LA 5031 Graph and Queries【名次树】【离线算法】

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题目大义

有一张n结点m条边的无向图，每个结点都有一个权值，你的任务是执行一系列操作，共3种。
1、D X 删除ID为x的边
2、Q X k 计算与x相连的边的第k大权值，如果不存在输出0
3、C X V 把X的权值改为V
题目链接什么的还是给一个 https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3032

本体实际上就是一个名次树+离线算法
离线算法：先读入所有的操作，删除所有边得到最终的图，然后反过来操作，在适合的时候加上边，在适合的时候改变点的权值，然后用名次树来维护每个连通分量，加边的时候就用并查集判一下就可以了，然后再将两棵树合并
名次树：实际上就是treap tree这种平衡树加上了一个名次功能（其他平衡树好像感觉也可以，但我弱鸡写不来）

都说到这里了就来顺便说一下treap tree吧，前面说过了，treap tree就是一种平衡树，而且应该是最简单也最方便的平衡树了，虽然不一定最快，但插入和删除都只有一种情况，写起来很舒服，它的基本操作就是基本的3种：旋转，插入，删除

旋转：旋转和其他的都差不多了，就是左旋右旋，想象成你的手去提一下就好了，然后再稍微做个小手术，让它还是保持二叉平衡树的基本性质就可以了

插入：实际上二叉平衡树的插入无论如何都只有一种，问题就在于插入后如何去平衡，treap tree采用的是随机数平衡法，再插入每个点的时候给每个点一个随机数，然后让这些随机数通过旋转保持大根堆的性质，这样来随机平衡它，实际效果很好

删除：删除的话子树处理是最大的问题，如果要删除的结点只有一颗子树就非常好办了，直接用那棵子树覆盖掉要删除的结点就好了，如果有多棵子树就将随机数较大的子树旋转上来，然后递归去删除即可，虽然要删除的结点在对立面会不满足大根堆的性质，但其它的节点都是满足的，所以删除后整棵树还是满足treap tree的基本性质

然后就是这道题相当金典，可以好好去推敲一下，我的代码在下面，因为网各种炸，vj和la都叫不上去，就不交了，样例过了，但不保证一定会对，同时附上刘汝佳的代码，我也是看了他的书的

我的代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
#include<stack>
#define N 20005
#define M 60005using namespace std;struct tree2
{tree2 *son[2];int num,n,r;int com(int x){if(x==n)return -1;return x>n?1:0;}
}*root[N],dizhi[N],*null;int d[N],t,tt;
int fa[M];
int n,m,kase;
int edge1[M],edge2[M],b[M];
int tot=0,cnt=0;void update(tree2 *&tree)
{tree->num=tree->son[0]->num+tree->son[1]->num+1;
}void rotate(tree2 *&tree,int d)//d=0左旋 d=1右旋
{tree2 *k=tree->son[d^1];tree->son[d^1]=k->son[d];k->son[d]=tree;update(tree);update(k);tree=k;
}int insert(tree2 *&tree,int x)
{if(tree==null){tree=&dizhi[t++];tree->n=x;tree->r=rand();tree->num=1;tree->son[0]=tree->son[1]=null;update(tree);return 1;}int d=tree->com(x);if(d==-1)d=1;insert(tree->son[d],x);if(tree->son[d]->r>tree->r)rotate(tree,d^1);update(tree);return 1;
}int remove(tree2 *&tree,int x)
{if(tree==null||tree==NULL)return -1;int d=tree->com(x);if(d==-1){if(tree->son[0]==null)tree=tree->son[1];else if(tree->son[1]=null)tree=tree->son[0];else{int d2=tree->son[0]->r>tree->son[1]->r?1:0;rotate(tree,d2);remove(tree->son[d2],x);}return 1;}remove(tree->son[d],x);
}bool find(tree2 *tree,int x)
{if(tree==null||tree==NULL)return 0;int d=tree->com(x);if(d==-1)return 1;return (tree->son[d],x);
}int kth(tree2 *tree,int k)
{if(tree==null||tree==NULL)return 0;if(k>tree->num||k<0)return 0;int s=tree->son[1]->num;if(k==s+1)return tree->n;if(k<=s)return kth(tree->son[1],k);return kth(tree->son[0],k-s-1);
}void mergeto(tree2 *&tree1,tree2 *&tree2)//将tree1全部插入tree2
{if(tree1->son[0]!=null)mergeto(tree1->son[0],tree2);if(tree1->son[1]!=null)mergeto(tree1->son[1],tree2);insert(tree2,tree1->n);tree1=null;
}int get_fa(int x)
{return x==fa[x]?x:fa[x]=get_fa(fa[x]);
}void addedge(int x)
{int s=edge1[x],v=edge2[x];int fa1=get_fa(s),fa2=get_fa(v);if(fa1==fa2)return ;if(root[s]->num>root[v]->num){mergeto(root[v],root[s]);fa[v]=fa[s];return ;}mergeto(root[s],root[v]);fa[s]=fa[v];
}void change_v(int x,int v)
{int f=get_fa(x);remove(root[f],d[x]);d[x]=v;insert(root[f],d[x]);
}#include<ctime>int main()
{while(scanf("%d%d",&n,&m)==2){memset(dizhi,0,sizeof(dizhi));memset(b,0,sizeof(b));t=0;null=&dizhi[t++];null->num=0;if(n==0&&m==0)return 0;for(int i=1;i<=n;i++)root[i]=null;stack<int>num1,num2,type;for(int i=1;i<=n;i++)scanf("%d",&d[i]);for(int i=1;i<=m;i++)scanf("%d%d",&edge1[i],&edge2[i]);for(;;){char c;scanf("%c",&c);if(c=='E')break;if(c=='D'){int x;scanf("%d",&x);b[x]=1;type.push(1);num1.push(x);num2.push(0);continue;}if(c=='Q'){int x,k;scanf("%d%d",&x,&k);type.push(2);num1.push(x);num2.push(k);continue;}if(c=='C'){int x,v;scanf("%d%d",&x,&v);type.push(3);num1.push(x);num2.push(d[x]);d[x]=v;}}for(int i=1;i<=n;i++)insert(root[i],d[i]);for(int i=1;i<=n;i++)fa[i]=i;for(int i=1;i<=m;i++)if(!b[i])addedge(i);int p=0;tot=0,cnt=0;while(!type.empty()){int temp=type.top();type.pop();int n1=num1.top();num1.pop();int n2=num2.top();num2.pop();if(temp==1)addedge(n1);if(temp==2){cnt++;tot+=kth(root[get_fa(n1)],n2);}if(temp==3)change_v(n1,n2);}printf("Case %d %.6lf\n",++kase,tot/(double)cnt);}return 0;
}

