3052: [wc2013]糖果公园

3052: [wc2013]糖果公园

Time Limit: 200 Sec  Memory Limit: 512 MB
Submit: 962  Solved: 464
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Description

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84
131
27
84

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树上莫队，，今天一天几乎都花在这上面了首先是按照bzoj1086的方法把树分块，每个块的大小控制在N^(2/3)对于每个询问，假设路径为(u,v)且u所在的块的编号不大于v把所有询问操作按照u为第一关键字，v为第二关键字，时间为第三关键字排序我们维护的信息是当前这条路径的ans(除去lca)转移的时候，设下一次的路径是(x,y)先暴力转移时间然后暴力从u走到x，路径上的点的信息取反同样地从v走到yans的话，，把lca的信息补上即可为什么这样做是对的呢？？？vfleaking大神讲的非常有道理
最后是时间复杂度(苟蒻也只会证明这个了。。GG)我们把树许多个分成了N^(2/3)的块于是块的数量就不超过N(1/3)显然，在同一块内，你怎么跑单次的复杂度都不超过O(N^(2/3))对于u所在的块确定是，v所在的块若在块内跑O(N^(2/3))，若要跑向右边的块，，总和是N^(2/3)*N^(1/3) = N至于每次确定两端点的块时，时间修改的总和是O(N)  右子树跑O(N^(1/3))次，左子树也是一样所以时间的修改是O(N^(5/3))，路径同理这个复杂度，，强行没有N^2反正可以过

写的时候Movetime函数写残了，，好蠢啊点取反再取回来是要特判的！！！得从cur for 到 target才对呀！！！！

#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<bitset>#include<algorithm>#include<cstring>#include<map>#include<stack>#include<set>#include<cmath>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 1E5 + 10;const int Siz = 1234;typedef long long LL;struct Mo{int pos,a,b;Mo(int _pos = 0,int _a = 0,int _b = 0) {pos = _pos; a = _a; b = _b;}}M[maxn];struct Qu{int L,R,t,beL,beR;Qu(int _L = 0,int _R = 0,int _beL = 0,int _beR = 0,int _t = 0) {L = _L; R = _R; beL = _beL; beR = _beR; t = _t;if (beL > beR) swap(L,R),swap(beL,beR);}bool operator < (const Qu &b) const {if (beL < b.beL) return 1;if (beL > b.beL) return 0;if (beR < b.beR) return 1;if (beR > b.beR) return 0;return t < b.t;}}Q[maxn]; int n,m,q,cnt,Root,top,t,s[maxn],co[maxn],fa[maxn][20],A,B,Lca,num[maxn],c[maxn],depth[maxn],belong[maxn];LL now,ans[maxn],va[maxn],w[maxn];bool vis[maxn];vector <int> v[maxn];int getint(){int ret = 0;char ch = getchar();while (ch < '0' || '9' < ch) ch = getchar();while ('0' <= ch && ch <= '9') ret = ret*10 + ch - '0',ch = getchar();return ret;}void dfs(int x,int from){for (int i = 1; i < 20; i++) fa[x][i] = fa[fa[x][i-1]][i-1];int k = top;for (int i = 0; i < v[x].size(); i++) {int to = v[x][i];if (to == from) continue;depth[to] = depth[x] + 1; fa[to][0] = x; dfs(to,x); if (top - k >= Siz) {++cnt;while (top != k) belong[s[top--]] = cnt; }}s[++top] = x;}void XorV(int x){if (vis[x]) {now -= va[num[x]]*w[co[num[x]]];vis[x] = 0; --co[num[x]];}else {++co[num[x]]; vis[x] = 1;now += va[num[x]]*w[co[num[x]]];}}void XorPath(int p,int q){if (depth[p] < depth[q]) swap(p,q);while (depth[p] > depth[q]) XorV(p),p = fa[p][0];if (p == q) return;while (p != q) {XorV(p); p = fa[p][0];XorV(q); q = fa[q][0];}}int LCA(int p,int q) {if (depth[p] < depth[q]) swap(p,q);int Log; for (Log = 0; depth[p] - (1<<Log) >= 1; Log++); --Log;for (int j = Log; j >= 0; j--)if (depth[p] - (1<<j) >= depth[q])p = fa[p][j];if (p == q) return p;for (int j = Log; j >= 0; j--)if (fa[p][j] != fa[q][j])p = fa[p][j],q = fa[q][j];return fa[p][0];}void Movetime(int cur,int target){if (cur < target) {for (int i = cur + 1; i <= target; i++) {if (!M[i].pos) continue;bool flag = 0; if (vis[M[i].pos]) XorV(M[i].pos),flag = 1;num[M[i].pos] = M[i].b; if (flag) XorV(M[i].pos);}}else {for (int i = cur; i > target; i--) {if (!M[i].pos) continue;bool flag = 0; if (vis[M[i].pos]) XorV(M[i].pos),flag = 1;num[M[i].pos] = M[i].a; if (flag) XorV(M[i].pos);}}}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);#endifn = getint(); m = getint(); q = getint();for (int i = 1; i <= m; i++) va[i] = getint();for (int i = 1; i <= n; i++) w[i] = getint();for (int i = 1; i < n; i++) {int x = getint(),y = getint();v[x].push_back(y);v[y].push_back(x);}for (int i = 1; i <= n; i++) num[i] = c[i] = getint();Root = n/2; depth[Root] = 1; dfs(Root,0); int tot = 0;while (top) belong[s[top--]] = cnt;for (int i = 1; i <= q; i++) {int typ = getint(),x = getint(),y = getint();if (typ) Q[++tot] = Qu(x,y,belong[x],belong[y],i);else M[i] = Mo(x,c[x],y),c[x] = y;}sort(Q + 1,Q + tot + 1);A = B = Lca = Root; t = 0;for (int i = 1; i <= tot; i++) {Movetime(t,Q[i].t); t = Q[i].t;XorPath(A,Q[i].L); A = Q[i].L;XorPath(B,Q[i].R); B = Q[i].R;Lca = LCA(A,B); XorV(Lca);ans[Q[i].t] = now; XorV(Lca);}for (int i = 1; i <= q; i++)if (ans[i])printf("%lld\n",ans[i]);return 0;}