集训考试1

T1:
题目大意:有n个点,使它们与某个点直接或间接相连
题解:裸Kruskal

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int M=3e5+10,N=5e4+10;
typedef long long ll;
int last[N],fa[N],
n,m,len=0;ll ans=0,tot=0;
struct Edge{
    int from,to,next,val;
    Edge(int from=0,int to=0,int next=0,int val=0):from(from),to(to),next(next),val(val){}
    bool operator <(const Edge& rhs)const{
        return rhs.val>val;
    }
}e[M];
void add_edge(int u,int v,int w){e[++len]=Edge(u,v,last[u],w);last[u]=len;}
int Find(int x){return fa[x]==x?fa[x]:fa[x]=Find(fa[x]);}
void Kruskal()
{
    fo(i,1,n+1)fa[i]=i;
    sort(e+1,e+1+tot);
    for(int i=1;i<=tot;i++) {
        int idx=Find(e[i].from),idy=Find(e[i].to);
        if(idx==idy)continue;
        ans+=e[i].val;
        fa[idx]=idy;
    } 
}
int main()
{
    freopen("connect.in","r",stdin);
    freopen("connect.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int w,i=1;i<=n;i++) {
        scanf("%d",&w);
        add_edge(i,n+1,w);tot++;
        add_edge(n+1,i,w);tot++;
    }
    for(int u,v,w,i=1;i<=m;i++) {
        scanf("%d%d%d",&u,&v,&w);
        add_edge(u,v,w);tot++;
        add_edge(v,u,w);tot++;
    }
    Kruskal();
    printf("%lld\n",ans);
    return 0;
}

T2:
求1到n的一条路径,使得路径上最大值最小,并且可以k次将一条路权值置为0
二分最大值,将大于最大值的边的边权置为1,走最短路

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=1e3+10,M=1e5+10,inf=0x7f7f7f;
int last[N],f[N][N],vis[N],
n,m,K,len=0,ans,
tot,dis[N];
struct Edge{int to,next,val;Edge(int to=0,int next=0,int val=0):to(to),next(next),val(val){}}e[M<<1];
void add_edge(int u,int v,int w){e[++len]=Edge(v,last[u],w);last[u]=len;}
bool check(int mxv)
{
   fo(i,1,n) dis[i]=inf;
   memset(vis,0,sizeof(vis));
   dis[1]=0;
   vis[1]=true;
   queue<int> q;
   q.push(1);
   while(!q.empty()){
      int cur=q.front();
      q.pop();
      vis[cur]=false;
      for(int i=last[cur];i;i=e[i].next){
          int id=e[i].to;
          int cost;cost= e[i].val<=mxv?0:1;
        if(dis[id]>cost+dis[cur]){
               dis[id]=cost+dis[cur];
          if(!vis[id]) {
               q.push(id);
               vis[id]=true;
          }
        }
      } 
   }
   if(dis[n]>K)return 0;
   else return 1;
}
void solve()
{
    int L=0,R=1e6;
    while(L<=R) {
        int mid=(L+R)>>1;
        if(check(mid)){ans=mid;R=mid-1;}
        else L=mid+1;
    }
}
int main()
{
    freopen("strike.in","r",stdin);
    freopen("strike.out","w",stdout);
    scanf("%d%d%d",&n,&m,&K);
    for(int u,v,w,i=1;i<=m;i++) {
        scanf("%d%d%d",&u,&v,&w);
        add_edge(u,v,w);add_edge(v,u,w);
    }
    solve();
    ans==0?printf("-1\n"):printf("%d\n",ans);
    return 0;
}

T3:
裸强连通分量计数