![[la P5031&hdu P3726] Graph and Queries](https://image.shishitao.com:8440/aHR0cHM6Ly9ia3FzaW1nLmlrYWZhbi5jb20vdXBsb2FkL2NoYXRncHQtcy5wbmc%2FIQ%3D%3D.png?!?w=700&webp=1 "[la P5031&hdu P3726] Graph and Queries")
[la P5031&hdu P3726] Graph and Queries
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem DescriptionYou are given an undirected graph with N vertexes and M edges. Every vertex in this graph has an integer value assigned to it at the beginning. You're also given a sequence of operations and you need to process them as requested. Here's a list of the possible operations that you might encounter:
1) Deletes an edge from the graph.
The format is [D X], where X is an integer from 1 to M, indicating the ID of the edge that you should delete. It is guaranteed that no edge will be deleted more than once.
2) Queries the weight of the vertex with K-th maximum value among all vertexes currently connected with vertex X (including X itself).
The format is [Q X K], where X is an integer from 1 to N, indicating the id of the vertex, and you may assume that K will always fit into a 32-bit signed integer. In case K is illegal, the value for that query will be considered as undefined, and you should return 0 as the answer to that query.
3) Changes the weight of a vertex.
The format is [C X V], where X is an integer from 1 to N, and V is an integer within the range [-106, 106].The operations end with one single character, E, which indicates that the current case has ended.
For simplicity, you only need to output one real number - the average answer of all queries.InputThere are multiple test cases in the input file. Each case starts with two integers N and M (1 <= N <= 2 * 104, 0 <= M <= 6 * 104), the number of vertexes in the graph. The next N lines describes the initial weight of each vertex (-106 <= weight[i] <= 106). The next part of each test case describes the edges in the graph at the beginning. Vertexes are numbered from 1 to N. The last part of each test case describes the operations to be performed on the graph. It is guaranteed that the number of query operations [Q X K] in each case will be in the range [1, 2 * 105], and there will be no more than 2 * 105 operations that change the values of the vertexes [C X V].There will be a blank line between two successive cases. A case with N = 0, M = 0 indicates the end of the input file and this case should not be processed by your program.
OutputFor each test case, output one real number – the average answer of all queries, in the format as indicated in the sample output. Please note that the result is rounded to six decimal places.
Sample Input
3 3
10
20
30
1 2
2 3
1 3
D 3
Q 1 2
Q 2 1
D 2
Q 3 2
C 1 50
Q 1 1
E
3 3
10
20
20
1 2
2 3
1 3
Q 1 1
Q 1 2
Q 1 3
E
0 0Sample Output
Case 1: 25.000000
Case 2: 16.666667
HintFor the first sample: D 3 -- deletes the 3rd edge in the graph (the remaining edges are (1, 2) and (2, 3)) Q 1 2 -- finds the vertex with the second largest value among all vertexes connected with 1. The answer is 20. Q 2 1 -- finds the vertex with the largest value among all vertexes connected with 2. The answer is 30. D 2 -- deletes the 2nd edge in the graph (the only edge left after this operation is (1, 2)) Q 3 2 -- finds the vertex with the second largest value among all vertexes connected with 3. The answer is 0 (Undefined). C 1 50 -- changes the value of vertex 1 to 50. Q 1 1 -- finds the vertex with the largest value among all vertex connected with 1. The answer is 50. E -- This is the end of the current test case. Four queries have been evaluated, and the answer to this case is (20 + 30 + 0 + 50) / 4 = 25.000. For the second sample, caution about the vertex with same weight: Q 1 1 – the answer is 20 Q 1 2 – the answer is 20 Q 1 3 – the answer is 10.
