正题

P2323

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题目大意

给你n个点和若干边，每条边有两种代价，问你选择n-1条边使得这n个点连通，且选择第一种代价的边不小于k，让你使代价最大值最小

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解题思路

二分答案，然后最小生成树，连接可以连得边，判断是否可行

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code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 20021
using namespace std;
int n,k,m,l,r,xx,yy,kn,nn,p[N],x[N],y[N],c1[N],c2[N],fa[10010];
int find(int x)
{return x==fa[x]?x:fa[x]=find(fa[x]);
}
bool solve(int num)
{memset(p,0,sizeof(p));nn=kn=0;for(int i=1;i<=n;++i)fa[i]=i;for(int i=1;i<=m;++i)//优先选第一种if(c1[i]<=num){xx=find(x[i]);yy=find(y[i]);if(xx!=yy){p[i]=1; fa[xx]=yy;nn++;kn++;}}if(kn<k)return false;if(nn>=n-1)return true;for(int i=1;i<=m;++i)if(c2[i]<=num&&num<c1[i]){xx=find(x[i]);yy=find(y[i]);if(xx!=yy){p[i]=2;fa[xx]=yy;nn++;}}if(nn>=n-1)return true;return false;
}
int main()
{scanf("%d%d%d",&n,&k,&m);m--;for(int i=1;i<=m;++i)scanf("%d%d%d%d",&x[i],&y[i],&c1[i],&c2[i]);l=0;r=30000;while(l<r){int mid=l+r>>1;if(solve(mid))r=mid;else l=mid+1;}printf("%d\n",l);solve(l);for(int i=1;i<=m;++i)if(p[i])printf("%d %d\n",i,p[i]);return 0;
}