给你一棵树，每次修改一个子树的所有值，然后单点查询。
按照DFS序把节点排列（即在DFS中出现的先后次序），同一个子树在序列中连续。
1 #include <cstdio> 2 using namespace std; 3 typedef long long ll; 4 int n,q,T,Tc,cnt,sum; 5 int col[210000],lzy[210000],sta[51000],fin[51000]; 6 int nxt[51000],to[51000],head[51000],ind[51000]; 7 void build(int k,int l,int r) 8 { 9 lzy[k] = -1; 10 if (l == r) 11 { 12 col[k] = -1; 13 return; 14 } 15 int mid = l + r >> 1; 16 build(k << 1,l,mid); 17 build(k << 1 | 1,mid + 1,r); 18 } 19 void down(int k,int l,int r) 20 { 21 if (l == r) 22 { 23 lzy[k] = -1; 24 return; 25 } 26 int mid = l + r >> 1; 27 col[k << 1] = lzy[k]; 28 col[k << 1 | 1] = lzy[k]; 29 lzy[k << 1] = lzy[k]; 30 lzy[k << 1 | 1] = lzy[k]; 31 lzy[k] = -1; 32 } 33 void change(int k,int l,int r,int x,int y,int c) 34 { 35 if (x <= l && r <= y) 36 { 37 lzy[k] = c; 38 col[k] = c; 39 return; 40 } 41 if (lzy[k] != -1) down(k,l,r); 42 int mid = l + r >> 1; 43 if (x <= mid) change(k << 1,l,mid,x,y,c); 44 if (y >= mid + 1) change(k << 1 | 1,mid + 1,r,x,y,c); 45 } 46 int query(int k,int l,int r,int x) 47 { 48 if (l == r) return col[k]; 49 if (lzy[k] != -1) down(k,l,r); 50 int mid = l + r >> 1; 51 if (x <= mid) return query(k << 1,l,mid,x); 52 if (x >= mid + 1) return query(k << 1 | 1,mid + 1,r,x); 53 } 54 void add(int x,int y) 55 { 56 nxt[++cnt] = head[x]; 57 to[cnt] = y; 58 head[x] = cnt; 59 } 60 void dfs(int x) 61 { 62 sta[x] = ++sum; 63 for (int i = head[x];i;i = nxt[i]) 64 dfs(to[i]); 65 fin[x] = sum; 66 } 67 int main() 68 { 69 for (scanf("%d",&T);T;T--) 70 { 71 printf("Case #%d: ",++Tc); 72 scanf("%d",&n); 73 int tx,ty; 74 for (int i = 1;i <= n - 1;i++) 75 { 76 scanf("%d%d",&tx,&ty); 77 add(ty,tx); 78 ind[tx]++; 79 } 80 for (int i = 1;i <= n;i++) 81 if (ind[i] == 0) 82 { 83 dfs(i); 84 break; 85 } 86 build(1,1,n); 87 scanf("%d",&q); 88 char str[10]; 89 for (int i = 1;i <= q;i++) 90 { 91 scanf("%s",str); 92 if (str[0] == 'T') 93 { 94 scanf("%d%d",&tx,&ty); 95 change(1,1,n,sta[tx],fin[tx],ty); 96 }else 97 { 98 scanf("%d",&tx); 99 printf("%d ",query(1,1,n,sta[tx])); 100 } 101 } 102 for (int i = 1;i <= n;i++) 103 ind[i] = head[i] = 0; 104 for (int i = 1;i <= cnt;i++) 105 nxt[i] = 0; 106 cnt = sum = 0; 107 } 108 return 0; 109 }