洛谷P4211 LCA

题意：多次询问，每次求点的标号在[l, r]之间的所有点到点z的lca的深度。

解：看到这题有没有想到某一道很熟悉的题？紫妹和幽香是17岁的少女，喜欢可爱的东西......

显然这就是开店的超级无敌弱化版......直接套用做法就行了。

记得对"爱你一生一世"取模。(滑稽)

 #include <cstdio>
 #include <algorithm>

 typedef long long LL;
 const int N = , M = ;

 struct Edge {
     int nex, v;
 }edge[N]; int tp;

 int e[N], top[N], num, fa[N], siz[N], son[N], d[N], n, pos[N];
 LL sum[M];
 int rt[N], tot, ls[M], rs[M], tag[M];

 inline void adde(int x, int y) {
     tp++;
     edge[tp].v = y;
     edge[tp].nex = e[x];
     e[x] = tp;
     return;
 }

 void DFS1(int x) { // get siz fa son d
     siz[x] = ;
     d[x] = d[fa[x]] + ;
     for(int i = e[x]; i; i = edge[i].nex) {
         int y = edge[i].v;
         fa[y] = x;
         DFS1(y);
         siz[x] += siz[y];
         if(siz[y] > siz[son[x]]) {
             son[x] = y;
         }
     }
     return;
 }

 void DFS2(int x, int f) { // get pos id top
     top[x] = f;
     pos[x] = ++num;
     if(son[x]) {
         DFS2(son[x], f);
     }
     for(int i = e[x]; i; i = edge[i].nex) {
         int y = edge[i].v;
         if(y == son[x]) {
             continue;
         }
         DFS2(y, y);
     }
     return;
 }

 void add(int x, int &y, int L, int R, int l, int r) {
     if(!y || x == y) {
         y = ++tot;
         sum[y] = sum[x];
         tag[y] = tag[x];
         ls[y] = ls[x];
         rs[y] = rs[x];
     }
     sum[y] += std::min(R, r) - std::max(L, l) + ;
     if(L <= l && r <= R) {
         tag[y]++;
         return;
     }
     int mid = (l + r) >> ;
     if(L <= mid) {
         add(ls[x], ls[y], L, R, l, mid);
     }
     if(mid < R) {
         add(rs[x], rs[y], L, R, mid + , r);
     }
     return;
 }

 LL ask(int x, int y, int L, int R, int l, int r, int vx, int vy) {
     if(L <= l && r <= R) {
         return sum[y] - sum[x] + 1ll * (vy - vx) * (r - l + );
     }
     vx += tag[x];
     vy += tag[y];
     int mid = (l + r) >> ;
     LL ans = ;
     if(L <= mid) {
         ans += ask(ls[x], ls[y], L, R, l, mid, vx, vy);
     }
     if(mid < R) {
         ans += ask(rs[x], rs[y], L, R, mid + , r, vx, vy);
     }
     return ans;
 }

 inline void add(int x, int time) {
     while(x) {
         add(rt[time - ], rt[time], pos[top[x]], pos[x], , n);
         x = fa[top[x]];
     }
     return;
 }

 inline LL ask(int x, int y, int z) {
     LL ans = ;
     while(z) {
         ans += ask(rt[x - ], rt[y], pos[top[z]], pos[z], , n, , );
         z = fa[top[z]];
     }
     return ans;
 }

 int main() {
     int q;
     scanf("%d%d", &n, &q);
     for(int i = , x; i <= n; i++) {
         scanf("%d", &x);
         adde(x + , i);
     }
     DFS1();
     DFS2(, );
     for(int i = ; i <= n; i++) {
         add(i, i);
     }
     for(int i = , x, y, z; i <= q; i++) {
         scanf("%d%d%d", &x, &y, &z);
         LL t = ask(x + , y + , z + );
         printf("%lld\n", t % );
     }
     return ;
 }

AC代码