CF 671D Roads in Yusland 线段树维护代价合并的思想 ★ ★ ★ ★
来源:互联网 发布:画图软件图标大全 编辑:程序博客网 时间:2024/06/06 09:26
题目大意
给定一颗
解题思路
这题有个特殊的性质,就是每个工人只会修一条指向祖先的路径。我们设
我们考虑把一些工人的属性叠加到一个工人上,那么我们最后查询时只要查询某些能维修完整棵树的最小值。还是用上面的思想,我们做到节点
因为我们要要实现对于一些工人的花费都加上某个值得操作,那么可以先把工人按
#include <cstring>#include <cstdio>#include <algorithm>#include <vector>using namespace std;typedef long long LL;const int MAXN = 3e5 + 5;const LL Inf = 1e15 + 7;struct Tree { LL Val, add;} Tr[MAXN * 4];LL F[MAXN];int N, M, time, Val[MAXN], Ord[MAXN], L[MAXN], R[MAXN];int tot, Last[MAXN], Out[MAXN], In[MAXN], Next[MAXN * 4], Go[MAXN * 4];void Link(int u, int v, int *Lst) { Next[++ tot] = Lst[u], Lst[u] = tot, Go[tot] = v; }void Dfs(int Now, int Pre) { L[Now] = time + 1; for (int p = In[Now]; p; p = Next[p]) Ord[Go[p]] = ++ time; for (int p = Last[Now]; p; p = Next[p]) { int v = Go[p]; if (v == Pre) continue; Dfs(v, Now); } R[Now] = time;}void Build(int Now, int l, int r) { Tr[Now].Val = Inf; if (l == r) return; int Mid = (l + r) >> 1; Build(Now * 2, l, Mid), Build(Now * 2 + 1, Mid + 1, r);}void Modify(int Now, int l, int r, int Side, LL Val) { if (l == r) { Tr[Now].Val = Val; return; } int Mid = (l + r) >> 1; if (Side <= Mid) Modify(Now * 2, l, Mid, Side, Val); else Modify(Now * 2 + 1, Mid + 1, r, Side, Val); Tr[Now].Val = min(Inf, min(Tr[Now * 2].Val, Tr[Now * 2 + 1].Val) + Tr[Now].add);}void Add(int Now, int l, int r, int lx, int rx, LL add) { if (lx > rx) return; if (l == lx && r == rx) { Tr[Now].Val = min(Tr[Now].Val + add, Inf); Tr[Now].add = min(Tr[Now].add + add, Inf); return; } int Mid = (l + r) >> 1; if (rx <= Mid) Add(Now * 2, l, Mid, lx, rx, add); else if (lx > Mid) Add(Now * 2 + 1, Mid + 1, r, lx, rx, add); else { Add(Now * 2, l, Mid, lx, Mid, add); Add(Now * 2 + 1, Mid + 1, r, Mid + 1, rx, add); } Tr[Now].Val = min(Inf, min(Tr[Now * 2].Val, Tr[Now * 2 + 1].Val) + Tr[Now].add);}LL Query(int Now, int l, int r, int lx, int rx) { if (lx > rx) return Inf; if (l == lx && r == rx) return Tr[Now].Val; int Mid = (l + r) >> 1; if (rx <= Mid) return min(Inf, Query(Now * 2, l, Mid, lx, rx) + Tr[Now].add); else if (lx > Mid) return min(Inf, Query(Now * 2 + 1, Mid + 1, r, lx, rx) + Tr[Now].add); else return min(Inf, min(Query(Now * 2, l, Mid, lx, Mid), Query(Now * 2 + 1, Mid + 1, r, Mid + 1, rx)) + Tr[Now].add);}void Solve(int Now, int Pre) { LL Ans = 0; for (int p = Last[Now]; p; p = Next[p]) { int v = Go[p]; if (v == Pre) continue; Solve(v, Now); Ans = min(Ans + F[v], Inf); } if (Now == 1) { F[1] = Ans; return; } for (int p = In[Now]; p; p = Next[p]) Modify(1, 1, M, Ord[Go[p]], Ans + Val[Go[p]]); for (int p = Out[Now]; p; p = Next[p]) Modify(1, 1, M, Ord[Go[p]], Inf); for (int p = Last[Now]; p; p = Next[p]) if (Go[p] != Pre) Add(1, 1, M, L[Go[p]], R[Go[p]], Ans - F[Go[p]]); F[Now] = Query(1, 1, M, L[Now], R[Now]);}int main() { scanf("%d %d", &N, &M); for (int i = 1; i < N; i ++) { int u, v; scanf("%d%d", &u, &v); Link(u, v, Last), Link(v, u, Last); } for (int i = 1; i <= M; i ++) { int l, r, t; scanf("%d%d%d", &l, &r, &Val[i]); Link(l, i, In), Link(r, i, Out); } Dfs(1, 0); Build(1, 1, M); Solve(1, 0); printf("%I64d", (F[1] == Inf) ? -1 : F[1]);}
0 0
- CF 671D Roads in Yusland 线段树维护代价合并的思想 ★ ★ ★ ★
- CF 671D Roads in Yusland 线段树维护代价合并的思想
- cf 671D Roads in Yusland
- Codeforces 671D Roads in Yusland dfs序+线段树
- [CF671D]Roads in Yusland
- CF 46 D Parking Lot(线段树区间合并)
- JZOJ4774 【GDOI2017模拟9.10】子串 线段树合并维护SAM的fail树信息(CF 666E类似)
- CF-46D-Parking Lot(线段树,区间合并,点表示线段)
- CF #343 div2 D Babaei and Birthday Cake dp+线段树维护+(离散)
- cf - 629D Babaei and Birthday Cake(DP+线段树维护)
- CF 243D Cubes(线段树)
- CF 228D Zigzag(线段树)
- CF 19D Points(线段树)
- CF 85D 五颗线段树
- CF 19D Points(线段树+set)
- CF-46D-Parking Lot(线段树)
- cf 689 D(线段树+二分)
- CF - 219D 线段树 + dfs序
- Android中去掉标题栏的几种方法(三种)
- HDU 4549 矩阵快速幂 + 快速幂取模 + 费马小定理
- DirectX导图(6):灯光
- JavaScript的那些坑之闭包
- 洛谷 P1772 [ZJOI2006] 物流运输
- CF 671D Roads in Yusland 线段树维护代价合并的思想 ★ ★ ★ ★
- python json详解
- Android之应用中执行Linux命令
- [LeetCode]--88. Merge Sorted Array
- 二维数组中的查找
- 一个留学生的自白:清华vsMIT
- JavaWeb之Cookie_Session
- DirectX导图(7):纹理
- Nginx负载均衡