[CodeChef - GERALD07 ] Chef and Graph Queries

Read problems statements in Mandarin Chineseand Russian.

Problem Statement

Chef has a undirected graph G. This graph consists of N vertices and M edges. Each vertex of the graph has an unique index from 1 to N, also each edge of the graph has an unique index from 1 to M.

Also Chef has Q pairs of integers: L_i, R_i (1 ≤ L_i ≤ R_i ≤ M). For each pair L_i, R_i, Chef wants to know: how many connected components will contain graph G if Chef erase all the edges from the graph, except the edges with indies X, where L_i ≤ X ≤ R_i. Please, help Chef with these queries.

Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains three integers N, M, Q. Each of the next M lines contains a pair of integers V_i, U_i - the current edge of graph G. Each of the next Q lines contains a pair of integers L_i, R_i - the current query.

Output

For each query of each test case print the required number of connected components.

Constraints

1 ≤ T ≤ 1000.
1 ≤ N, M, Q ≤ 200000.
1 ≤ U_i, V_i ≤ N.
1 ≤ L_i ≤ R_i ≤ M.
Sum of all values of N for test cases is not greater than 200000. Sum of all values of M for test cases is not greater than 200000. Sum of all values of Q for test cases is not greater than 200000.
Graph G can contain self-loops and multiple edges.

Example

破题坑了一晚上

题目大意？见上面超链接的中文题面。

树 LCT 动态维护生成树

将所有询问按右端点递增顺序排序，从1到m依次加边。

用LCT维护当前的生成树，生成树中保留的是编号尽量大的边。

同时用线段树维护生成树中保留了哪些边（权值线段树，存编号）。

查询时，在线段树上查询$[L,R]$范围内有多少条边，若查询结果为$ x $，则连通块数为$n-x$

注意有自环和重边，这使得维护生成树的时候不能一律删编号最靠前的边，而要先查询生成树中有没有重边。

细节蛮多的。

注意多组数据，初始化结点的时候要初始化到n+m

 #include<iostream>

 #include<algorithm>

 #include<cstring>

 #include<cstdio>

 #include<cmath>

 #define LL long long

 using namespace std;

 const int INF=0x3f3f3f3f;

 const int mxn=;

 int read(){

     int x=,f=;char ch=getchar();

     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}

     return x*f;

 }

 struct edge{

     int x,y;

 }e[mxn];

 struct query{

     int L,R;

     int id;

     bool operator < (const query &b)const{

         return R<b.R;

     }

 }q[mxn];

 int ans[mxn];

 //

 struct SGT{

     int smm[mxn<<];

     #define ls rt<<1

     #define rs rt<<1|1

     void clear(int ed){

         ed=ed<<;

         for(register int i=;i<=ed;i++)smm[i]=;

     }

     void pushup(int rt){smm[rt]=smm[ls]+smm[rs];return;}

     void update(int p,int v,int l,int r,int rt){

         if(l==r){

             smm[rt]+=v;

             return;

         }

         int mid=(l+r)>>;

         if(p<=mid)update(p,v,l,mid,ls);

         else update(p,v,mid+,r,rs);

         pushup(rt);

         return;

     }

     int query(int L,int R,int l,int r,int rt){

         if(L<=l && r<=R){

             return smm[rt];

         }

         int mid=(l+r)>>;

         if(R<=mid)return query(L,R,l,mid,ls);

         if(L>mid)return query(L,R,mid+,r,rs);

         return query(L,R,l,mid,ls)+query(L,R,mid+,r,rs);

     }

     #undef ls

     #undef rs

 }SG;

 int n,m,Q;

 //

 struct node{

     int ch[],fa;

     int val,mn,minpos;

     bool rev;

 }t[mxn];

 int cnt=;

 void rever(int x){

     int &lc=t[x].ch[],&rc=t[x].ch[];

     swap(lc,rc);

     t[lc].rev^=;t[rc].rev^=;

     return;

 }

 void PD(int x){

     if(t[x].rev){

         rever(x);

         t[x].rev=;

     }

     return;

 }

 void pushup(int x){

     int lc=t[x].ch[],rc=t[x].ch[];

     t[x].mn=t[x].val;t[x].minpos=x;

     if(t[lc].mn<t[x].mn){

         t[x].mn=t[lc].mn; t[x].minpos=t[lc].minpos;

     }

     if(t[rc].mn<t[x].mn){

         t[x].mn=t[rc].mn; t[x].minpos=t[rc].minpos;

     }

     return;

