首先肯定是要先选好模板,并且把模板理解透彻
这个树链剖分的模板题 洛谷出的很好 https://www.luogu.org/problem/P3384
这个题目涉及了树链剖分的各种各样的操作
模板
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <algorithm> #include <cstdlib> #include <vector> #include <stack> #include <map> #include <string> #define inf 0x3f3f3f3f #define inf64 0x3f3f3f3f3f3f3f3f using namespace std; const int maxn = 2e5 + 10; int f[maxn];//f 保存u的父亲节点 int dep[maxn];//dep保存节点u 的深度 int siz[maxn];//siz保存以u为根的子节点的个数 int son[maxn];//son 保存u的重儿子 int rk[maxn];//rk当前dfs序在树中所对应的节点 int top[maxn];// top保存当前结点所在链的顶端结点 int id[maxn];//dfs的执行顺序 int a[maxn]; int mod, n; int sum[maxn * 4], lazy[maxn * 4]; //------------------线段树部分---------------// void push_up(int id) { sum[id] = (sum[id << 1] + sum[id << 1 | 1]) % mod; // printf("sum[%d]=%d sum[%d]=%d ", id << 1, sum[id << 1], id << 1 | 1, sum[id << 1 | 1]); // printf("sum[%d]=%d ", id, sum[id]); } void build(int id,int l,int r) { lazy[id] = 0; if(l==r) { sum[id] = a[rk[l]] % mod; // printf("id=%d sum=%d ", id, sum[id]); return; } int mid = (l + r) >> 1; build(id << 1, l, mid); build(id << 1 | 1, mid + 1, r); push_up(id); } void push_down(int id,int len1,int len2) { if (lazy[id] == 0) return; sum[id << 1] += lazy[id] % mod * len1%mod; sum[id << 1] %= mod; lazy[id << 1] += lazy[id] % mod; lazy[id << 1] %= mod; sum[id << 1 | 1] += lazy[id] % mod * len2%mod; sum[id << 1 | 1] %= mod; lazy[id << 1 | 1] += lazy[id] % mod; lazy[id << 1 | 1] %= mod; lazy[id] = 0; } void update(int id,int l,int r,int x,int y,int val) { // printf("id=%d l=%d r=%d x=%d y=%d val=%d ", id, l, r, x, y, val); if(x<=l&&y>=r) { // printf("id=%d sum=%d ", id, sum[id]); sum[id] += val * (r - l + 1) % mod; sum[id] %= mod; lazy[id] += val; lazy[id] %= mod; // printf("%d ", sum[id]); return; } int mid = (l + r) >> 1; push_down(id, mid - l + 1, r - mid); if (x <= mid) update(id << 1, l, mid, x, y, val); if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, val); push_up(id); } int query(int id,int l,int r,int x,int y) { if (x <= l && y >= r) return sum[id]; int mid = (l + r) >> 1, ans = 0; push_down(id, mid - l + 1, r - mid); if (x <= mid) ans = (ans + query(id << 1, l, mid, x, y)) % mod; if (y > mid) ans = (ans + query(id << 1 | 1, mid + 1, r, x, y)) % mod; return ans; } //------------------------树链剖分-------------------// // int f[maxn];//f 保存u的父亲节点 // int dep[maxn];//dep保存节点u 的深度 // int siz[maxn];//siz保存以u为根的子节点的个数 // int son[maxn];//son 保存u的重儿子 // int rk[maxn];//rk当前dfs序在树中所对应的节点 // int top[maxn];// top保存当前结点所在链的顶端结点 // int id[maxn];//dfs的执行顺序 struct node { int v, nxt; node(int v=0,int nxt=0):v(v),nxt(nxt){} }ex[maxn]; int head[maxn], cnt = 0, tot; void init() { cnt = 0, tot = 0; memset(son, 0, sizeof(son)); memset(head, -1, sizeof(head)); } void add(int u,int v) { ex[cnt] = node(v, head[u]); head[u] = cnt++; ex[cnt] = node(u, head[v]); head[v] = cnt++; } void dfs1(int u,int fa,int depth) { f[u] = fa; dep[u] = depth; siz[u] = 1; for(int i=head[u];i!