【20200212】【lyk】CF1299D Around the World

最新推荐文章于 2022-03-16 15:42:44 发布

weixin_38926546

最新推荐文章于 2022-03-16 15:42:44 发布

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本文链接：https://blog.youkuaiyun.com/weixin_38926546/article/details/104288481

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题意

给你一个图，所有过1号点的简单环大小不超过3，可以任意删去与1号点相连的边，求有多少种删法使任意通过一号点的平凡环边权异或和不为0（边权小于32）.平凡环：从某个点出发，经过任意多点最终回到起始点，过程中边和点都可以重复经过，但是必须有某条边只经过了奇数次。

思路

首先可以看出大概是以1号点为根分成若干枝，每个枝有1个或两个相邻点与1有边。每次从1出发走任意多个分支回到1，但是每个分支只能由一个简单环产生作用。到此可以确定思路为dp。考虑状态：首先可以dfs每个分支，只需考虑返祖边代表的环（任意一个简单环一定可以由若干返祖边代表的环的异或和得到）。于是可以得到每个分支所能贡献的异或和的种类（可以理解为返祖边所构成的向量空间），有两个与1相连的点的分支有2种状态，其他1种。考虑5维向量空间的子空间只有374种（一开始以为是 $2^{32}$ ,思考了一下大概 $32^4$ ,后来我跑了一下按照线性基状压，发现要大概3W个状态，看了题解发现如果按照向量空间种类计算只有374个不同状态），可以提前预处理转移，预处理用map的话是O( $1024S^2logS$ )。总复杂度O( $SN+1024S^2logS$ )

#include <bits/stdc++.h>

using namespace std;
struct Edge {
	int to,next,w;
}bus[303030];
int tic;
int n,m,vis[101010],dis[101010],st[101010],z = 1,aa[101010],bb[101010],cc[101010],ppp[101010],dfn[101010];
int root[101010],p[101010],t[101010];
vector<int> ex[101010];
bool ban[101010];
long long status[1010],f[2][404] = {0};
int nxt[1010][32] = {0},mo[505][505];
int len = 0,tail = 0;
map<long long,int> M;
const int mod = 1000000007;
inline void dfs(int now,int last) {
	int to;
	for (int i = st[now];i;i = bus[i].next) {
		to = bus[i].to;
		if (to == 1 || i == (last^1))
			continue;
		if (vis[to] == tic) { 
			if (dfn[to] < dfn[now]) {
	 			if ((dis[now]^dis[to]^bus[i].w) == 0)
					ban[tic] = 1;
				ex[tic].push_back(dis[now]^dis[to]^bus[i].w);
			}
		}
		else {
			vis[to] = tic;
			dfn[bus[i].to] = dfn[now]+1;
			dis[bus[i].to] = (dis[now]^bus[i].w);
			dfs(bus[i].to,i);
		}
			
