HDU 6162 && 2017 多校训练：Ch's gift（树链剖分）

题意：

有一棵n个节点的树，m次查询，每次查询x, y, a, b表示节点x到y的路径上所有在[a, b]范围内的权值和

树链剖分+线段树

询问离线，按a和b分别排序求出所有小于a和所有小于b+1的权值和，然后相减就可以了

因为排了序，所以更新当然只用单点逐一更新

#include<stdio.h>#include<vector>#include<string.h>#include<algorithm>using namespace std;#define LL long longvector<LL> G[100005];typedef struct{LL id;LL x, y;LL l, r;}Quer;Quer s[100005], cnt[100005];LL tre[445005];LL n, val[100005], son[100005], fa[100005], siz[100005], dep[100005];LL k, top[100005], rak[100005], id[100005], anl[100005], anr[100005];bool comp1(Quer a, Quer b){if(a.l<b.l)return 1;return 0;}bool compr(Quer a, Quer b){if(a.r<b.r)return 1;return 0;}void Sechs(LL u, LL p){LL v, i;fa[u] = p;dep[u] = dep[p]+1;for(i=0;i<G[u].size();i++){v = G[u][i];if(v==p)continue;Sechs(v, u);siz[u] += siz[v]+1;if(son[u]==0 || siz[v]>siz[son[u]])son[u] = v;}}void Sechr(LL u, LL p){LL v, i;top[u] = p;rak[u] = ++k, id[k] = u;if(son[u]==0)return;Sechr(son[u], p);for(i=0;i<G[u].size();i++){v = G[u][i];if(v==son[u] || v==fa[u])continue;Sechr(v, v);}}void Update(LL l, LL r, LL x, LL a, LL b){LL m;if(l==r){tre[x] += b;return;}m = (l+r)/2;if(a<=m)Update(l, m, x*2, a, b);elseUpdate(m+1, r, x*2+1, a, b);tre[x] = tre[x*2]+tre[x*2+1];}LL Query(LL l, LL r, LL x, LL a, LL b){LL m, sum = 0;if(l>=a && r<=b)return tre[x];m = (l+r)/2;if(a<=m)sum += Query(l, m, x*2, a, b);if(b>=m+1)sum += Query(m+1, r, x*2+1, a, b);return sum;}LL TreQuery(LL x, LL y){LL sum, p1, p2;p1 = top[x], p2 = top[y], sum = 0;while(p1!=p2){if(dep[p1]<dep[p2])swap(p1, p2), swap(x, y);sum += Query(1, n, 1, rak[p1], rak[x]);x = fa[p1], p1 = top[x];}if(dep[x]>dep[y])swap(x, y);sum += Query(1, n, 1, rak[x], rak[y]);return sum;}int main(void){LL i, j, x, y, q;while(scanf("%lld%lld", &n, &q)!=EOF){for(i=1;i<=n;i++)G[i].clear();for(i=1;i<=n;i++){scanf("%lld", &val[i]);cnt[i].id = i;cnt[i].r = val[i];}sort(cnt+1, cnt+n+1, compr);for(i=1;i<=n-1;i++){scanf("%lld%lld", &x, &y);G[x].push_back(y);G[y].push_back(x);}memset(siz, 0, sizeof(siz));memset(son, 0, sizeof(son));k = 0;Sechs(1, 0);Sechr(1, 1);for(i=1;i<=q;i++){scanf("%lld%lld%lld%lld", &s[i].x, &s[i].y, &s[i].l, &s[i].r);s[i].id = i;}sort(s+1, s+q+1, comp1);memset(tre, 0, sizeof(tre));j = 1;for(i=1;i<=q;i++){while(cnt[j].r<s[i].l && j<=n){Update(1, n, 1, rak[cnt[j].id], cnt[j].r);j += 1;}anl[s[i].id] = TreQuery(s[i].x, s[i].y);}sort(s+1, s+q+1, compr);memset(tre, 0, sizeof(tre));j = 1;for(i=1;i<=q;i++){while(cnt[j].r<=s[i].r && j<=n){Update(1, n, 1, rak[cnt[j].id], cnt[j].r);j += 1;}anr[s[i].id] = TreQuery(s[i].x, s[i].y);}printf("%lld", anr[1]-anl[1]);for(i=2;i<=q;i++)printf(" %lld", anr[i]-anl[i]);printf("\n");}return 0;}