=-1;i=ex[i].nxt) { int v = ex[i].v; if (v == fa) continue; dfs1(v, u, depth + 1); siz[u] += siz[v]; if (siz[v] > siz[son[u]]) son[u] = v; } } void dfs2(int u,int t) { top[u] = t; id[u] = ++tot;//标记dfs序 rk[tot] = u;//序号tot对应的结点u if (!son[u]) return; dfs2(son[u], t); /*我们选择优先进入重儿子来保证一条重链上各个节点dfs序连续, 一个点和它的重儿子处于同一条重链,所以重儿子所在重链的顶端还是t*/ for(int i=head[u];i!=-1;i=ex[i].nxt) { int v = ex[i].v; if (v != son[u] && v != f[u]) dfs2(v, v);//一个点位于轻链底端,那么它的top必然是它本身 } } void update2(int x,int y,int z)//修改x到y路径的值 { while(top[x]!=top[y])//不在同一条链上 { if (dep[top[x]] < dep[top[y]]) swap(x, y);//x为深度大的链 update(1, 1, n, id[top[x]], id[x], z);//x为深度大的链 x = f[top[x]];//深度大的向上跳 } if (dep[x] > dep[y]) swap(x, y); //这里x和y在同一条链 update(1, 1, n, id[x], id[y], z); //x和y这条链的更新 } int query2(int x,int y) { int ret = 0; while(top[x]!=top[y]) { if (dep[top[x]] < dep[top[y]]) swap(x, y); ret = (ret + query(1, 1, n, id[top[x]], id[x])) % mod; x = f[top[x]]; } if (dep[x] > dep[y]) swap(x, y); ret = (ret + query(1, 1, n, id[x], id[y])) % mod; return ret; } //------------------树链剖分结束-------------------// int main() { init(); int m, r; scanf("%d%d%d%d", &n, &m, &r, &mod); for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] %= mod; for (int i = 1; i < n; i++) { int u, v; scanf("%d%d", &u, &v); add(u, v); } dfs1(r, 0, 1), dfs2(r, r); build(1, 1, n); while(m--) { int opt, x, y, z; scanf("%d", &opt); if(opt==1) { scanf("%d%d%d", &x, &y, &z); update2(x,y,z); } if(opt==2) { scanf("%d%d", &x, &y); int ans = query2(x,y); printf("%d ", ans); } if(opt==3) { scanf("%d%d", &x, &z); update(1, 1, n, id[x], id[x] + siz[x] - 1, z); } if(opt==4) { scanf("%d", &x); int ans = query(1, 1, n, id[x], id[x] + siz[x] - 1); printf("%d ", ans); } } }
然后就是一个简单的树链剖分的练习
树链剖分注意初始化和数组稍微要开大一点,不然容易re
https://vjudge.net/problem/CodeForces-343D
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <algorithm> #include <cstdlib> #include <vector> #include <stack> #include <map> #include <string> #define inf 0x3f3f3f3f #define inf64 0x3f3f3f3f3f3f3f3f using namespace std; const int maxn = 5e5 + 10; int f[maxn], dep[maxn], siz[maxn], son[maxn], rk[maxn], top[maxn], id[maxn]; int head[maxn], cnt, tot, sum[maxn * 4], lazy[maxn * 4], n; struct node { int v, nxt; node(int v=0,int nxt=0):v(v),nxt(nxt){} }ex[maxn*4]; void init() { memset(son, 0, sizeof(son)); memset(head, -1, sizeof(head)); cnt = tot = 0; } void push_up(int id) { sum[id] = sum[id << 1] + sum[id << 1 | 1]; } void build(int id,int l,int r) { lazy[id] = -1,sum[id] = 0; if (l == r) return; int mid = (l + r) >> 