题意：一棵n个结点的树，每个点有点权，有如下操作：
1 x y c：给x到y的链上所有结点点权加上c
2 x y：查询x到y链上的点权和
3 x c：给x及其子树上的所有结点点权加上c
4 x：查询x及其子树所有结点的点权和
第一遍dfs：预处理，子树大小，节点深度，顺便标记结点的父亲结点编号
void dfs(int u, int fa, int cnt){ sum[u] = 1; //sum[u]表示u的子树大小，首先只有自己 rk[u] = cnt; //深度为cnt f[u] = fa; //标记父节点 for (int i = head[u]; i; i = e[i].nxt){ int v = e[i].v; if (v == fa) continue; dfs(v, u, cnt+1); sum[u] += sum[v]; //加上儿子v的子树大小 if (sum[v] > sum[son[u]]) son[u] = v; //如果v的子树比当前重儿子大，更新v为重儿子 } }
第二遍dfs：处理每个点的顶端，按dfs序给每个点分配一个编号，使其能够映射到线段树上
void dfs2(int u, int t){ top[u] = t;//top是能沿重链走到的最上方的点 idx[u] = ++cnt; //打上时间戳 a[cnt] = b[u]; //映射到线段树的区间上 if (!son[u]) return; //这是叶节点 dfs2(son[u], t); //沿重链走 for (int i = head[u]; i; i = e[i].nxt){ int v = e[i].v; if (!idx[v]) dfs2(v, v); //处理轻儿子 } }
怎样利用树链剖分的轻重链（主要是重链）加速链上的修改？
类似LCA的思想，x，y较深的一个往上走
1. 将深的那个结点往上提，直至x，y在同一条重链，每次都对走过的链上结点修改。
2. 修改x，y之间的链，这条链是重链
这部分代码：
void TreeAdd(int x, int y, LL c){ while (top[x] != top[y]){ //不在同一重链 if (rk[top[x]] < rk[top[y]]) swap(x, y); upd(idx[top[x]], idx[x], 1, c); x = f[top[x]]; //走到顶端的父亲 } if (rk[x] > rk[y]) swap(x, y); upd(idx[x], idx[y], 1, c); //修改现在所在的这一条重链 }
然后就只是普通的线段树操作而已了
完整代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #define LL long long 6 #define debug(x) cout << "[" << x << "]" << endl 7 #define lid id << 1 8 #define rid id << 1 | 1 9 using namespace std; 10 11 const int mx = 1e5+10; 12 13 struct tree{ 14 int l, r; 15 LL sum, lazy; 16 }tree[mx<<2]; 17 18 struct edge{ 19 int v, nxt; 20 }e[mx<<1]; 21 int tot = 1, mod; 22 int head[mx], sum[mx], rk[mx], son[mx], f[mx], top[mx], idx[mx]; 23 LL a[mx], b[mx]; 24 25 void add(int u, int v){ 26 e[tot].v = v; 27 e[tot].nxt = head[u]; 28 head[u] = tot++; 29 } 30 31 void dfs(int u, int fa, int cnt){ 32 sum[u] = 1; 33 rk[u] = cnt; 34 f[u] = fa; 35 for (int i = head[u]; i; i = e[i].nxt){ 36 int v = e[i].v; 37 if (v == fa) continue; 38 dfs(v, u, cnt+1); 39 sum[u] += sum[v]; 40 if (sum[v] > sum[son[u]]) son[u] = v; 41 } 42 } 43 44 int cnt = 0; 45 void dfs2(int u, int t){ 46 top[u] = t; 47 idx[u] = ++cnt; 48 a[cnt] = b[u]; 49 if (!son[u]) return; 50 dfs2(son[u], t); 51 for (int i = head[u]; i; i = e[i].nxt){ 52 int v = e[i].v; 53 if (!idx[v]) dfs2(v, v); 54 } 55 } 56 57 void push_up(int id){ 58 tree[id].sum = (tree[lid].sum + tree[rid].sum)%mod; 59 } 60 61 void build(int l, int r, int id){ 62 tree[id].l = l; 63 tree[id].r = r; 64 tree[id].lazy = 0; 65 if (l == r) { 66 tree[id].sum = a[l]%mod; 67 return; 68 } 69 int mid = (l+r)>>1; 70 build(l, mid, lid); 