Source
这题很经典，但又有坑——
首先，这题涉及到删边。
删边不方便，这时我们自然会想到删边的反面——加边。
那我们就可以倒着来操作，方便D操作。
但同时，每个点的权值也要从终状态到始状态来变化。
然后，我们就可以用并查集来实现加边操作，就把一条边加上了。
当然，最终状态也存在一些边，刚开始也要连上。
然后，Q操作要求我们查询连通块内第k大，如果k不合法则返回0。
第k大可以用一个平衡树来维护，我用了treap名次树，方便一些。
但是，注意这里的第k大可能有些定义上的模糊，我最后几发就被坑了。
似乎没有人是加了一个“相同点的个数”这个域A掉了的，我也是改用lrj的写法才A了的，这个需要注意一下。
然后第k大不合法，有可能是超出连通块大小，也有可能k<1。
对于C操作，就相当于再某个连通块中先删掉一个点权为w1的点，再加入一个点权为为w2的点，w1，w2就不多说了。
但是无论如何，都要注意，点权的变化顺序是怎样的。
如果就这样写——就T了。
为什么呢？可能会有数据卡“加边”这一个操作。在这里我们需要合并两颗平衡树。
怎么合并？我们采用启发式合并。就是从节点数小的向节点数大的树并。
怎么证明这样操作的时间复杂度是对的？假设树T1有n1个节点，T2有n2个节点，且n1>n2。
那么显然，我们会把T2并到T1里面。复杂度接近于O(n2logn1)。
由于对于原来T2里的每个点，它所在的树的大小至少增大了一倍，所以对于任意节点，它会被移动不超过logn次。
所以总复杂度是O(n(logn)^2)的。
然后。。通过这个题目，学到了treap的启发式合并和删除。。
code：
%:pragma GCC optimize() #include<bits/stdc++.h> #define LL long long using namespace std; ,M=,Q=; int n,m,q,cas,w[N],fa[N]; char ch; LL ans,cnt; bool vis[M]; struct edg {int x,y;}a[M]; struct opt {char t; int x,k;}o[Q]; class node { private: public: ]; node() {ch[]=ch[]=;} inline void newnode(node* &cu,int v) { cu=; } inline void update(node* &cu) { cu->s=; ]!=) cu->s+=cu->ch[]->s; ]!=) cu->s+=cu->ch[]->s; } inline void rotate(node* &cu,bool dr) { node* tmp=cu->ch[dr^]->ch[dr]; cu->ch[dr^]->ch[dr]=cu; cu=cu->ch[dr^]; cu->ch[dr]->ch[dr^]=tmp; update(cu->ch[dr]); update(cu); } inline void insert(node* &cu,int v) { ) {newnode(cu,v); return;} bool p=v<=cu->v; insert(cu->ch[p],v); ); else update(cu); } inline void remove(node* &cu,int v) { ) return; if (v==cu->v) { ]==&&cu->ch[]==) {cu=; return;} ]==||cu->ch[]==) { cu=cu->ch[]==?cu->ch[]:cu->ch[]; return; } ]->k>cu->ch[]->k; rotate(cu,p^); remove(cu->ch[p^],v); update(cu); return; } int p=v<=cu->v; remove(cu->ch[p],v); update(cu); } inline int kth(node* cu,int k) { ||cu==) ; ]!=)?cu->ch[]->s:; ) return cu->v; ) ],k); ],k-s-); } inline void merge(node* &major,node* &minor) { insert(major,minor->v); ]!=) merge(major,minor->ch[]); ]!=) merge(major,minor->ch[]); ; } inline void clear(node* &cu) { ]!=) clear(cu->ch[]); ]!=) clear(cu->ch[]); ; } }t,*root[N]; inline int read() { ,f=; ch=getchar(); ') {if (ch=='-') f=-f; ch=getchar();} +ch-',ch=getchar(); return x*f; } inline char readch() { while (ch<'A'||ch>'Z') ch=getchar(); return ch; } inline int get(int x) { return fa[x]==x?x:fa[x]=get(fa[x]); } int main() { srand(),cas=; while (scanf("%d%d",&n,&m)!=EOF&&n) { ans=cnt=q=,memset(vis,,sizeof vis); ,x; i<=n; i++) w[i]=read(); ; i<=m; i++) a[i].x=read(),a[i].y=read(); for ( ; ; ) { o[++q].t=readch(); if (o[q].t=='E') break; o[q].x=read(); if (o[q].t!='D') o[q].k=read(); } ,v; i<q; i++) { ; if (o[i].t=='C') swap(w[o[i].x],o[i].k); } ; i<=n; i++) { fa[i]=i; ) t.clear(root[i]); } ,x,y; i<=m; i++) if (!vis[i]) { x=get(a[i].x),y=get(a[i].y); if (x==y) continue; else fa[x]=y; } ; i<=n; i++) t.insert(root[get(i)],w[i]); ; i--) { if (o[i].t=='D') { x=get(a[o[i].x].x),y=get(a[o[i].x].y); if (x==y) continue; if (root[x]->s>root[y]->s) fa[y]=x,t.merge(root[x],root[y]); else fa[x]=y,t.merge(root[y],root[x]); }else if (o[i].t=='Q'){ cnt++,ans+=(LL)t.kth(root[get(o[i].x)],o[i].k); } else if (o[i].t=='C') { t.remove(root[x=get(o[i].x)],w[o[i].x]); t.insert(root[x],o[i].k); w[o[i].x]=o[i].k; } } printf("Case %d: %.6lf\n",++cas,1.0*ans/cnt); } ; }