标签:
我直接来讲在线好了
这是一个很巧妙的方法,把边作为一个点
做一遍最小生成树,当加如一条边时,我们把这条边两点x,y的并查集的根i,j的父亲都设为这条边代表的点k,由k向i,j连边
这样我们就构建出一棵树,这棵树的叶子都是原来节点
且每棵子树都是在子树根所代表的边的限制下的最小连通块
这样我们就可以通过dfs序(只用对叶子标号)+主席树来维护k大了并通过倍增找到限制
这两题都是一副卡pascal过不了的样子……QAQ
另外网上的一些标称(bzoj3551)似乎没有考虑一个点没有边可走,但询问k=1的情况,可以加数据cha
1 type node=record 2 po,next:longint; 3 end; 4 way=record 5 x,y,z:longint; 6 end; 7 point=record 8 l,r,s:longint; 9 end; 10 11 var e:array[0..400010] of node; 12 w:array[0..500010] of way; 13 v:array[0..400010] of boolean; 14 anc:array[0..200010,0..20] of longint; 15 d,l,r,p,fa,c,a,b,h:array[0..200010] of longint; 16 tree:array[0..200010*20] of point; 17 num,j,s,size,i,len,t,n,m,q,x,y,ans,k:longint; 18 19 function min(a,b:longint):longint; 20 begin 21 if a>b then exit(b) else exit(a); 22 end; 23 24 function max(a,b:longint):longint; 25 begin 26 if a>b then exit(a) else exit(b); 27 end; 28 29 function getf(x:longint):longint; 30 begin 31 if fa[x]<>x then fa[x]:=getf(fa[x]); 32 exit(fa[x]); 33 end; 34 35 procedure swap(var a,b:longint); 36 var c:longint; 37 begin 38 c:=a; 39 a:=b; 40 b:=c; 41 end; 42 43 procedure sort(l,r:longint); 44 var i,j,x:longint; 45 begin 46 i:=l; 47 j:=r; 48 x:=a[(l+r) shr 1]; 49 repeat 50 while a[i]<x do inc(i); 51 while x<a[j] do dec(j); 52 if i<=j then 53 begin 54 swap(a[i],a[j]); 55 swap(b[i],b[j]); 56 inc(i); 57 dec(j); 58 end; 59 until i>j; 60 if l<j then sort(l,j); 61 if i<r then sort(i,r); 62 end; 63 64 procedure qsort(l,r:longint); 65 var i,j,x:longint; 66 y:way; 67 begin 68 i:=l; 69 j:=r; 70 x:=w[(l+r) shr 1].z; 71 repeat 72 while w[i].z<x do inc(i); 73 while x<w[j].z do dec(j); 74 if i<=j then 75 begin 76 y:=w[i]; w[i]:=w[j]; w[j]:=y; 77 inc(i); 78 dec(j); 79 end; 80 until i>j; 81 if l<j then qsort(l,j); 82 if i<r then qsort(i,r); 83 end; 84 85 procedure dfs(x:longint); 86 var i,y:longint; 87 begin 88 v[x]:=true; 89 if x<=n then 90 begin 91 inc(num); 92 b[num]:=x; 93 l[x]:=num; 94 r[x]:=num; 95 exit; 96 end; 97 i:=p[x]; 98 l[x]:=n; 99 r[x]:=0; 100 while i<>0 do 101 begin 102 y:=e[i].po; 103 anc[y,0]:=x; 104 dfs(y); 105 l[x]:=min(l[x],l[y]); 106 r[x]:=max(r[x],r[y]); 107 i:=e[i].next; 108 end; 109 end; 110 111 function find(x,y:longint):longint; 112 var i:longint; 113 begin 114 for i:=size downto 0 do 115 if a[anc[x,i]]<=y then x:=anc[x,i]; 116 exit(x); 117 end; 118 119 function build(l,r:longint):longint; 120 var m,q:longint; 121 begin 122 inc(t); 123 q:=t; 124 if l<>r then 125 begin 126 m:=(l+r) shr 1; 127 