71 build(mid+1, r, rid); 72 push_up(id); 73 } 74 75 void push_down(int id){ 76 if (tree[id].lazy != 0){ 77 LL x = tree[id].lazy; 78 tree[lid].lazy += x; 79 tree[lid].lazy %= mod; 80 tree[rid].lazy += x; 81 tree[rid].lazy %= mod; 82 tree[lid].sum += x*(tree[lid].r-tree[lid].l+1); 83 tree[lid].sum %= mod; 84 tree[rid].sum += x*(tree[rid].r-tree[rid].l+1); 85 tree[rid].sum %= mod; 86 tree[id].lazy = 0; 87 } 88 } 89 90 void upd(int l, int r, int id, LL c){ 91 if (tree[id].l == l && tree[id].r == r){ 92 tree[id].sum += c*(r-l+1); 93 tree[id].sum %= mod; 94 tree[id].lazy += c; 95 tree[id].lazy %= mod; 96 return; 97 } 98 push_down(id); 99 int mid = (tree[id].l + tree[id].r)>>1; 100 if (r <= mid) upd(l, r, lid, c); 101 else if (mid < l) upd(l, r, rid, c); 102 else { 103 upd(l, mid, lid, c); 104 upd(mid+1, r, rid, c); 105 } 106 push_up(id); 107 } 108 109 LL query(int l, int r, int id){ 110 if (tree[id].l == l && tree[id].r == r) return tree[id].sum%mod; 111 push_down(id); 112 int mid = (tree[id].l + tree[id].r)>>1; 113 if (r <= mid) return query(l, r, lid)%mod; 114 else if (mid < l) return query(l, r, rid)%mod; 115 return (query(l, mid, lid) + query(mid+1, r, rid))%mod; 116 } 117 118 void TreeAdd(int x, int y, LL c){ 119 while (top[x] != top[y]){ 120 if (rk[top[x]] < rk[top[y]]) swap(x, y); 121 upd(idx[top[x]], idx[x], 1, c); 122 x = f[top[x]]; 123 } 124 if (rk[x] > rk[y]) swap(x, y); 125 upd(idx[x], idx[y], 1, c); 126 } 127 128 LL TreeSum(int x, int y){ 129 LL ans = 0; 130 while (top[x] != top[y]){ 131 if (rk[top[x]] < rk[top[y]]) swap(x, y); 132 ans += query(idx[top[x]], idx[x], 1); 133 ans %= mod; 134 x = f[top[x]]; 135 } 136 if (rk[x] > rk[y]) swap(x, y); 137 ans += query(idx[x], idx[y], 1); 138 ans %= mod; 139 return ans; 140 } 141 142 int main(){ 143 int n, m, r, u, v; 144 scanf("%d%d%d%d", &n, &m, &r, &mod); 145 for (int i = 1; i <= n; i++) scanf("%lld", &b[i]); 146 for (int i = 2; i <= n; i++){ 147 scanf("%d%d", &u, &v); 148 add(u, v); 149 add(v, u); 150 } 151 dfs(r, 0, 1); 152 dfs2(r, r); 153 build(1, n, 1); 154 int op, x, y; 155 LL c; 156 while (m--){ 157 scanf("%d%d", &op, &x); 158 if (op == 1) { 159 scanf("%d%lld", &y, &c); 160 TreeAdd(x, y, c); 161 } 162 else if (op == 2){ 163 scanf("%d", &y); 164 printf("%lld ", TreeSum(x, y)); 165 } 166 else if (op == 3){ 167 scanf("%lld", &c); 168 upd(idx[x], idx[x]+sum[x]-1, 1, c%mod); 169 } 170 else printf("%lld ", query(idx[x], idx[x]+sum[x]-1, 1)); 171 } 172 return 0; 173 }