	}
}
inline void AddE(int from,int to,int w) {
	bus[++z].to = to;
	bus[z].w = w;
	bus[z].next = st[from];
	st[from] = z;
}
inline void init() {
	memset(nxt,-1,sizeof(nxt));
	for (int i = 1;i < 32; i++) {
		status[++tail] = ((1LL<<i)|1);
		M[status[tail]] = tail;
	}
	for (int i = 1;i <= 4; i++) {
		len = tail;
		for (int j = 1;j <= len; j++) {
			for (int k = 1;k < 32; k++) {
				if (!((1LL<<k)&status[j])) {
					long long tmp = status[j];
					for (int ii = 0;ii < 32; ii++) {
						if ((1LL<<ii)&status[j]) {
							tmp |= (1LL<<(ii^k));
						}
					}
					if (!M[tmp]) {
						M[tmp] = ++tail;
						status[tail] = tmp;
					}
					nxt[j][k] = M[tmp];
				}
			}
		}
	}
}
inline int calc(int A,int B) {
	long long tmp = 0;
	for (int i = 0;i < 32; i++)
		if ((1LL<<i)&status[A]) {
			for (int j = 0;j < 32; j++)
				if ((1LL<<j)&status[B]) {
					if (tmp&(1LL<<(i^j)))
						return -1;
					tmp |= (1LL<<(i^j));
				}	
		}
	return M[tmp];
}
int main() {
	init();
	status[374] = 1;
	tail = 374;
	M[1] = 374;
	scanf("%d%d",&n,&m);
	for (int i = 1;i <= m; i++) {
		scanf("%d%d%d",&aa[i],&bb[i],&cc[i]);
		if (aa[i] == 1) {
			root[bb[i]] = 1;
			dis[bb[i]] = cc[i];
		}
		if (bb[i] == 1)
			root[aa[i]] = 1,dis[aa[i]] = cc[i];
	}
	for (int i = 1;i <= m; i++) {
		if (root[aa[i]] && root[bb[i]]) {
			p[aa[i]] = bb[i];
			p[bb[i]] = aa[i];
			ppp[aa[i]] = ppp[bb[i]] = (cc[i]^dis[aa[i]]^dis[bb[i]]);
		}
		AddE(aa[i],bb[i],cc[i]);
		AddE(bb[i],aa[i],cc[i]);
	}
	for (int i = 2;i <= n; i++)
		if (root[i]) {
			tic = i; 
			vis[i] = i;
			dfn[i] = 0;
			dfs(i,-10);
			t[i] = 374;
			if (ex[i].size()) {
				for (int j = ex[i].size()-1;~j; j--) {
					if (!ex[i][j]){
						ban[i] = 1;
						break;
					}
					if (!mo[ex[i][j]][t[i]])
						mo[ex[i][j]][t[i]] = mo[t[i]][ex[i][j]] = calc(t[i],ex[i][j]);
					if (mo[ex[i][j]][t[i]] < 0) {
						ban[i] = 1;
						break; 
					}
					t[i] = mo[ex[i][j]][t[i]];
				}
			}
		}
	int I = 0;
	f[I][374] = 1;
	for (int i = 2;i <= n; i++) {
		if (ban[i])
			continue;
		if (!root[i])
			continue;
		ban[i] = ban[p[i]] = 1;
		int pp = t[p[i]],pppp = ppp[i];
		if (!mo[t[i]][pppp])
			mo[t[i]][pppp] = mo[pppp][t[i]] = calc(pppp,t[i]);
		pppp = mo[pppp][t[i]];
		if (ppp[i] == 0)
			pppp = -1;
		for (int j = 0;j <= 400; j++)
			f[I^1][j] = 0;
		for (int j = 1;j <= 374; j++) {
			f[I][j] %= mod;
			f[I^1][j] += f[I][j];
			if (!mo[j][t[i]])
				mo[j][t[i]] = mo[t[i]][j] = calc(j,t[i]);
			if (mo[j][t[i]] > 0) {
				f[I^1][mo[j][t[i]]] += f[I][j];
			}
			if (p[i]) {
				if (!mo[j][pp])
					mo[j][pp] = mo[pp][j] = calc(j,pp);
				if (mo[j][pp] > 0)
					f[I^1][mo[j][pp]] += f[I][j];
				if (pppp > 0) {
					if (!mo[j][pppp])
						mo[j][pppp] = mo[pppp][j] = calc(j,pppp);
					if (mo[j][pppp] > 0)
						f[I^1][mo[j][pppp]] += f[I][j];
				}
			}
		}
		I ^= 1;
	}
	for (int j = 2;j <= 374; j++)
		f[I][1] += f[I][j];
	f[I][1] %= mod;
	printf("%lld",f[I][1]); 
	return 0;
}
/*
6 8
1 2 0
2 3 1
2 4 3
2 6 2
3 4 8
3 5 4
5 4 5
5 6 6
*/