tree[q].l:=build(l,m); 128 tree[q].r:=build(m+1,r); 129 end; 130 exit(q); 131 end; 132 133 function add(l,r,last,x:longint):longint; 134 var m,q:longint; 135 begin 136 inc(t); 137 q:=t; 138 if l=r then tree[q].s:=tree[last].s+1 139 else begin 140 m:=(l+r) shr 1; 141 if x<=m then 142 begin 143 tree[q].r:=tree[last].r; 144 tree[q].l:=add(l,m,tree[last].l,x); 145 end 146 else begin 147 tree[q].l:=tree[last].l; 148 tree[q].r:=add(m+1,r,tree[last].r,x); 149 end; 150 tree[q].s:=tree[tree[q].l].s+tree[tree[q].r].s; 151 end; 152 exit(q); 153 end; 154 155 function ask(l,r,x,y,k:longint):longint; 156 var m,s:longint; 157 begin 158 if l=r then exit(c[l]) 159 else begin 160 m:=(l+r) shr 1; 161 s:=tree[tree[y].r].s-tree[tree[x].r].s; 162 if s>=k then exit(ask(m+1,r,tree[x].r,tree[y].r,k)) 163 else exit(ask(l,m,tree[x].l,tree[y].l,k-s)); 164 end; 165 end; 166 167 procedure ins(x,y:longint); 168 begin 169 inc(len); 170 e[len].po:=y; 171 e[len].next:=p[x]; 172 p[x]:=len; 173 end; 174 175 begin 176 readln(n,m,q); 177 for i:=1 to n do 178 begin 179 read(d[i]); 180 a[i]:=d[i]; 181 b[i]:=i; 182 end; 183 sort(1,n); 184 s:=1; 185 d[b[1]]:=1; 186 c[1]:=a[1]; 187 for i:=2 to n do 188 begin 189 if a[i]<>a[i-1] then 190 begin 191 inc(s); 192 c[s]:=a[i]; 193 end; 194 d[b[i]]:=s; 195 end; 196 for i:=1 to 2*n do 197 fa[i]:=i; 198 for i:=1 to m do 199 readln(w[i].x,w[i].y,w[i].z); 200 201 qsort(1,m); 202 t:=n; 203 fillchar(a,sizeof(a),0); 204 for i:=1 to m do 205 begin 206 x:=getf(w[i].x); 207 y:=getf(w[i].y); 208 if x<>y then 209 begin 210 inc(t); 211 fa[x]:=t; 212 fa[y]:=t; 213 a[t]:=w[i].z; 214 ins(t,x); 215 ins(t,y); 216 if t=2*n-1 then break; 217 end; 218 end; 219 220 a[0]:=2147483647; 221 len:=t; 222 for i:=1 to len do 223 if not v[i] then dfs(getf(i)); 224 size:=trunc(ln(len)/ln(2)+0.1); 225 for j:=1 to size do 226 for i:=1 to len do 227 begin 228 x:=anc[i,j-1]; 229 anc[i,j]:=anc[x,j-1]; 230 end; 231 t:=0; 232 h[0]:=build(1,s); 233 for i:=1 to num do 234 h[i]:=add(1,s,h[i-1],d[b[i]]); 235 ans:=-1; 236 for i:=1 to q do 237 begin 238 readln(x,y,k); 239 { if ans<>-1 then 240 begin 241 x:=x xor ans; 242 y:=y xor ans; 243 k:=k xor ans; 244 end; } 245 x:=find(x,y); 246 if x<=n then 247 begin 248 if k=1 then ans:=c[d[x]] 249 else ans:=-1; 250 end 251 else if tree[h[r[x]]].s-tree[h[l[x]-1]].s<k then ans:=-1 252 else ans:=ask(1,s,h[l[x]-1],h[r[x]],k); 253 writeln(ans); 254 end; 255 end.
标签:
原文地址:http://www.cnblogs.com/phile/p/4590756.html