c/c++语言开发共享洛谷P3313 [SDOI2014]旅行(树链剖分 动态开节点线段树)

题意

题目链接

sol

树链剖分板子 + 动态开节点线段树板子

#include<bits/stdc++.h>  #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second //#define int long long  #define ll long long  #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << 'n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int ch(int x, int l, int r) {     return x >= l && x <= r; } int n, q, c[maxn], w[maxn], dep[maxn], fa[maxn], son[maxn], siz[maxn], top[maxn], id[maxn], cnt; vector<int> v[maxn]; void dfs1(int x, int _fa) {     dep[x] = dep[_fa] + 1; siz[x] = 1; fa[x] = _fa;      for(auto &to : v[x]) {         if(to == _fa) continue;         dfs1(to, x);         siz[x] += siz[to];         if(siz[to] > siz[son[x]]) son[x] = to;     } } void dfs2(int x, int topf) {     top[x] = topf; id[x] = ++cnt;     if(!son[x]) return ;     dfs2(son[x], topf);     for(auto &to : v[x]) {         if(top[to]) continue;         dfs2(to, to);     } } const int ss = (3e6 + 10); int root[maxn], ls[ss], rs[ss], mx[ss], sum[ss], tot; void update(int k) {     mx[k] = max(mx[ls[k]], mx[rs[k]]);     sum[k] = sum[ls[k]] + sum[rs[k]]; } void modify(int &k, int l, int r, int p, int v) {     if(!k) k = ++tot;     if(l == r) {sum[k] = v, mx[k] = v; return ;}     int mid = l + r >> 1;     if(p <= mid) modify(ls[k], l, mid, p, v);     if(p  > mid) modify(rs[k], mid + 1, r, p, v);     update(k); } int query(int k, int l, int r, int ql, int qr, int opt) {     if(ql <= l && r <= qr) return opt == 0 ? mx[k] : sum[k];     int mid = l + r >> 1;     if(ql > mid) return query(rs[k], mid + 1, r, ql, qr, opt);     else if(qr <= mid) return query(ls[k], l, mid, ql, qr, opt);     else return opt == 0 ? max(query(ls[k], l, mid, ql, qr, opt), query(rs[k], mid + 1, r, ql, qr, opt))                           : query(ls[k], l, mid, ql, qr, opt) + query(rs[k], mid + 1, r, ql, qr, opt); } int treemax(int x, int y) {     int ans = -inf, p = x;     while(top[x] != top[y]) {         if(dep[top[x]] < dep[top[y]]) swap(x, y);         chmax(ans, query(root[c[p]], 1, n, id[top[x]], id[x], 0));         x = fa[top[x]];     }     if(dep[x] < dep[y]) swap(x, y);     chmax(ans, query(root[c[p]], 1, n, id[y], id[x], 0));     return ans; } int treesum(int x, int y) {     int ans = 0, p = x;     while(top[x] != top[y]) {         if(dep[top[x]] < dep[top[y]]) swap(x, y);         ans += query(root[c[p]], 1, n, id[top[x]], id[x], 1);         x = fa[top[x]];     }     if(dep[x] < dep[y]) swap(x, y);     ans += query(root[c[p]], 1, n, id[y], id[x], 1);     return ans; } signed main() {     n = read(); q = read();     for(int i = 1; i <= n; i++) w[i] = read(), c[i] = read();     for(int i = 1; i <= n - 1; i++) {         int x = read(), y = read();         v[x].push_back(y);         v[y].push_back(x);     }     dfs1(1, 0);     dfs2(1, 1);     for(int i = 1; i <= n; i++)          modify(root[c[i]], 1, n, id[i], w[i]);     while(q--) {         char s[4];         scanf("%s", s);         int x = read(), c = read();         if(s[0] == 'c' && s[1] == 'c') {             modify(root[c[x]], 1, n, id[x], 0);             modify(root[c], 1, n, id[x], w[x]);             c[x] = c;         }         if(s[0] == 'c' && s[1] == 'w') {             modify(root[c[x]], 1, n, id[x], c);             w[x] = c;         }         if(s[0] == 'q' && s[1] == 'm') {             printf("%dn", treemax(x, c));           }         if(s[0] == 'q' && s[1] == 's') {             printf("%dn", treesum(x, c));         }     }     return 0; } /* 5 6 3 1 2 3 1 2 3 3 5 1 1 2 1 3 3 4 3 5 qs 1 5 cc 3 1 qs 1 5 cw 3 3 qs 1 5 qm 2 4 */