百度空间马上要下架的说，赶快把最后一点题解补完，然后搬家
这是一道不错的题，首先注意询问是满足区间减法的，我们把他变成前缀和表示
设我们询问[1,r]中的点和z的LCA深度和，假设我们确定一个根，不难发现一个有趣的事情
点z和点i的LCA深度=z和i到根公共路径（LCA到根的路径）上点的个数！
也就是说，当我们查询时，我们只要知道，[1,r]上的点到根路径上的点在z到根上出现的次数和
所以不难想到离线的做法，按照编号的顺序，每次做到点i，就把点i到根路径上点权值+1
询问的时候我们只要查询z到根路径上点权值和即可，显然我们可以用树链剖分+线段树维护
1 const mo=201314; 2 type node=record 3 po,next:longint; 4 end; 5 que=record 6 x,y,p,op:longint; 7 end; 8 9 var p,c,top,ans,fa,size:array[0..50010] of longint; 10 tree,lazy:array[0..50010*4] of longint; 11 q:array[0..100010] of que; 12 e:array[0..50010] of node; 13 t,l,r,x,i,j,n,m:longint; 14 15 procedure swap(var a,b:que); 16 var c:que; 17 begin 18 c:=a; 19 a:=b; 20 b:=c; 21 end; 22 23 procedure add(x,y:longint); 24 begin 25 e[i].po:=y; 26 e[i].next:=p[x]; 27 p[x]:=i; 28 end; 29 30 procedure dfs1(x:longint); 31 var i,y:longint; 32 begin 33 size[x]:=1; 34 i:=p[x]; 35 while i<>0 do 36 begin 37 y:=e[i].po; 38 fa[y]:=x; 39 dfs1(y); 40 size[x]:=size[x]+size[y]; 41 i:=e[i].next; 42 end; 43 end; 44 45 procedure dfs2(x:longint); 46 var i,y,q:longint; 47 begin 48 inc(t); 49 c[x]:=t; 50 q:=n+1; 51 i:=p[x]; 52 while i<>0 do 53 begin 54 y:=e[i].po; 55 if size[y]>size[q] then q:=y; 56 i:=e[i].next; 57 end; 58 if q<>n+1 then 59 begin 60 top[q]:=top[x]; 61 dfs2(q); 62 end; 63 i:=p[x]; 64 while i<>0 do 65 begin 66 y:=e[i].po; 67 if y<>q then 68 begin 69 top[y]:=y; 70 dfs2(y); 71 end; 72 i:=e[i].next; 73 end; 74 end; 75 76 procedure sort(l,r:longint); 77 var i,j,y:longint; 78 begin 79 i:=l; 80 j:=r; 81 y:=q[(l+r) shr 1].y; 82 repeat 83 while q[i].y<y do inc(i); 84 while y<q[j].y do dec(j); 85 if not(i>j) then 86 begin 87 swap(q[i],q[j]); 88 inc(i); 89 dec(j); 90 end; 91 until i>j; 92 if l<j then sort(l,j); 93 if i<r then sort(i,r); 94 end; 95 96 procedure push(i,l,r:longint); 97 var m:longint; 98 begin 99 m:=(l+r) shr 1; 100 tree[i*2]:=tree[i*2]+lazy[i]*(m+1-l); 101 tree[i*2+1]:=tree[i*2+1]+lazy[i]*(r-m); 102 inc(lazy[i*2],lazy[i]); 103 inc(lazy[i*2+1],lazy[i]); 104 lazy[i]:=0; 105 end; 106 107 procedure work(i,l,r,x,y:longint); 108 var m:longint; 109 begin 110 if (x<=l) and (y>=r) then 111 begin 112 tree[i]:=tree[i]+r-l+1; 113 lazy[i]:=lazy[i]+1; 114 end 115 else begin 116 if lazy[i]>0 then push(i,l,r); 117 m:=(l+r) shr 1; 118 if x<=m then work(i*2,l,m,x,y); 119 if y>m then work(i*2+1,m+1,r,x,y); 120 tree[i]:=(tree[i*2]+tree[i*2+1]) mod mo; 121 end; 122 end; 123 124 function getans(i,l,r,x,y:longint):longint; 125 var m,s:longint; 126 begin 127 if (x<=l) and (y>=r) then exit(tree[i]) 128 else begin 129 if lazy[i]>0 then push(i,l,r); 130 m:=(l+r) shr 1; 131 s:=0; 132 if x<=m then s:=s+getans(i*2,l,m,x,y); 133 if y>m then s:=s+getans(i*2+1,m+1,r,x,y); 134 exit(s); 135 end; 136 end; 137 138 procedure ins(x:longint); 139 begin 140 while top[x]<>-1 do 141 begin 142 work(1,1,n,c[top[x]],c[x]); 143 x:=fa[top[x]]; 144 end; 145 work(1,1,n,1,c[x]); 146 end; 147 148 function ask(x:longint):longint; 149 begin 150 ask:=0; 151 while top[x]<>-1 do 152 begin 153 ask:=ask+getans(1,1,n,c[top[x]],c[x]); 154 x:=fa[top[x]]; 155 end; 156 ask:=(ask+getans(1,1,n,1,c[x])) mod mo; 157 end; 158 159 begin 160 readln(n,m); 161 for i:=1 to n-1 do 162 begin 163 read(x); 164 add(x,i); 165 end; 166 dfs1(0); 167 top[0]:=-1; 168 dfs2(0); 169 t:=0; 170 for i:=1 to m do 171 begin 172 readln(l,r,x); 173 inc(t); 174 q[t].x:=x; q[t].y:=l-1; 175 q[t].p:=i; q[t].op:=-1; 176 inc(t); 177 q[t].x:=x; q[t].y:=r; 178 q[t].p:=i; q[t].op:=1; 179 end; 180 sort(1,t); 181 j:=0; 182 for i:=1 to t do 183 begin 184 while (j<n) and (j<=q[i].y) do 185 begin 186 ins(j); 187 inc(j); 188 end; 189 ans[q[i].p]:=(ans[q[i].p]+q[i].op*ask(q[i].x)+mo) mod mo; 190 end; 191 for i:=1 to m do 192 writeln(ans[i]); 193 end.
