BZOJ 1196 公路修建问题(二分+最小生成树)

题目要求求出图中的一颗生成树，使得最大的边权最小，且满足一级公路的个数>=k。

考虑二分最大边，问题就变为给出的图的生成树中，是否满足所有的边<=val,且一级公路的个数>=k。

所以我们把边按一级公路权值排序，优先选择能构成生成树的一级公路。这样贪心的构造。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=10005;
//Code begin...

struct Edge{int u, v, w1, w2;}edge[N<<1];
int n, m, k, fa[N];
int find(int x)
{
    int s, temp;
    for (s=x; fa[s]>=0; s=fa[s]) ;
    while (s!=x) temp=fa[x], fa[x]=s, x=temp;
    return s;
}
void union_set(int x, int y)
{
    int temp=fa[x]+fa[y];
    if (fa[x]>fa[y]) fa[x]=y, fa[y]=temp;
    else fa[y]=x, fa[x]=temp;
}
bool check(int x)
{
    int num=0;
    mem(fa,-1);
    FO(i,1,m) {
        int u=find(edge[i].u), v=find(edge[i].v);
        if (edge[i].w2>x||u==v) continue;
        if (edge[i].w1<=x) ++num;
        union_set(u,v);
    }
    return num>=k&&fa[find(1)]==-n;
}
bool comp(Edge a, Edge b){return a.w1<b.w1;}
int main ()
{
    int u, v, w1, w2;
    n=Scan(); k=Scan(); m=Scan();
    FO(i,1,m) edge[i].u=Scan(), edge[i].v=Scan(), edge[i].w1=Scan(), edge[i].w2=Scan();
    sort(edge+1,edge+m,comp);
    int l=0, r=30001, mid;
    while (l<r) {
        mid=(l+r)>>1;
        if (check(mid)) r=mid;
        else l=mid+1;
    }
    printf("%d\n",r);
    return 0;
}