Bzoj 2243: segmentation of the dyeing tree chain !, Bzoj2243

Bzoj 2243: segmentation of the dyeing tree chain !, Bzoj2243

Question: Chinese.

Idea: A good tree chain is split. After the tree is split, the line segment tree should record the left endpoint l, right endpoint r, left endpoint color lc, right endpoint color rc, range-based updated tag, Range

The number of color segments. When merging intervals, note that if the color of the right side of the left subtree is the same as that of the right subtree, the number must be reduced by one. However, there is a problem.

The color of the last link at the intersection edge may be the same. If the color is the same, the answer must be reduced by one. Therefore, when calculating the answer, we need to record the left end of the last split chain.

The color of the vertex, Which is compared with the color of the right endpoint of the current link (because the longer the processed line segment tree is near the root point, the more left, if they are the same, the answer is subtracted.

1. Because u and v are both in the upward direction, the ans1 and ans2 variables must be recorded to store the "last left endpoint color ". Note that when

When top [u] = top [v] is already on the same heavy chain, the color of the two ends must be compared with the corresponding ans, and the same answer must be reduced by one. For details, see the code:

/*************************************** * ****************** File name: bzoj2243.cpp author: kereo create time: january 23, 2015 Friday ******************************** * ***********************/# include <iostream> # include <cstdio> # include <cstring> # include <queue> # include <set> # include <map> # include <vector> # include <stack> # include <cmath> # include <string> # include <algorithm> using namespace std; typedef long l Ong ll; const int sigma_size = 26; const int N = 100 + 50; const int MAXN = 100000 + 50; const int inf = 0x3fffffff; const double eps = 1e-8; const int mod = 100000000 + 7; # define L (x) (x <1) # define R (x) (x <1 | 1) # define PII pair <int, int> # define mk (x, y) make_pair (x), (y) int n, m, edge_cnt, cnt, Lc, Rc; char str [N]; int col [MAXN], head [MAXN], sz [MAXN], dep [MAXN], fa [MAXN], son [MAXN], top [MAXN], pos [MAXN]; struct Edge {int u, v, next;} edg E [MAXN <1]; struct node {int l, r; int num, tag, lc, rc;} segtree [MAXN <2]; void init () {edge_cnt = cnt = 0; memset (head,-1, sizeof (head);} void addedge (int u, int v) {edge [edge_cnt]. u = u; edge [edge_cnt]. v = v; edge [edge_cnt]. next = head [u]; head [u] = edge_cnt ++;} void dfs1 (int u, int pre, int depth) {sz [u] = 1; fa [u] = pre; son [u] = 0; dep [u] = depth; for (int I = head [u]; I! =-1; I = edge [I]. next) {int v = edge [I]. v; if (v = pre) continue; dfs1 (v, u, depth + 1); sz [u] + = sz [v]; if (sz [son [u] <sz [v]) son [u] = v ;}} void dfs2 (int u, int tp) {pos [u] = ++ cnt; top [u] = tp; if (son [u]! = 0) dfs2 (son [u], top [u]); for (int I = head [u]; I! =-1; I = edge [I]. next) {int v = edge [I]. v; if (v = fa [u] | v = son [u]) continue; dfs2 (v, v) ;}} void push_down (int rt) {if (segtree [rt]. tag) {segtree [L (rt)]. tag = segtree [R (rt)]. tag = segtree [rt]. tag; segtree [L (rt)]. num = segtree [R (rt)]. num = 1; segtree [L (rt)]. lc = segtree [L (rt)]. rc = segtree [rt]. lc; segtree [R (rt)]. lc = segtree [R (rt)]. rc = segtree [rt]. lc; segtree [rt]. tag = 0 ;}} void push_up (int rt) {segtree [rt]. lc = segtree [L (rt)]. lc; segtree [rt]. rc = segtree [R (rt)]. rc; int ans = segtree [L (rt)]. num + segtree [R (rt)]. num; if (segtree [L (rt)]. rc = segtree [R (rt)]. lc) ans --; segtree [rt]. num = ans;} void build (int rt, int l, int r) {segtree [rt]. l = l; segtree [rt]. r = r; segtree [rt]. num = 0; if (l = r) return; int mid = (l + r)> 1; build (L (rt), l, mid ); build (R (rt), mid + 1, r);} void update (int rt, int l, int r, int x) {if (segtree [rt]. l = l & segtree [Rt]. r = r) {segtree [rt]. num = segtree [rt]. tag = 1; segtree [rt]. lc = segtree [rt]. rc = x; return;} push_down (rt); int mid = (segtree [rt]. l + segtree [rt]. r)> 1; if (r <= mid) update (L (rt), l, r, x); else if (l> mid) update (R (rt), l, r, x); else {update (L (rt), l, mid, x); update (R (rt ), mid + 1, r, x);} push_up (rt);} int query (int rt, int l, int r, int L, int R) {if (segtree [rt]. l = L) Lc = segtree [rt]. lc; if (segtree [rt]. r = R) Rc = s Egtree [rt]. rc; if (segtree [rt]. l = l & segtree [rt]. r = r) return segtree [rt]. num; push_down (rt); int mid = (segtree [rt]. l + segtree [rt]. r)> 1; if (r <= mid) return query (L (rt), l, r, L, R); else if (l> mid) return query (R (rt), l, r, L, R); else {int ans = query (L (rt), l, mid, L, R) + query (R (rt), mid + 1, r, L, R); if (segtree [L (rt)]. rc = segtree [R (rt)]. lc) ans --; return ans;} push_up (rt);} int solve (int u, int v, int id, int c) {Int ans = 0; if (id = 1) {while (top [u]! = Top [v]) {if (dep [top [u] <dep [top [v]) swap (u, v); update (1, pos [top [u], pos [u], c); u = fa [top [u];} if (dep [u]> dep [v]) swap (u, v); update (1, pos [u], pos [v], c );} else {// printf ("u = % d v = % d pos [u] = % d pos [v] = % d \ n", u, v, pos [u], pos [v]); int ans1 =-1, ans2 =-1; // record the color of the left end of the last chain while (top [u]! = Top [v]) {if (dep [top [u] <dep [top [v]) {swap (u, v); swap (ans1, ans2);} ans + = query (1, pos [top [u], pos [u], pos [top [u], pos [u]); if (Rc = ans1) ans --; // printf ("u = % d top [u] = % d Lc = % d Rc = % d ans = % d \ n", u, top [u], lc, Rc, ans); ans1 = Lc; u = fa [top [u];} if (dep [u] <dep [v]) {swap (u, v); swap (ans1, ans2);} ans + = query (1, pos [v], pos [u], pos [v], pos [u]); if (Rc = ans1) ans --; if (Lc = ans2) ans --; // printf ("u = % d v = % d Lc = % d R C = % d ans = % d \ n ", u, v, Lc, Rc, ans);} return ans;} int main () {// freopen ("in.txt", "r", stdin); while (~ Scanf ("% d", & n, & m) {init (); for (int I = 1; I <= n; I ++) scanf ("% d", & col [I]); for (int I = 1; I <n; I ++) {int u, v; scanf ("% d", & u, & v); addedge (u, v); addedge (v, u);} dfs1 (1, 1 ); dfs2 (1, 1); build (1, 1, n); // <F9> printf ("% d \ n", son [1], son [2]); /* for (int I = 1; I <= n; I ++) printf ("top [I] = % d pos [I] % d \ n ", top [I], pos [I]); printf ("--------------------------------------- \ n"); printf ("\ n"); */for (int I = 1; I <= n; I ++) update (1, pos [I], pos [I], col [I]); while (m --) {scanf ("% s", str); int u, v; if (str [0] = 'C') {int C; scanf ("% d", & u, & v, & c); solve (u, v, 1, c) ;} else {int u, v; scanf ("% d", & u, & v); printf ("% d \ n", solve (u, v, 2, 0 )); // printf ("---------------------------------------- \ n") ;}} return 0 ;}