Problem

给你一张图,点的权值,边和几个操作:

D x: 删除第x条边

Q x y: 询问包含x的联通块中权值第y大的权值

C x y: 将x这个点的权值改为y

Solution

一看就要离线处理,把所有操作都倒过来

然后删除操作变为加边操作

Notice

记得: 是改完以后再把点一个一个加入Treap中!!

Code

非旋转Treap

#pragma GCC optimize(2)
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 3000000;
const double eps = 1e-6, phi = acos(-1.0);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
struct Node
{
int num1, num2, type;
}Q[N + 5];
int point = 0, fa[N + 5], Root[N + 5], T[N + 5], From[N + 5], To[N + 5], Flag[N + 5];
struct node
{
int Val[N + 5], Level[N + 5], Size[N + 5], Son[2][N + 5], Num[N + 5];
inline void up(int u)
{
Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + 1;
}
int Newnode(int v)
{
int u = ++point;
Val[u] = v, Level[u] = rand();
Son[0][u] = Son[1][u] = 0, Size[u] = 1;
return u;
}
int Merge(int X, int Y)
{
if (X * Y == 0) return X + Y;
if (Level[X] < Level[Y])
{
Son[1][X] = Merge(Son[1][X], Y);
up(X); return X;
}
else
{
Son[0][Y] = Merge(X, Son[0][Y]);
up(Y); return Y;
}
}
void Split(int u, int t, int &x, int &y)
{
if (!u)
{
x = y = 0;
return;
}
if (Val[u] <= t) x = u, Split(Son[1][u], t, Son[1][u], y);
else y = u, Split(Son[0][u], t, x, Son[0][u]);
up(u);
}
int Find_num(int u, int v)
{
if (!u) return 0;
if (v <= Size[Son[0][u]]) return Find_num(Son[0][u], v);
else if (v <= Size[Son[0][u]] + 1) return u;
else return Find_num(Son[1][u], v - Size[Son[0][u]] - 1);
}
void Insert(int &u, int v)
{
int t = Newnode(v), x, y;
Split(u, v, x, y);
u = Merge(Merge(x, t), y);
}
void Delete(int &u, int v)
{
int x, y, z;
Split(u, v, x, z), Split(x, v - 1, x, y);
u = Merge(Merge(x, Merge(Son[0][y], Son[1][y])), z);
}
}Treap;
int Find(int x)
{
if (fa[x] != x) fa[x] = Find(fa[x]);
return fa[x];
}
void Union(int u, int v)
{
if (Treap.Size[Root[u]] < Treap.Size[Root[v]]) swap(u, v);
while (Treap.Size[Root[v]])
{
int t = Treap.Find_num(Root[v], 1);
Treap.Insert(Root[u], Treap.Val[t]);
Treap.Delete(Root[v], Treap.Val[t]);
}
fa[v] = u;
Root[v] = 0;
}
int sqz()
{
int n, m, cas = 0;
while (~scanf("%d %d", &n, &m) && (n || m))
{
point = 0;
rep(i, 1, n) T[i] = read(), fa[i] = i, Root[i] = 0;
rep(i, 1, m) From[i] = read(), To[i] = read(), Flag[i] = 0;
int q = 0; char op[5];
while (scanf("%s", op) && op[0] != 'E')
{
q++;
if (op[0] == 'D')
Q[q].num1 = read(), Flag[Q[q].num1] = 1, Q[q].type = 0;
else
{
Q[q].num1 = read(), Q[q].num2 = read();
if (op[0] == 'C') swap(T[Q[q].num1], Q[q].num2), Q[q].type = 1;
else Q[q].type = 2;
}
}
rep(i, 1, n) Treap.Insert(Root[i], T[i]);
rep(i, 1, m)
if (!Flag[i])
{
int u = Find(From[i]), v = Find(To[i]);
if (u != v) Union(u, v);
}
ll ans = 0; int tot = 0;
per(i, q, 1)
{
if (Q[i].type == 0)
{
int u = Find(From[Q[i].num1]), v = Find(To[Q[i].num1]);
if (u != v) Union(u, v);
}
else if (Q[i].type == 1)
{
int u = Find(Q[i].num1);
Treap.Delete(Root[u], T[Q[i].num1]);
Treap.Insert(Root[u], Q[i].num2);
T[Q[i].num1] = Q[i].num2;
}
else
{
int u = Find(Q[i].num1);
int t = Treap.Find_num(Root[u], Treap.Size[Root[u]] - Q[i].num2 + 1);
if (t != -INF) ans += Treap.Val[t];
tot++;
}
}
printf("Case %d: %.6f\n", ++cas, ans * 1.0 / tot);
}
return 0;
}

