[BZOJ2908] 又是nand （树链刨分）

题意：定义位运算与非：a nand b = not(a and b)。这个运算不满足交换律，结合律。给出一棵树，支持询问0依次nand这条路径上所有点权得到的结构，或者单点修改。

Claris讲题的时候我就把这题秒啦哈哈哈。。线段树上记录0/1从左往右、从右往左经过这个点的时候会变成什么。。然后线段树上的询问分为从左到右，从右到左两种。树上的询问分为从下到上，从上到下两种。树上从下到上好办，不停向上爬即可，从上往下可以先递归再合并。复杂度nlog^3n。

线段树+树剖实现这种有方向的树上路径询问好恶心啊！！！想起了原来NOIP模拟赛那个无聊的树上有向路径询问的题，巴蜀的lgz大神当场写了个5k的树链剖分还写对了（可惜评测鸡捉急只得了50分），真是膜膜膜orz。。。

#include<cstdio>#include<cstring>#include<algorithm>#define rep(i,a,b) for(int i=a;i<=b;++i)#define erp(i,a,b) for(int i=a;i>=b;--i)#define bit(a, i) ((a)>>i&1)#define lch(i) ((i)<<1)#define rch(i) ((i)<<1|1)const int MAXN = 100005;using namespace std;typedef unsigned int u32;template<typename T>inline void get(T&r){register char c; r=0;do c=getchar(); while(c<'0'||c>'9');do r=r*10+c-'0', c=getchar(); while(c>='0'&&c<='9');}int N, M, K;u32 v1[MAXN], v2[MAXN];struct Ed {int to; Ed*nxt;} Edges[MAXN*2], *ecnt=Edges, *adj[MAXN];void adde(int a, int b){(++ecnt)->to = b;ecnt->nxt = adj[a];adj[a] = ecnt;}int fa[MAXN], sz[MAXN], dep[MAXN], id[MAXN];int htp[MAXN], hsn[MAXN], tmr, dfi[MAXN];void DFS1(int u){dep[u] = dep[fa[u]]+1;sz[u] = 1;for (Ed*p = adj[u]; p; p=p->nxt){if (p->to == fa[u]) continue;fa[p->to] = u;DFS1(p->to);sz[u] += sz[p->to];if (sz[p->to]>sz[hsn[u]]) hsn[u] = p->to;}}void DFS2(int u){v2[dfi[u] = ++tmr] = v1[u];id[dfi[u]] = u;if (hsn[u]) htp[hsn[u]] = htp[u], DFS2(hsn[u]);for (Ed*p = adj[u]; p; p=p->nxt)if (p->to != fa[u] && p->to != hsn[u])htp[p->to] = p->to, DFS2(p->to);}inline int lca(int u, int v){while (htp[u] != htp[v])if (dep[htp[u]]>dep[htp[v]]) u = fa[htp[u]];else v = fa[htp[v]];return dep[u]<dep[v] ? u : v;}inline int find(int u, int d) //求u的深度为d的祖先  {if (d<=0) return 0;while (dep[htp[u]] > d) u = fa[htp[u]];return id[dfi[u] - (dep[u]-d)];}namespace seg{// 0 nand 0 =  1,   0 nand 1  =  1// 1 nand 0  =  1,   1 nand 1  =  0int pos[MAXN];struct Node {int l, r;u32 vl[2], vr[2];void calc(u32 t){vl[0] = vl[1] = 0;vr[0] = vr[1] = 0;for (int j = 0; j<K; ++j){vl[0] |= 1<<j, vr[0] |= 1<<j;if (!bit(t, j)) vl[1] |= 1<<j, vr[1] |= 1<<j;}}} a[MAXN * 4];inline void pushup(int i){Node&L=a[lch(i)], &R = a[rch(i)];a[i].vl[0] = a[i].vl[1] = 0;a[i].vr[0] = a[i].vr[1] = 0;for (int j = 0; j<K; ++j){a[i].vl[0] |= bit(R.vl[bit(L.vl[0], j)], j) << j;a[i].vl[1] |= bit(R.vl[bit(L.vl[1], j)], j) << j;a[i].vr[0] |= bit(L.vr[bit(R.vr[0], j)], j) << j;a[i].vr[1] |= bit(L.vr[bit(R.vr[1], j)], j) << j;}}void build(int i, int L, int R){a[i].l = L, a[i].r = R;if (L == R) { a[i].calc(v2[L]); pos[L] = i; return; }int mid = (L+R)>>1;build(lch(i), L, mid), build(rch(i), mid+1, R);pushup(i);}u32 ask1(int i, int l, int r, u32 pre) //线段树中从右往左走{if (a[i].l >= l && a[i].r <= r){u32 ans = 0;for (int j = 0; j<K; ++j)ans |= bit(a[i].vr[bit(pre, j)], j) << j;return ans;}int mid = (a[i].l + a[i].r)>>1;if (r <= mid) return ask1(lch(i), l, r, pre);if (l > mid) return ask1(rch(i), l, r, pre);u32 rig = ask1(rch(i), mid+1, r, pre);return ask1(lch(i), l, mid, rig);}u32 ask2(int i, int l, int r, u32 pre){if (a[i].l >= l && a[i].r <= r){u32 ans = 0;for (int j = 0; j<K; ++j)ans |= bit(a[i].vl[bit(pre, j)], j) << j;return ans;}int mid = (a[i].l + a[i].r)>>1;if (r <= mid) return ask2(lch(i), l, r, pre);if (l > mid) return ask2(rch(i), l, r, pre);u32 lef = ask2(lch(i), l, mid, pre);return ask2(rch(i), mid+1, r, lef);}void update(int i, u32 t){a[pos[i]].calc(t);for (i=pos[i]>>1; i>0; i>>=1) pushup(i);}}u32 askdu(int u, int v, u32 pre)  // go from u up to v{using namespace seg;for (; htp[u] != htp[v]; u = fa[htp[u]])pre = ask1(1, dfi[htp[u]], dfi[u], pre);return ask1(1, dfi[v], dfi[u], pre);}u32 askud(int u, int v, u32 pre) // go from u down to v{using namespace seg;if (htp[u] == htp[v]) return ask2(1, dfi[u], dfi[v], pre);u32 ha = askud(u, fa[htp[v]], pre);return ask2(1, dfi[htp[v]], dfi[v], ha);}u32 quary(int u, int v){if (u == v) return askdu(u, u, 0);int z = lca(u, v);if (v == z) return askdu(u, v, 0);if (u == z) return askud(u, v, 0);u32 pre = askdu(u, z, 0);return askud(find(v, dep[z]+1), v, pre);}void modify(int u, u32 x){using namespace seg;update(dfi[u], v2[dfi[u]] = x);}int main(){get(N), get(M), get(K);rep(i, 1, N) get(v1[i]);char op[20];int u, v;rep(i, 1, N-1){get(u), get(v);adde(u, v), adde(v, u);}DFS1(1), DFS2(htp[1] = 1);seg::build(1, 1, N);while (M --){scanf("%s", op);get(u), get(v);if (op[0] == 'Q') printf("%u\n", quary(u, v));else modify(u, v);}return 0;}