HDU6251

题解: 基环树  首先我们考虑只有树的情况下的 联通块为sigma(f[i])-n+1 f[i]表示这个节点相连的不同颜色边的数目 对于基环树 我们只要考虑加入的这条边与其两个节点之间的边颜色的关系即可 然后大力分类讨论

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=2e5+10;
const double eps=1e-8;
#define ll long long
using namespace std;
struct edge{int t,v;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y,int vul){o->t=y;o->v=vul;o->next=h[x];h[x]=o++;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}

typedef struct node{
    int u,v,key;
}node;
node ed[MAXN];
int f[MAXN][21],a[MAXN],num[MAXN],dep[MAXN];
multiset<int>s[MAXN],ss;
multiset<int>::iterator ite,ip;
void dfs(int x,int pre,int deep){
    f[x][0]=pre;dep[x]=deep+1;
    link(x){
        if(j->t!=pre){
            a[j->t]=j->v;
            ite=s[j->t].lower_bound(j->v);
            if(ite==s[j->t].end()||(*ite)!=j->v)num[j->t]++;
            s[j->t].insert(j->v);
            ite=s[x].lower_bound(j->v);
            if(ite==s[x].end()||(*ite)!=j->v)num[x]++;
            s[x].insert(j->v);
            dfs(j->t,x,deep+1);
        }
    }
}
void dfs1(int x){
    for(int i=1;i<=20;i++)f[x][i]=f[f[x][i-1]][i-1];
    link(x){
        if(j->t!=f[x][0])dfs1(j->t);
    }
}
int Lca(int u,int v){
    if(dep[u]<dep[v])swap(u,v);
    int tmp=dep[u]-dep[v];
    for(int i=0;i<=20;i++)if(tmp&(1<<i))u=f[u][i];
    if(u==v)return u;
    for(int i=20;i>=0;i--){
        if(f[u][i]!=f[v][i])u=f[u][i],v=f[v][i];
    }
    return f[u][0];
}
bool vis[MAXN];
int fa[MAXN];
int find1(int x){
    if(fa[x]==x)return x;
    else return fa[x]=find1(fa[x]);
}
void debug(int x){
    for(ite=s[x].begin();ite!=s[x].end();ite++)cout<<(*ite)<<" ";
        cout<<endl;
}
int main(){
    int _=read();int ca=0;
    while(_--){
        memset(vis,0,sizeof(vis));memset(num,0,sizeof(num));
        memset(e,0,sizeof(e));memset(h,0,sizeof(h));o=e;
        int n,m;n=read();m=read();ss.clear();
        inc(i,1,n)ed[i].u=read(),ed[i].v=read(),ed[i].key=read(),fa[i]=i,s[i].clear();
        inc(i,1,n){
            int t1=find1(ed[i].u);int t2=find1(ed[i].v);
            if(t1==t2){swap(ed[i],ed[n]);break;}
            fa[t1]=t2;
        }
        inc(i,1,n-1)add(ed[i].u,ed[i].v,ed[i].key),add(ed[i].v,ed[i].u,ed[i].key);
        dfs(1,0,0);dfs1(1);ll ans=0;
        inc(i,1,n)ans+=num[i];
        //cout<<ans<<endl;
        ans-=(n-1);
        int lca=Lca(ed[n].u,ed[n].v);
        int u=ed[n].u;int v=ed[n].v;int key;
        while(u!=lca){ss.insert(a[u]);vis[u]=1;u=f[u][0];}
        while(v!=lca){ss.insert(a[v]);vis[v]=1;v=f[v][0];}
        printf("Case #%d:\n",++ca);
        while(m--){
            u=read();v=read();key=read();
            if(min(u,v)==min(ed[n].u,ed[n].v)&&max(v,u)==max(ed[n].v,ed[n].u))ed[n].key=key;
            else{
                int t1;bool flag=0;
                if(dep[u]>dep[v])t1=a[u],flag=vis[u],a[u]=key;else t1=a[v],flag=vis[v],a[v]=key;
                //cout<<flag<<"==="<<u<<" "<<v<<endl;
                if(flag){
                    ite=ss.lower_bound(t1);ss.erase(ite);
                    ss.insert(key);
                }
                //cout<<u<<" "<<t1<<endl;
                //debug(u);
                ite=s[u].lower_bound(t1);s[u].erase(ite);
                ite=s[u].lower_bound(t1);
                //cout<<u<<"===="<<v<<" "<<t1<<endl;
                //cout<<ans<<"::::"<<endl;
                if(ite==s[u].end()||(*ite)!=t1)ans--;
                ite=s[u].lower_bound(key);
                if(ite==s[u].end()||(*ite)!=key)ans++;
                s[u].insert(key);
                ite=s[v].lower_bound(t1);s[v].erase(ite);
                ite=s[v].lower_bound(t1);
                if(ite==s[v].end()||(*ite)!=t1)ans--;
                ite=s[v].lower_bound(key);
                if(ite==s[v].end()||(*ite)!=key)ans++;              
                s[v].insert(key);
                //cout<<ans<<"==="<<endl;
            }
            int tt=ed[n].key;
            ite=ss.begin();ip=ss.end();ip--;
            ll sum=ans;
            //cout<<(*ite)<<" "<<(*ip)<<" "<<tt<<endl;
            if((*ite)==tt&&(*ip)==tt)printf("%lld\n",sum);
            else{
                bool flag1=0,flag2=0;
                ite=s[ed[n].u].lower_bound(tt);
                if(ite==s[ed[n].u].end()||(*ite)!=tt)flag1=0;
                else flag1=1;
                ite=s[ed[n].v].lower_bound(tt);
                if(ite==s[ed[n].v].end()||(*ite)!=tt)flag2=0;
                else flag2=1;
                if(flag1&&flag2)sum--;
                else if(!flag1&&!flag2)sum++;
                printf("%lld\n",sum);       
            }
        }
    }
    return 0;
}