旋转Treap

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 3000000;
const double eps = 1e-6, phi = acos(-1.0);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
struct Node
{
int num1, num2, type;
}Q[N + 5];
int point = 0, fa[N + 5], Root[N + 5], T[N + 5], From[N + 5], To[N + 5], Flag[N + 5];
struct node
{
int Val[N + 5], Level[N + 5], Size[N + 5], Son[2][N + 5], Num[N + 5];
inline void up(int u)
{
Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + Num[u];
}
inline void Newnode(int &u, int v)
{
u = ++point;
Level[u] = rand(), Val[u] = v;
Size[u] = Num[u] = 1, Son[0][u] = Son[1][u] = 0;
}
inline void Lturn(int &x)
{
int y = Son[1][x]; Son[1][x] = Son[0][y], Son[0][y] = x;
up(x); up(y); x = y;
}
inline void Rturn(int &x)
{
int y = Son[0][x]; Son[0][x] = Son[1][y], Son[1][y] = x;
up(x); up(y); x = y;
} void Insert(int &u, int t)
{
if (u == 0)
{
Newnode(u, t);
return;
}
Size[u]++;
if (t == Val[u]) Num[u]++;
else if (t > Val[u])
{
Insert(Son[0][u], t);
if (Level[Son[0][u]] < Level[u]) Rturn(u);
}
else if (t < Val[u])
{
Insert(Son[1][u], t);
if (Level[Son[1][u]] < Level[u]) Lturn(u);
}
}
void Delete(int &u, int t)
{
if (!u) return;
if (Val[u] == t)
{
if (Num[u] > 1)
{
Num[u]--, Size[u]--;
return;
}
if (Son[0][u] * Son[1][u] == 0) u = Son[0][u] + Son[1][u];
else if (Level[Son[0][u]] < Level[Son[1][u]]) Rturn(u), Delete(u, t);
else Lturn(u), Delete(u, t);
}
else if (t > Val[u]) Size[u]--, Delete(Son[0][u], t);
else Size[u]--, Delete(Son[1][u], t);
} int Find_num(int u, int t)
{
if (!u) return -INF;
if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
else if (t <= Size[Son[0][u]] + Num[u]) return Val[u];
else return Find_num(Son[1][u], t - Size[Son[0][u]] - Num[u]);
}
}Treap;
int Find(int x)
{
if (fa[x] != x) fa[x] = Find(fa[x]);
return fa[x];
}
void Union(int u, int v)
{
if (Treap.Size[Root[u]] < Treap.Size[Root[v]]) swap(u, v);
while (Treap.Size[Root[v]])
{
int t = Treap.Find_num(Root[v], 1);
Treap.Insert(Root[u], t);
Treap.Delete(Root[v], t);
}
fa[v] = u;
Root[v] = 0;
}
int sqz()
{
int n, m, cas = 0;
while (~scanf("%d %d", &n, &m) && (n || m))
{
point = 0;
rep(i, 1, n) T[i] = read(), fa[i] = i, Root[i] = 0;
rep(i, 1, m) From[i] = read(), To[i] = read(), Flag[i] = 0;
int q = 0; char op[5];
while (scanf("%s", op) && op[0] != 'E')
{
q++;
if (op[0] == 'D')
Q[q].num1 = read(), Flag[Q[q].num1] = 1, Q[q].type = 0;
else
{
Q[q].num1 = read(), Q[q].num2 = read();
if (op[0] == 'C') swap(T[Q[q].num1], Q[q].num2), Q[q].type = 1;
else Q[q].type = 2;
}
}
rep(i, 1, n) Treap.Insert(Root[i], T[i]);
rep(i, 1, m)
if (!Flag[i])
{
int u = Find(From[i]), v = Find(To[i]);
if (u != v) Union(u, v);
}
ll ans = 0; int tot = 0;
per(i, q, 1)
{
if (Q[i].type == 0)
{
int u = Find(From[Q[i].num1]), v = Find(To[Q[i].num1]);
if (u != v) Union(u, v);
}
else if (Q[i].type == 1)
{
int u = Find(Q[i].num1);
Treap.Delete(Root[u], T[Q[i].num1]);
Treap.Insert(Root[u], Q[i].num2);
T[Q[i].num1] = Q[i].num2;
}
else
{
int u = Find(Q[i].num1);
int t = Treap.Find_num(Root[u], Q[i].num2);
if (t != -INF) ans += t;
tot++;
}
}
printf("Case %d: %.6f\n", ++cas, ans * 1.0 / tot);
}
return 0;
}

[HDU3726]Graph and Queries的更多相关文章

  1. HDU 3726 Graph and Queries 平衡树+前向星+并查集+离线操作+逆向思维 数据结构大综合题

    Graph and Queries Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Other ...

