) ) continue; 190 if ( !yget( y1 , l ) ) continue; 191 if ( !yget( y1+y2 , r ) ) continue; 192 if ( tree.ask(1,g,l,1) != num[0] ) continue; 193 if ( tree.ask(1,g,r,1) != num[0]+num[1] ) continue; 194 if ( tree.ask(g+1,h,l,1) != num[3] ) continue; 195 if ( tree.ask(g+1,h,r,1) != num[3]+num[4] ) continue; 196 if ( tree.ask(h+1,xcnt,l,1) != num[6] ) continue; 197 if ( tree.ask(h+1,xcnt,r,1) != num[6]+num[7] ) continue; 198 flag = true; 199 printf("%.6f %.6f\n" , xx[g]-0.6 , xx[h]-0.4 ); 200 printf("%.6f %.6f\n" , yy[l]-0.6 , yy[r]-0.4 ); 201 break; 202 } while ( next_permutation(num,num+9) ); 203 if ( !flag ) printf("-1\n"); 204 }
