题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1036
题意：
一棵树上有n个节点，编号分别为1到n，每个节点都有一个权值w。
我们将以下面的形式来要求你对这棵树完成一些操作：
I. CHANGE u t : 把结点u的权值改为t
II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值
III. QSUM u v: 询问从点u到点v的路径上的节点的权值和
注意：从点u到点v的路径上的节点包括u和v本身
思路：
权值在点上的树链剖分+线段树单点更新+线段树成段询问。
1 #include <iostream> 2 #include <cstring> 3 #include <string> 4 #include <cstdio> 5 #include <vector> 6 using namespace std; 7 int n, q; 8 #define maxn 30010 9 #define lson l, m, rt<<1 10 #define rson m+1, r, rt<<1|1 11 int siz[maxn], top[maxn], fa[maxn], son[maxn], dep[maxn]; 12 int w[maxn], fw[maxn]; 13 int A[maxn]; 14 vector <int> mp[maxn]; 15 int pos; 16 int sum[maxn<<2], mmax[maxn<<2]; 17 int col[maxn<<2]; 18 void PushUp(int rt) 19 { 20 sum[rt] = sum[rt<<1] + sum[rt<<1|1]; 21 mmax[rt] = max(mmax[rt<<1], mmax[rt<<1|1]); 22 } 23 void build(int l, int r, int rt) 24 { 25 mmax[rt] = -0x3f3f3f3f; 26 sum[rt] = 0; 27 if(l == r) 28 { 29 mmax[rt] = A[fw[l]]; 30 sum[rt] = A[fw[l]]; 31 return; 32 } 33 int m = (l+r)>>1; 34 build(lson); 35 build(rson); 36 PushUp(rt); 37 } 38 void update(int p, int val, int l, int r, int rt) 39 { 40 if(l == r) 41 { 42 sum[rt] = val; mmax[rt] = val; col[rt] = 0; 43 return; 44 } 45 int m = (l+r)>>1; 46 if(p <= m) update(p, val, lson); 47 else update(p, val, rson); 48 PushUp(rt); 49 } 50 int query_max(int L, int R, int l, int r, int rt) 51 { 52 if(L <= l && R >= r) 53 { 54 return mmax[rt]; 55 } 56 int m = (l+r)>>1; 57 int ret = -0x3f3f3f3f; 58 if(L <= m) ret = max(ret, query_max(L, R, lson)); 59 if(R > m) ret = max(ret, query_max(L, R, rson)); 60 return ret; 61 } 62 int query_sum(int L, int R, int l, int r, int rt) 63 { 64 if(L <= l && R >= r) 65 { 66 return sum[rt]; 67 } 68 int m = (l+r)>>1; 69 int ret = 0; 70 if(L <= m) ret += query_sum(L, R, lson); 71 if(R > m) ret += query_sum(L, R, rson); 72 return ret; 73 } 74 int dfs1(int u, int pre, int deep) 75 { 76 siz[u] = 1; dep[u] = deep; fa[u] = pre; 77 int mmax = 0; 78 for(int i = 0; i < mp[u].size(); i++) 79 { 80 if(mp[u][i] != pre) 81 { 82 int temp = dfs1(mp[u][i], u, deep+1); 83 siz[u] += temp; 84 if(son[u] == -1 || temp >= mmax) 85 { 86 son[u] = mp[u][i]; 87 mmax = temp; 88 } 89 } 90 } 91 return siz[u]; 92 } 93 void dfs2(int u, int val) 94 { 95 top[u] = val; 96 if(son[u] != -1) 97 { 98 w[u] = ++pos; 99 fw[w[u]] = u; 100 dfs2(son[u], val); 101 } 102 else if(son[u] == -1) 103 { 104 w[u] = ++pos; 105 fw[w[u]] = u; 106 return; 107 } 108 for(int i = 0; i < mp[u].size(); i++) 109 { 110 if(mp[u][i] != son[u] && mp[u][i] != fa[u]) dfs2(mp[u][i], mp[u][i]); 111 } 112 } 113 int find_sum(int u, int v) 114 { 115 int f1 = top[u], f2 = top[v]; 116 int temp = 0; 117 while(f1 != f2) 118 { 119 if(dep[f1] < dep[f2]) 120 { 121 swap(f1, f2); 122 swap(u, v); 123 } 124 temp += query_sum(w[f1], w[u], 1, pos, 1); 125 u = fa[f1]; f1 = top[u]; 126 } 127 if(dep[u] > dep[v]) swap(u, v); 128 temp += query_sum(w[u], w[v], 1, pos, 1); 129 return temp; 130 } 131 int find_max(int u, int v) 132 { 133 int f1 = top[u], f2 = top[v]; 134 int temp = -0x3f3f3f3f; 135 while(f1 != f2) 136 { 137 if(dep[f1] < dep[f2]) 138 { 139 swap(f1, f2); 140 swap(u, v); 141 } 142 temp = max(temp, query_max(w[f1], w[u], 1, pos, 1)); 143 u = fa[f1]; f1 = top[u]; 144 } 145 if(dep[u] > dep[v]) swap(u, v); 146 temp = max(temp, query_max(w[u], w[v], 1, pos, 1)); 147 return temp; 148 } 149 int main() 150 { 151 // freopen("in.txt", "r", stdin); 152 // freopen("out.txt", "w", stdout); 153 while(~scanf("%d", &n)) 154 { 155 for(int i = 1; i <= n; i++) mp[i].clear(); 156 pos = 0; 157 memset(son, -1, sizeof(son)); 158 for(int i = 1; i <= n-1; i++) 159 { 160 int a, b; scanf("%d%d", &a, &b); 161 mp[a].push_back(b); 162 mp[b].push_back(a); 163 } 164 for(int i = 1; i <= n; i++) scanf("%d", &A[i]); 165 dfs1(1, -1, 1); 166 dfs2(1, 1); 167 build(1, pos, 1); 168 169 scanf("%d", &q); 170 char op[10]; 171 while(q--) 172 { 173 scanf("%s", op); int u, v; 174 scanf("%d%d", &u, &v); 175 if(op[0] == 'C') update(w[u], v, 1, pos, 1); 176 else if(op[1] == 'M') printf("%d ", find_max(u, v)); 177 else if(op[1] == 'S') printf("%d ", find_sum(u, v)); 178 179 } 180 181 } 182 return 0; 183 }