1; build(id << 1, l, mid); build(id << 1 | 1, mid + 1, r); push_up(id); } void push_down(int id,int len1,int len2) { if (lazy[id] == -1) return; sum[id << 1] = len1 * lazy[id]; sum[id << 1 | 1] = len2 * lazy[id]; lazy[id << 1] = lazy[id << 1 | 1] = lazy[id]; lazy[id] = -1; } void update(int id,int l,int r,int x,int y,int val) { if(x<=l&&y>=r) { sum[id] = val * (r - l + 1); lazy[id] = val; return; } int mid = (l + r) >> 1; push_down(id, mid - l + 1, r - mid); if (x <= mid) update(id << 1, l, mid, x, y, val); if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, val); push_up(id); } int query(int id,int l,int r,int pos) { if (l == r) return sum[id]; int mid = (l + r) >> 1; int ans = 0; push_down(id, mid - l + 1, r - mid); if (pos <= mid) ans += query(id << 1, l, mid, pos); else ans += query(id << 1 | 1, mid + 1, r, pos); return ans; } void add(int u,int v) { ex[cnt] = node(v, head[u]); head[u] = cnt++; ex[cnt] = node(u, head[v]); head[v] = cnt++; } void dfs1(int u,int fa,int depth) { f[u] = fa, dep[u] = depth, siz[u] = 1; for(int i=head[u];i!=-1;i=ex[i].nxt) { int v = ex[i].v; if (v == fa) continue; dfs1(v, u, depth + 1); siz[u] += siz[v]; if (siz[v] > siz[son[u]]) son[u] = v; } } void dfs2(int u,int t) { top[u] = t;//标记这个结点的顶端 id[u] = ++tot;//标记这个结点的标号 rk[tot] = u;//标号对应的结点 if (!son[u]) return; dfs2(son[u], t); for(int i=head[u];i!=-1;i=ex[i].nxt) { int v = ex[i].v; if (v == u) continue; if (v != son[u] && v != f[u]) dfs2(v, v); } } void update1(int x,int y,int z) { while(top[x]!=top[y]) { if (dep[top[x]] < dep[top[y]]) swap(x, y); update(1, 1, n, id[top[x]], id[x], z); x = f[top[x]]; } if (dep[x] > dep[y]) swap(x, y); update(1, 1, n, id[x], id[y], z); } int main() { init(); int m; scanf("%d", &n); for(int i=1;i<n;i++) { int u, v; scanf("%d%d", &u, &v); add(u, v); } dfs1(1, 0, 1), dfs2(1, 1); build(1, 1, n); scanf("%d", &m); while(m--) { int opt, v; scanf("%d%d", &opt, &v); if (opt == 1) update(1, 1, n, id[v], id[v] + siz[v] - 1, 1); if (opt == 2) update1(v, 1, 0); if (opt == 3) printf("%d ", query(1, 1, n, id[v])); } return 0; }
然后就是学校oj的一个也很裸的树链剖分
https://10.64.70.166/problem/1005
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <algorithm> #include <cstdlib> #include <vector> #include <stack> #include <map> #include <string> #define inf 0x3f3f3f3f #define inf64 0x3f3f3f3f3f3f3f3f using namespace std; typedef long long ll; const int maxn = 2e5 + 10; int f[maxn], dep[maxn], siz[maxn], son[maxn], rk[maxn], top[maxn], ver[maxn]; int head[maxn], cnt, tot, n, a[maxn], m; ll sum[maxn * 4], lazy[maxn * 4]; struct node { int v, nxt; node(int v=0,int nxt=0):v(v),nxt(nxt){} }ex[maxn*4]; void init() { memset(son, 0, sizeof(son)); memset(head, -1, sizeof(head)); cnt = tot = 0; } void