BZOJ2322 [BeiJing2011]梦想封印 【set + 线性基】

题目链接

BZOJ2322

题解

鉴于BZOJ2115,要完成此题,就简单得多了
对图做一遍\(dfs\),形成\(dfs\)树,从根到每个点的路径形成一个权值,而每个返祖边形成一个环
我们从根出发去走一个环再回到根,最终会异或上环的权值而又回到根
所以环是可以任意选的
我们把环的权值丢进线性基,记线性基中有\(tot\)各位置,那么环的权值异或和方案数就是\(2^{tot}\)
我们将剩余路径权值放入线性基中消元后去重,剩余的路径权值是和线性基中的权值线性无关的
但路径只能选一个,记有\(x\)个线性无关的非\(0\)路径
除去\(0\),那么答案就是

\[(x + 1)2^{tot} - 1 \]

但这题动态删边,常规操作,离线改为加边
加边后有\(3\)种情况:

  1. 两端点都没访问过,那么不会产生任何影响
  2. 有一个端点访问过,那么现在就可以继续\(dfs\)到另一个没访问过的端点
  3. 两端点都访问过,就形成一个环,更新线性基即可

每次线性基被更新时,都要取出所有路径消一次元
每次路径加入集合前,都要放入线性基消元
集合去重用\(set\)即可

由于线性基只会被更新\(O(logw)\)次,路径只有\(O(nlogn)\)
所以复杂度为\(O(nlognlogw)\)

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 50005,maxm = 100005,INF = 0x3f3f3f3f;
inline LL read(){
	LL out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
	return flag ? out : -out;
}
const int N = 63;
set<LL> S;
set<LL>::iterator it;
LL bin[maxn],d[maxn],A[70],c[maxn],ci,ansi,ans[maxn];
int tot,flag;
int n,m,Q,pos[maxn],out[maxn],q[maxn],vis[maxn];
int h[maxn],ne = 1,fa[maxn];
struct EDGE{int to,nxt,id; LL w;}ed[maxn << 1];
inline void build(int u,int v,int i,LL w){
	ed[++ne] = (EDGE){v,h[u],i,w}; h[u] = pos[i] = ne;
	ed[++ne] = (EDGE){u,h[v],i,w}; h[v] = ne;
}
bool ins(LL x){
	for (int i = N; ~i; i--){
		if (x & bin[i]){
			if (!A[i]){A[i] = x; tot++; return true;}
			else x ^= A[i];
		}
	}
	return false;
}
LL check(LL x){
	for (int i = N; ~i; i--) if ((x & bin[i]) && A[i]) x ^= A[i];
	return x;
}
void dfs(int u){
	vis[u] = true; LL x;
	if (x = check(d[u])) S.insert(x);
	Redge(u) if (!out[ed[k].id] && (to = ed[k].to) != fa[u]){
		if (!vis[to]) fa[to] = u,d[to] = d[u] ^ ed[k].w,dfs(to);
		else if (ins(d[to] ^ d[u] ^ ed[k].w)) flag = true;
	}
}
void upd(){
	if (!flag) return;
	it = S.begin(); ci = 0;
	while (it != S.end()) c[++ci] = *it,it++;
	S.clear(); LL x;
	for (int i = 1; i <= ci; i++)
		if (x = check(c[i])) S.insert(x);
}
void rebuild(int k){
	int u = ed[pos[k] ^ 1].to,v = ed[pos[k]].to; LL w = ed[pos[k]].w;
	out[k] = false; flag = false;
	if (!vis[u] && !vis[v]) return;
	else if (vis[u]) fa[v] = u,d[v] = d[u] ^ w,dfs(v),upd();
	else if (vis[v]) fa[u] = v,d[u] = d[v] ^ w,dfs(u),upd();
	else ins(d[v] ^ d[u] ^ w),upd();
}
void print(){ans[ansi--] = 1ll * (S.size() + 1) * bin[tot] - 1;}
void work(){
	upd(); print();
	for (int i = Q; i; i--){
		rebuild(q[i]);
		print();
	}
}
int main(){
	bin[0] = 1; for (int i = 1; i <= N; i++) bin[i] = bin[i - 1] << 1ll;
	n = read(); m = read(); Q = read();
	int a,b; LL w;
	REP(i,m){
		a = read(); b = read(); w = read();
		build(a,b,i,w);
	}
	REP(i,Q) out[q[i] = read()] = true;
	ansi = Q; dfs(1);
	work();
	for (int i = 0; i <= Q; i++) printf("%lld\n",ans[i]);
	return 0;
}

posted @ 2018-07-12 10:35  Mychael  阅读(264)  评论(0编辑  收藏  举报