G. MinOr Tree
作者:互联网
题目链接
G. MinOr Tree
求最小或生成树
解题思路
贪心,并查集
按位考虑,从高位开始,用并查集合并当前边该位为 \(0\)的点,再判断所有点是否组成一棵生成树,如果可以说明该位可以为 \(0\),再删除那些该位为 \(1\) 的边,否则该位必须为 \(1\),删除该位为 \(0\) 的边
- 时间复杂度:\(O(t\times (n+m)\times logn)\)
代码
// Problem: G. MinOr Tree
// Contest: Codeforces - Codeforces Round #764 (Div. 3)
// URL: https://codeforces.com/contest/1624/problem/G
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
// %%%Skyqwq
#include <bits/stdc++.h>
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
template <typename T> void inline read(T &x) {
int f = 1; x = 0; char s = getchar();
while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
x *= f;
}
const int N=2e5+5;
struct T
{
int x,y,w;
}tr[N];
int n,t,m,fa[N],v[N];
int get(int x)
{
return fa[x]==x?x:fa[x]=get(fa[x]);
}
int main()
{
for(cin>>t;t;t--)
{
memset(v,0,sizeof v);
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)fa[i]=i;
for(int i=1;i<=m;i++)scanf("%d%d%d",&tr[i].x,&tr[i].y,&tr[i].w);
int res=0;
for(int i=30;~i;i--)
{
for(int j=1;j<=n;j++)fa[j]=j;
for(int j=1;j<=m;j++)
{
if(v[j]||(tr[j].w>>i&1))continue;
fa[get(tr[j].x)]=get(tr[j].y);
}
int f=0;
for(int j=1;j<=n;j++)f+=(j==fa[j]);
if(f==1)
{
for(int j=1;j<=m;j++)
if(tr[j].w>>i&1)v[j]=true;
}
else
{
for(int j=1;j<=m;j++)
if(tr[j].w>>i&1==0)v[j]=true;
res|=1<<i;
}
}
printf("%d\n",res);
}
return 0;
}
标签:typedef,get,int,Tree,long,fa,MinOr,define 来源: https://www.cnblogs.com/zyyun/p/15965248.html