BNUOJ 52318 Be Friends prim+Trie

题目链接：

https://acm.bnu.edu.cn/v3/problem_show.php?pid=52318

B. Be Friends

Case Time Limit: 2500ms
Memory Limit: 524288KB

题意

给你n个点，每个点有一个权值a[i]，对于两个点u,v,他们之间的边的权值为a[u]^a[v]，现在让你求一颗最小生成树。

题解

Trie可以求离点u最近的点v（既u^v最小）,利用这一点，我们用prim来求最小生成树, 可以用优先队列维护一下离我们已经扩展的集合的最近的点，最近的点是可以用Trie处理出来的。

代码

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxnode=2e6+10;
const int maxn=1e5+10;

struct Node {
int u,v;
bool operator < (const Node& tmp) const {
return (u^v)>(tmp.utmp.v);
}
Node(int u,int v):u(u),v(v) {}
};

///ch[0]为超级节点，不止是代表第一个点，很多点会连到上面，所以它的cntv必须为0，代表着终结。
struct Trie {
int ch[maxnode][2];
//cntv统计子树下的单词节点个数，val记录单词节点
int cntv[maxnode],val[maxnode];
int sz,cnt;
Trie() {
sz=1;
clr(ch[0],0);
cnt=0;
}

void insert(int x) {
cnt++;
int arr[44]= {0},tot=0,tmp=x;
while(tmp) {
arr[++tot]=tmp&1;
tmp>>=1;
}

int u=0;
for(int i=33; i>=1; i--) {
int c=arr[i];
if(!ch[u][c]) {
clr(ch[sz],0);
cntv[sz]=0;
val[sz]=-10086;
ch[u][c]=sz++;
}
u=ch[u][c];
cntv[u]++;
}
val[u]=x;
}

int query(int x) {
if(cnt==0) return -1;

int arr[44]= {0},tot=0,tmp=x;
while(tmp) {
arr[++tot]=tmp&1;
tmp>>=1;
}

int u=0;
for(int i=33; i>=1; i--) {
int c=arr[i];
if(cntv[ch[u][c]]) {
u=ch[u][c];
} else {
u=ch[u][c^1];
}
}
return val[u];
}

void del(int x) {
cnt--;
int arr[44]= {0},tot=0,tmp=x;
while(tmp) {
arr[++tot]=tmp&1;
tmp>>=1;
}

int u=0;
for(int i=33; i>=1; i--) {
int c=arr[i];
cntv[ch[u][c]]--;
u=ch[u][c];
}
}

} trie;

int arr[maxn],n;
map<int,bool> mp;

int main() {
scf("%d",&n);
for(int i=1; i<=n; i++) scf("%d",&arr[i]);
sort(arr+1,arr+n+1);
n=unique(arr+1,arr+n+1)-arr-1;

for(int i=2; i<=n; i++) {
trie.insert(arr[i]);
}

priority_queue pq;

if(n==1) {
prf("0\n");
return 0;
}

///prim求最小生成树
int tmp=trie.query(arr[1]);
pq.push(Node(arr[1],tmp));
mp[arr[1]]=true;

LL ans=0;
for(int i=2; i<=n; i++) {

while(mp[pq.top().v]){
int u=pq.top().u;
pq.pop();
int v=trie.query(u);
if(v>=0){
// prf("f(%d,%d)\n",u,v);
pq.push(Node(u,v));
}
}

int u=pq.top().u,v=pq.top().v;
pq.pop();
trie.del(v);
mp[v]=true;
ans+=u^v;

int u2=trie.query(u);
if(u2>=0){
pq.push(Node(u,u2));
}

int v2=trie.query(v);
if(v2>=0){
pq.push(Node(v,v2));
}
}

prf("%lld\n",ans);

return 0;
}