考場思路:
二分+最小生成樹驗證,明顯添的邊越多分成的組就越少,滿足單調性
考場zz的把數組開了1000,直接卡到20pts
還有一種思路:
最小生成樹
發現在不添邊的時候,圖中有n個獨立集合
每成功加進去一條邊,集合個數 -1;
這樣在成功添n - k條邊時,集合個數爲k;
則成功添進去(剛好破壞k個集合)的第n - k + 1條邊即爲所求答案
大概代碼比較簡單這裏就不上了
二分代碼↓
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXN = 1000000 + 50;
struct edge{
int f,t;
double v;
}l[MAXN];
bool cmp(edge a,edge b){
return a.v < b.v;
}
int n,k;
struct zt{
double x,y;
}node[MAXN];
int tot,cnt;
double lin[MAXN];
int L,R,mid;
int fa[MAXN],rank[MAXN];
bool used[MAXN];
void init(int n){
for(int i = 1;i <= n;i ++){
fa[i] = i;used[i] = 0;rank[i] = 0;
}
}
int find(int x){
return fa[x] == x ? x : fa[x] = find(fa[x]);
}
void merge(int a,int b){
a = find(a);
b = find(b);
if(rank[a] < rank[b])swap(a,b);
fa[b] = a;
if(rank[a] == rank[b])rank[a] ++;
}
bool same(int a,int b){
return find(a) == find(b);
}
bool C(int x){
init(n);
cnt = 0;
for(int i = 1;i < x;i ++){
if(!same(l[i].f,l[i].t)){
merge(l[i].f,l[i].t);
}
}
for(int i = 1;i <= n;i ++){
int t = find(i);
if(!used[t]){
used[t] = true;
cnt ++;
}
}
if(cnt >= k)return true;
else return false;
}
int main(){
freopen("ship.in","r",stdin);
freopen("ship.out","w",stdout);
scanf("%d%d",&n,&k);
for(int i = 1;i <= n;i ++){
scanf("%lf%lf",&node[i].x,&node[i].y);
}
for(int i = 1;i <= n;i ++){
for(int j = 1;j < i;j ++){
double v = sqrt((node[i].x - node[j].x)*(node[i].x - node[j].x)
+ (node[i].y - node[j].y)*(node[i].y - node[j].y));
l[++tot] = (edge){i,j,v};
lin[tot] = v;
}
}
sort(l + 1,l + tot + 1,cmp);
//sort(lin + 1,lin + tot + 1);
L = -1,R = tot + 1;
while(R - L > 1){
mid = L + R >> 1;
if(C(mid)){
L = mid;
// for(int i = 1;i <= n;i ++)printf("%d ",find(i));printf("cnt=%d\n",cnt);
}
else R = mid;
}
printf("%.2lf",l[L].v);
fclose(stdin);
fclose(stdout);
return 0;
}