刘汝佳代码：

// LA5031/UVa1479 Graph and Queries
// Rujia Liu
#include<cstdlib>struct Node {Node *ch[2]; // 左右子树int r; // 随机优先级int v; // 值int s; // 结点总数Node(int v):v(v) { ch[0] = ch[1] = NULL; r = rand(); s = 1; }int cmp(int x) const {if (x == v) return -1;return x < v ? 0 : 1;}void maintain() {s = 1;if(ch[0] != NULL) s += ch[0]->s;if(ch[1] != NULL) s += ch[1]->s;}
};void rotate(Node* &o, int d) {Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o->maintain(); k->maintain(); o = k;
}void insert(Node* &o, int x) {if(o == NULL) o = new Node(x);else {int d = (x < o->v ? 0 : 1); // 不要用cmp函数，因为可能会有相同结点insert(o->ch[d], x);if(o->ch[d]->r > o->r) rotate(o, d^1);}o->maintain();
}void remove(Node* &o, int x) {int d = o->cmp(x);int ret = 0;if(d == -1) {Node* u = o;if(o->ch[0] != NULL && o->ch[1] != NULL) {int d2 = (o->ch[0]->r > o->ch[1]->r ? 1 : 0);rotate(o, d2); remove(o->ch[d2], x);} else {if(o->ch[0] == NULL) o = o->ch[1]; else o = o->ch[0];delete u;}} elseremove(o->ch[d], x);if(o != NULL) o->maintain();
}#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;const int maxc = 500000 + 10;
struct Command {char type;int x, p; // 根据type, p代表k或者v
} commands[maxc];const int maxn = 20000 + 10;
const int maxm = 60000 + 10;
int n, m, weight[maxn], from[maxm], to[maxm], removed[maxm];// 并查集相关
int pa[maxn];     
int findset(int x) { return pa[x] != x ? pa[x] = findset(pa[x]) : x; } // 名次树相关
Node* root[maxn]; // Treapint kth(Node* o, int k) { // 第k大的值if(o == NULL || k <= 0 || k > o->s) return 0;int s = (o->ch[1] == NULL ? 0 : o->ch[1]->s);if(k == s+1) return o->v;else if(k <= s) return kth(o->ch[1], k);else return kth(o->ch[0], k-s-1);
}void mergeto(Node* &src, Node* &dest) {if(src->ch[0] != NULL) mergeto(src->ch[0], dest);if(src->ch[1] != NULL) mergeto(src->ch[1], dest);insert(dest, src->v);delete src;src = NULL;
}void removetree(Node* &x) {if(x->ch[0] != NULL) removetree(x->ch[0]);if(x->ch[1] != NULL) removetree(x->ch[1]);delete x;x = NULL;
}// 主程序相关
void add_edge(int x) {int u = findset(from[x]), v = findset(to[x]);if(u != v) {if(root[u]->s < root[v]->s) { pa[u] = v; mergeto(root[u], root[v]); }else { pa[v] = u; mergeto(root[v], root[u]); }}
}int query_cnt;
long long query_tot;
void query(int x, int k) {query_cnt++;query_tot += kth(root[findset(x)], k);
}void change_weight(int x, int v) {int u = findset(x);remove(root[u], weight[x]);insert(root[u], v);weight[x] = v;
}int main() {int kase = 0;while(scanf("%d%d", &n, &m) == 2 && n) {for(int i = 1; i <= n; i++) scanf("%d", &weight[i]);for(int i = 1; i <= m; i++) scanf("%d%d", &from[i], &to[i]);memset(removed, 0, sizeof(removed));// 读命令int c = 0;for(;;) {char type;int x, p = 0, v = 0;scanf(" %c", &type);if(type == 'E') break;scanf("%d", &x);if(type == 'D') removed[x] = 1;if(type == 'Q') scanf("%d", &p);if(type == 'C') {scanf("%d", &v);p = weight[x];weight[x] = v;}commands[c++] = (Command){ type, x, p };}// 最终的图for(int i = 1; i <= n; i++) {pa[i] = i; if(root[i] != NULL) removetree(root[i]);root[i] = new Node(weight[i]);}for(int i = 1; i <= m; i++) if(!removed[i]) add_edge(i);// 反向操作query_tot = query_cnt = 0;for(int i = c-1; i >= 0; i--) {if(commands[i].type == 'D') add_edge(commands[i].x);if(commands[i].type == 'Q') query(commands[i].x, commands[i].p);if(commands[i].type == 'C') change_weight(commands[i].x, commands[i].p);}printf("Case %d: %.6lf\n", ++kase, query_tot / (double)query_cnt);}return 0;
}