首页 > 代码库 > 2243: [SDOI2011]染色树链剖分
2243: [SDOI2011]染色树链剖分
对于线段树的操作,维护左端值,维护右端值,维护种类数,更新的时候,如果左儿子的右端==右儿子的左端,种类数减一,剩下的就是细节了。
#include<iostream>#include<cstdio>#include<cstring>#include<map>#include<vector>#include<stdlib.h>using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int maxn = 222222;int deep[maxn];struct Node{ int next; int to;}e[maxn * 2];int len, z; int head[maxn];int top[maxn], son[maxn], father[maxn], size[maxn], pos[maxn], val[maxn], vis[maxn];int color[maxn << 2], sum[maxn << 2], lsum[maxn << 2], rsum[maxn << 2];void add(int from, int to){ e[len].to = to; e[len].next = head[from]; head[from] = len++;}void init(int x){ size[x] = 1; son[x] = 0; for (int i = head[x]; i != -1; i = e[i].next){ int cc = e[i].to; if (cc == father[x]) continue; father[cc] = x; deep[cc] = deep[x] + 1; init(cc); size[x] += size[cc]; if (size[son[x]] < size[cc]) son[x] = cc; }}void dfs(int x, int tp){ pos[x] = ++z; vis[z] = x; top[x] = tp; if (son[x]) dfs(son[x], tp); for (int i = head[x]; i != -1; i = e[i].next){ int cc = e[i].to; if (cc == son[x] || cc == father[x]) continue; dfs(cc, cc); }}void up(int rt){ lsum[rt] = lsum[rt << 1]; rsum[rt] = rsum[rt << 1 | 1]; if (rsum[rt << 1] == lsum[rt << 1 | 1]) sum[rt] = sum[rt << 1] + sum[rt << 1 | 1] - 1; else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void build(int l, int r, int rt){ color[rt] = -1; if (l == r){ color[rt] = lsum[rt] = rsum[rt] = val[vis[l]]; sum[rt] = 1; return; } int mid = (l + r) >> 1; build(lson); build(rson); up(rt);}void down(int rt){ if (~color[rt]){ sum[rt << 1] = sum[rt << 1 | 1] = 1; lsum[rt << 1] = color[rt]; rsum[rt << 1] = color[rt]; color[rt << 1] = color[rt << 1 | 1] = color[rt]; lsum[rt << 1 | 1] = color[rt]; rsum[rt << 1 | 1] = color[rt]; color[rt] = -1; }}void update(int L, int R, int ans, int l, int r, int rt){ if (L <= l&&r <= R){ color[rt] = ans; sum[rt] = 1; lsum[rt] = ans; rsum[rt] = ans; return; } int mid = (l + r) >> 1; down(rt); if (L <= mid) update(L, R, ans, lson); if (R > mid) update(L, R, ans, rson); up(rt);}int ask(int L, int R, int l, int r, int rt){ if (L <= l&&r <= R) return sum[rt]; int ans = 0; int mid = (l + r) >> 1; down(rt); if (L <= mid) ans += ask(L, R, lson); if (R > mid) ans += ask(L, R, rson); if (L <= mid&&R > mid&&rsum[rt << 1] == lsum[rt << 1 | 1]) ans--; return ans;}void gao(int x, int y, int ans){ int fx = top[x]; int fy = top[y]; while (fx != fy){ if (deep[fx] < deep[fy]){ swap(x, y); swap(fx, fy); } update(pos[fx], pos[x], ans, 1, z, 1); x = father[fx]; fx = top[x]; } if (deep[x]>deep[y]) swap(x, y); update(pos[x], pos[y], ans, 1, z, 1);}int ask1(int key, int l, int r, int rt){ if (l == r) return color[rt]; down(rt); int mid = (l + r) >> 1; if (key <= mid) return ask1(key, lson); else return ask1(key, rson);}int gao1(int x, int y){ int ans = 0; int fx = top[x]; int fy = top[y]; while (fx != fy){ if (deep[fx] < deep[fy]){ swap(x, y); swap(fx, fy); } ans += ask(pos[fx], pos[x], 1, z, 1); x = father[fx];//手戳 ,写成了father[x],看了老半天 int t = ask1(pos[x], 1, z, 1); int t1 = ask1(pos[fx], 1, z, 1); if (t == t1) ans--; fx = top[x]; } if (deep[x]>deep[y]) swap(x, y); ans += ask(pos[x], pos[y], 1, z, 1);//pos[x] 顺手写成了pos[son[x]],看了好久 return ans;}int main(){ char str[100]; int n, m; int a, b, c; while (cin >> n >> m){ z = len = 0; memset(head, -1, sizeof(head)); for (int i = 1; i <= n; i++){ scanf("%d", &val[i]); } for (int i = 1; i <= n - 1; i++){ scanf("%d%d", &a, &b); add(a, b); add(b, a); } deep[1] = 1; init(1); dfs(1, 1); build(1, z, 1); for (int i = 0; i < m; i++){ scanf("%s", str); if (str[0] == ‘C‘){ scanf("%d%d%d", &a, &b, &c); gao(a, b, c); } else{ scanf("%d%d", &a, &b); int t = gao1(a, b); printf("%d\n", t); //printf("%d %d %d %d jijiji\n",ask1(pos[1],1,z,1),ask1(pos[2],1,z,1),ask1(pos[3],1,z,1),ask1(pos[5],1,z,1)); } } } return 0;}
2243: [SDOI2011]染色树链剖分
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。