[AtCoder Beginner Contest 133]F - Colorful Tree

时间:2019-09-20
本文章向大家介绍[AtCoder Beginner Contest 133]F - Colorful Tree,主要包括[AtCoder Beginner Contest 133]F - Colorful Tree使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

题目链接

写完题去网上逛一圈发现全都是离线LCA,Orz。

大致题意是一颗树上边有边权和颜色,每次询问会先把颜色为x的边的边权变为y,再询问u到v的边权和。注意,每次询问的修改只针对当前询问。

由于题目是树上距离,所以树剖大致是可以做的。

树剖完后将每条边的边权转点权,赋给深度较高的节点。

每次查询,我们只要知道两点的权值和sum1,颜色为x的边的权值和sum2以及颜色为x的边的个数num,则答案为$sum1-sum2+num*y$。

所以就建权值线段树,以颜色为权值,同时维护颜色个数以及这种颜色的权值和。

就是普普通通的树剖主席树啦QAQ

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 const int maxm = 1e6 + 10;
  4 const int maxn = 3e5+10;
  5 typedef long long ll;
  6 struct node {
  7     int s, e, c, w, next;
  8 }edge[maxn];
  9 int head[maxn], len;
 10 int n, q;
 11 void init() {
 12     memset(head, -1, sizeof(head));
 13     len = 0;
 14 }
 15 void add(int s, int e, int c, int w) {
 16     edge[len].s = s;
 17     edge[len].e = e;
 18     edge[len].w = w;
 19     edge[len].c = c;
 20     edge[len].next = head[s];
 21     head[s] = len++;
 22 }
 23 int son[maxn], top[maxn], tid[maxn], fat[maxn], siz[maxn], dep[maxn], rak[maxn];
 24 int dfx;
 25 void dfs1(int x, int fa, int d) {
 26     siz[x] = 1, son[x] = -1, fat[x] = fa, dep[x] = d;
 27     for (int i = head[x]; i != -1; i = edge[i].next) {
 28         int y = edge[i].e;
 29         if (y == fa)
 30             continue;
 31         dfs1(y, x, d + 1);
 32         siz[x] += siz[y];
 33         if (son[x] == -1 || siz[y] > siz[son[x]])
 34             son[x] = y;
 35     }
 36 }
 37 void dfs2(int x, int c) {
 38     top[x] = c; tid[x] = ++dfx; rak[dfx] = x;
 39     if (son[x] == -1)
 40         return;
 41     dfs2(son[x], c);
 42     for (int i = head[x]; i != -1; i = edge[i].next) {
 43         int y = edge[i].e;
 44         if (y == fat[x] || y == son[x])
 45             continue;
 46         dfs2(y, y);
 47     }
 48 }
 49 int cnt, root[maxn], ls[maxn * 40], rs[maxn * 40], num[maxn * 40], val[maxn * 40];
 50 struct C {
 51     int color, dis;
 52 }a[maxn];
 53 void build(C k, int l, int r, int& i) {
 54     num[++cnt] = num[i] + 1, val[cnt] = val[i] + k.dis, ls[cnt] = ls[i], rs[cnt] = rs[i];
 55     i = cnt;
 56     if (l == r)
 57         return;
 58     int mid = l + r >> 1;
 59     if (k.color <= mid)
 60         build(k, l, mid, ls[i]);
 61     else
 62         build(k, mid + 1, r, rs[i]);
 63 }
 64 C query(int u, int v, int k, int l, int r) {
 65     if (l == r) {
 66         return { num[v] - num[u],val[v] - val[u] };
 67     }
 68         
 69     int mid = l + r >> 1;
 70     if (k <= mid)
 71         return query(ls[u], ls[v], k, l, mid);
 72     else
 73         return query(rs[u], rs[v], k, mid + 1, r);
 74 }
 75 int solve(int cr, int dis, int x, int y) {
 76     C ans = { 0,0 };
 77     int sum = 0;
 78     while (top[x] != top[y]) {
 79         if (dep[top[x]] < dep[top[y]])
 80             swap(x, y);
 81         C t = query(root[tid[top[x]] - 1], root[tid[x]], cr, 1, n);
 82         sum += val[root[tid[x]]] - val[root[tid[top[x]] - 1]];
 83         ans.color += t.color, ans.dis += t.dis;
 84         x = fat[top[x]];
 85     }
 86     
 87     if (x != y) {
 88         if (dep[x] < dep[y])
 89             swap(x, y);
 90         C t = query(root[tid[son[y]] - 1], root[tid[x]], cr, 1, n);
 91         sum += val[root[tid[x]]] - val[root[tid[son[y]] - 1]];
 92         ans.color += t.color, ans.dis += t.dis;
 93     }
 94     return sum - ans.dis + ans.color * dis;
 95 }
 96 int main() {
 97     scanf("%d%d", &n, &q);
 98     init();
 99     for (int i = 1, x, y, c, d; i < n; i++) {
100         scanf("%d%d%d%d", &x, &y, &c, &d);
101         add(x, y, c, d);
102         add(y, x, c, d);
103     }
104     dfs1(1, 0, 1);
105     dfs2(1, 1);
106     for (int i = 0; i < len; i += 2) {
107         int x = edge[i].e;
108         int y = edge[i].s;
109         if (dep[x] < dep[y])
110             swap(x, y);
111         a[tid[x]] = { edge[i].c, edge[i].w };
112     }
113     for (int i = 2; i <= dfx; i++) {
114         root[i] = root[i - 1];
115         build(a[i], 1, n, root[i]);
116     }
117     while (q--) {
118         int x, y, u, v;
119         scanf("%d%d%d%d", &x, &y, &u, &v);
120         printf("%d\n", solve(x, y, u, v));
121     }
122 }

原文地址:https://www.cnblogs.com/sainsist/p/11557047.html