　　

Inkopolis

Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 414    Accepted Submission(s): 101

Problem Description

Inkopolis is the city in which Inklings live in, it can be regarded as an undirected connected graph with exactly N vertices and N edges between vertices. It is guaranteed that the graph doesn’t contain duplicated edges or self loops. Inklings can splatter a special type of colored ink to decorate the roads they live in. Inklings are capricious so they often change the color of the roads to celebrate the upcoming Splatfest.
The Splatfest lasts for exactly M days, on each day Inklings will splatter ink on exactly one road, the color on this road will be coverd by the new color of the ink. At the end of each day, they wonder how many different colored regions are there in the Inkopolis. A colored region is a set of connected roads with same color, to be clear, two roads are in the same colored region if they have the same color and share a common vertex.

 

 

Input

The first line of the input gives the number of test cases, T. T test cases follow.
For each test case, the first line contains two integers N and M, where N is the number of vertexes and roads in Inkopolis, and M is the number of days that Splatfest lasts.
Following N lines describe the roads between the vertexes. Each line contains 3 integers x, y, c, representing there is a road with color c between the xth and the ythvertex.
Next following M lines describe the operations on each day. Each line contains 3 integers x, y, c, representing an operation that Inklings splatter the ink with color c to the road between the xth and the yth vertex, it is guaranteed that there is such a road.
1≤T≤100
3≤N≤2×105
1≤M≤2×105
1≤x,y,c≤n
∑N,∑M≤106

 

 

Output

For each test case, output one line containing “Case #x:” first, where x is the test case number (starting from 1).
The following M lines each consists of an integer which is the number of colored regional in Inkopolis after each day.

 

 

Sample Input

2 4 3 4 2 4 2 3 3 3 4 2 1 4 1 3 4 2 2 3 4 3 4 3 4 4 1 2 1 2 3 1 3 4 1 4 1 1 1 2 2 3 4 2 2 3 2 4 1 4

 

 

Sample Output

Case #1: 4 3 3 Case #2: 2 4 2 2