push_up(int id) { sum[id] = sum[id << 1] + sum[id << 1 | 1]; } void build(int id,int l,int r) { lazy[id] = 0; if (l == r) { sum[id] = a[rk[l]]; return; } int mid = (l + r) >> 1; build(id << 1, l, mid); build(id << 1 | 1, mid + 1, r); push_up(id); } void push_down(int id,int len1,int len2) { if (lazy[id] == 0) return; sum[id << 1] += len1 * lazy[id]; sum[id << 1 | 1] += len2 * lazy[id]; lazy[id << 1] += lazy[id]; lazy[id << 1 | 1] += lazy[id]; lazy[id] = 0; } void update(int id,int l,int r,int x,int y,int val) { if(x<=l&&y>=r) { sum[id] += val * (r - l + 1); lazy[id] += val; return; } int mid = (l + r) >> 1; push_down(id, mid - l + 1, r - mid); if (x <= mid) update(id << 1, l, mid, x, y, val); if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, val); push_up(id); } ll query(int id,int l,int r,int pos) { if (l == r) return sum[id]; int mid = (l + r) >> 1; ll ans = 0; push_down(id, mid - l + 1, r - mid); if (pos <= mid) ans += query(id << 1, l, mid, pos); else ans += query(id << 1 | 1, mid + 1, r, pos); return ans; } void add(int u,int v) { ex[cnt] = node(v, head[u]); head[u] = cnt++; ex[cnt] = node(u, head[v]); head[v] = cnt++; } void dfs1(int u,int fa,int depth) { f[u] = fa, dep[u] = depth, siz[u] = 1; for(int i=head[u];i!=-1;i=ex[i].nxt) { int v = ex[i].v; if (v == fa) continue; dfs1(v, u, depth + 1); siz[u] += siz[v]; if (siz[v] > siz[son[u]]) son[u] = v; } } void dfs2(int u,int t) { top[u] = t;//标记这个结点的顶端 ver[u] = ++tot;//标记这个结点的标号 rk[tot] = u;//标号对应的结点 if (!son[u]) return; dfs2(son[u], t); for(int i=head[u];i!=-1;i=ex[i].nxt) { int v = ex[i].v; if (v == u) continue; if (v != son[u] && v != f[u]) dfs2(v, v); } } void update1(int x,int y,int z) { while(top[x]!=top[y]) { if (dep[top[x]] < dep[top[y]]) swap(x, y); update(1, 1, n, ver[top[x]], ver[x], z); x = f[top[x]]; } if (dep[x] > dep[y]) swap(x, y); update(1, 1, n, ver[x], ver[y], z); } int opt[maxn], ux[maxn], vx[maxn], xx[maxn]; int main() { while(scanf("%d%d",&n,&m)!=EOF) { init(); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i=1;i<n;i++) { int u, v; scanf("%d%d", &u, &v); add(u, v); } dfs1(1, 0, 1), dfs2(1, 1); build(1, 1, n); for(int i=1;i<=m;i++) { scanf("%d", &opt[i]); if(opt[i]==1) { scanf("%d%d", &ux[i], &xx[i]); update(1, 1, n, ver[ux[i]], ver[ux[i]] + siz[ux[i]] - 1, xx[i]); } if(opt[i]==2) { scanf("%d%d%d", &ux[i], &vx[i], &xx[i]); update1(ux[i], vx[i], xx[i]); } if(opt[i]==3) { int t; scanf("%d", &t); if(opt[t]==1) { ux[i] = ux[t], xx[i] = -xx[t]; update(1, 1, n, ver[ux[i]], ver[ux[i]] + siz[ux[i]] - 1, xx[i]); } if(opt[t]==2) { ux[i] = ux[t], vx[i] = vx[t], xx[i] = -xx[t]; update1(ux[i], vx[i], xx[i]); } } if(opt[i]==4) { scanf("%d", &ux[i]); ll ans = query(1, 1, n, ver[ux[i]]); printf("%lld ", ans); } } } }