 }

 inline bool isroot(int x){

     return (t[t[x].fa].ch[]^x)&&(t[t[x].fa].ch[]^x);

 }

 void rotate(int x){

     int y=t[x].fa,z=t[y].fa,lc,rc;

     if(t[y].ch[]==x)lc=;else lc=;rc=lc^;

     if(!isroot(y)){

         t[z].ch[t[z].ch[]==y]=x;

     }

     t[x].fa=z;t[y].fa=x;

     t[t[x].ch[rc]].fa=y;

     t[y].ch[lc]=t[x].ch[rc];

     t[x].ch[rc]=y;

     pushup(y);

     return;

 }

 int st[mxn],top=;

 void Splay(int x){

     st[top=]=x;

     for(int i=x;t[i].fa;i=t[i].fa){

         st[++top]=t[i].fa;

     }

     while(top)PD(st[top--]);

     while(!isroot(x)){

         int y=t[x].fa,z=t[y].fa;

         if(!isroot(y)){

             if((t[y].ch[]==x)^(t[z].ch[]==y))rotate(x);

                 else rotate(y);

         }

         rotate(x);

     }

     pushup(x);

     return;

 }

 //

 void access(int x){

     for(int y=;x;x=t[x].fa){

         Splay(x);

         t[x].ch[]=y;

         pushup(x);

         y=x;

     }

     return;

 }

 int find(int x){

     access(x);Splay(x);

     while(t[x].ch[]){

         x=t[x].ch[];

     }

     return x;

 }

 void mkroot(int x){

     access(x);Splay(x);

     t[x].rev^=;

     return;

 }

 void link(int x,int y){

     mkroot(x); t[x].fa=y;

     return;

 }

 void cut(int x,int y){

     mkroot(x);

     access(y);Splay(y);

     if(t[y].ch[]==x){t[x].fa=;t[y].ch[]=;}

     pushup(y);

     return;

 }

 //

 void solve(){

     int hd=;

     cnt=n;

     for(int i=;i<=m;i++){//按序加边

         int x=e[i].x,y=e[i].y;

         if(x!=y && find(x)==find(y)){

             bool flag=;

             mkroot(x);access(y);Splay(y);

             //

 /*            printf("-----\n");

             for(int j=1;j<=cnt;j++){

                 printf("#%d: lc:%d rc:%d fa:%d\n",j,t[j].ch[0],t[j].ch[1],t[j].fa);

             }

             printf("-----\n");*/

             //

             int res=t[y].ch[];

             while(t[res].ch[])res=t[res].ch[];

             res=res-n;

             if(res> && e[res].x==x && e[res].y==y){

                 flag=;

                 cut(x,res+n);

                 cut(y,res+n);

                 SG.update(res,-,,m,);

             }

             //

             if(!flag){

                 mkroot(x);access(y);Splay(y);

                 int tar=t[y].minpos;

                 cut(e[tar-n].x,tar);

                 cut(e[tar-n].y,tar);

                 SG.update(tar-n,-,,m,);

             }

         }

         ++cnt;

         if(x!=y){

             t[cnt].val=t[cnt].mn=i;

             t[cnt].minpos=cnt;

             link(x,cnt);

             link(cnt,y);

             SG.update(i,,,m,);

         }

         while(hd<=Q && q[hd].R<=i){

             if(q[hd].R==i){

                 ans[q[hd].id]=n-SG.query(q[hd].L,q[hd].R,,m,);

             }

             hd++;

         }

     }

     return;

 }

 void init(int n,int m){

     int ed=n+m;

     t[].minpos=;t[].mn=INF;

     for(register int x=;x<=ed;x++){

         t[x].ch[]=t[x].ch[]=t[x].fa=;

         t[x].minpos=;t[x].mn=INF;

         t[x].val=INF;

         t[x].rev=;

     }

     SG.clear(ed);

     return;

 }

 int main(){

 //    freopen("in.txt","r",stdin);

     int i;

     int T=read();

     while(T--){

         n=read();m=read();Q=read();

         init(n,m);

         for(i=;i<=m;i++){

             e[i].x=read();e[i].y=read();

             if(e[i].x>e[i].y)swap(e[i].x,e[i].y);

         }

         for(i=;i<=Q;i++){

             q[i].L=read();q[i].R=read();

             q[i].id=i;

         }

         sort(q+,q+Q+);

         solve();

         for(i=;i<=Q;i++){

             printf("%d\n",ans[i]);

         }

     }

     return ;

 }