andyc_03的博客

题目描述 Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms. Each of the N (1 ≤ N ≤ 1,000) farms (co

大于等于c的建边，跑kruskal即可 #include<bits/stdc++.h> using namespace std; const int maxn=2005; int n,c,tot; int x[maxn],y[maxn],f[maxn]; struct edge { int u,v,w; }e[maxn*maxn]; bool cmp(edge aa ,edge bb) { return aa.w<bb.w; } int find(int x) { if(x.

贪心的考虑，由于一级公路比二级公路消耗大，还要求至少有k条一级的，那么我们就正好取k条一级的公路kruskal，剩下的n-k-1条就直接按照二级价格kruskal即可 注意多次排序的关键字，以及输出的顺序 #include<bits/stdc++.h> using namespace std; const int maxn=1e4+5; int n,k,m,f[maxn],ans; struct edge { int u,v,w1,w2,num; }e[maxn<<1].

由于要求两个部落之间的最小距离最大，可以每次先连接两个最近的节点，就和最小生成树的思路一样了 代码 #include<bits/stdc++.h> using namespace std; struct node { int u,v; double w; }a[1000000]; int n,f[10000],cnt,k,l; int x[10000],y[10000]; void add(int uu,int vv,double ww) {a[++cnt]..