  2. HDU 3726 Graph and Queries (离线处理+splay tree)

    Graph and Queries Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Other ...

  3. [la P5031&hdu P3726] Graph and Queries

    [la P5031&hdu P3726] Graph and Queries Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: ...

  4. HDU 3726 Graph and Queries treap树

    题目来源:HDU 3726 Graph and Queries 题意:见白书 思路:刚学treap 參考白皮书 #include <cstdio> #include <cstring ...

  5. HDU 3726 Graph and Queries(平衡二叉树)(2010 Asia Tianjin Regional Contest)

    Description You are given an undirected graph with N vertexes and M edges. Every vertex in this grap ...

  6. UVALive5031 Graph and Queries(Treap)

    反向操作,先求出最终状态,再反向操作. 然后就是Treap 的合并,求第K大值. #include<cstdio> #include<iostream> #include< ...

  7. UVa 1479 (Treap 名次树) Graph and Queries

    这题写起来真累.. 名次树就是多了一个附加信息记录以该节点为根的树的总结点的个数,由于BST的性质再根据这个附加信息,我们可以很容易找到这棵树中第k大的值是多少. 所以在这道题中用一棵名次树来维护一个 ...

  8. uvalive 5031 Graph and Queries 名次树+Treap

    题意:给你个点m条边的无向图,每个节点都有一个整数权值.你的任务是执行一系列操作.操作分为3种... 思路:本题一点要逆向来做,正向每次如果删边,复杂度太高.逆向到一定顺序的时候添加一条边更容易.详见 ...

  9. 【HDOJ】3726 Graph and Queries

    Treap的基础题目,Treap是个挺不错的数据结构. /* */ #include <iostream> #include <string> #include <map ...

随机推荐

  1. 类似于placehoder效果的图标展示

    在做app开发的时候往往会有那个注册登录啊,什么的页面,里面就会包含这那种类似于placeholder的效果的图标,当时我也是和ios和安卓混合开发一款app里面的页面全是我写,最开始就是登陆啊,注册 ...

  2. Python&HDF5目录

    最近一直没更新python&量化的博客,是因为忙于看HDF5的书,写VNPY框架,学scrapy爬虫. 本来写博客的目的就是为了当作一种教材,当遇到不会的问题过来找答案. 对于HDF5下面这本 ...

  3. 用basicTrendline画一元线性回归直线的置信区间

    感慨统计学都还给老师了..恶补! R安装包的时候貌似需要用管理员权限启动,否则安装不了,国内镜像卡得渣渣,还是国外镜像真香~选择hongkong就好了. install.packages(" ...

  4. You Don't Know JS: Async & Performance(第一章, 异步:now & later)

    Chapter 1: Asynchrony: Now & Later 在一门语言中,比如JavaScript, 最重要但仍然常常被误解的编程部分是如何在一个完整的时间周期表示和操作程序行为. ...

  5. HDOJ-2175 汉诺塔IX

    题目大意:基于汉诺塔原型,第一根柱子上有n个盘子,从上至下编号从1依次递增至n.在最佳移动方案中,第m次所移动的盘子的编号. 解题思路:模拟必然是会超时的.但根据汉诺塔的递归原理,容易发现,对于n阶汉 ...

  6. node初学者笔记

    helloworld 编辑一个js文件——在该文件所属目录打开命令行cmd——输入'node -v可查看版本——输入'node  00-hellowolrd.js(你的js名字)' 或者直接在文件所属 ...

  7. php字符串转成数组

    /* 4.$m = “woxihuanphp”,编程实现:将字符串分割为单个字符存放到一个数组中,并打印数组? */ $m='woxihuanphp'; echo $res=trim(chunk_sp ...

  8. 数组<-->变量

    /** * *数组与变量之间转换 **/ $name='jb51'; $email='jb51@jb51.net'; $info=compact('name','email'); print_r($i ...

  9. Matlab-9:中心差分方法解常微分算例(SOR完整版)

    函数文件: function [x,n,flag]=sor(A,b,eps,M,max1) %sor函数为用松弛迭代法求解线性方程组 %A为线性方程组的系数矩阵 %b为线性方程组的常数向量 %eps为 ...

  10. element-ui radio 再次点击取消选中

    <el-radio-group v-model="radio2"> <el-radio @click.native.prevent="clickitem ...