bzoj 2115 Xor
题目大意:
求一条从$1 \rightarrow n$的路径是异或和最大
思路:
先随便求一棵生成树,然后求出所有环,对于所有环都可以去转一圈只取到这个环的贡献
那么就是线性基裸题了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #include<queue> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #define ll long long 12 #define db double 13 #define inf 2139062143 14 #define MAXN 50100 15 #define MOD 998244353 16 #define maxi 123456789123456789LL 17 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i) 18 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i) 19 #define ren for(register int i=fst[x];i;i=nxt[i]) 20 #define pb(i,x) vec[i].push_back(x) 21 #define pls(a,b) (a+b)%MOD 22 #define mns(a,b) (a-b+MOD)%MOD 23 #define mul(a,b) (1LL*(a)*(b))%MOD 24 using namespace std; 25 inline ll read() 26 { 27 ll x=0,f=1;char ch=getchar(); 28 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 29 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 30 return x*f; 31 } 32 ll p[70],dis[MAXN],val[MAXN<<2]; 33 int n,m,fst[MAXN],to[MAXN<<2],nxt[MAXN<<2],cnt,vis[MAXN]; 34 void add(int u,int v,ll w) {nxt[++cnt]=fst[u],fst[u]=cnt,to[cnt]=v,val[cnt]=w;} 35 void ins(ll x) {dwn(i,63,0) if((x>>i)&1){if(!p[i]){p[i]=x;break;}x^=p[i];};} 36 ll query(ll x) {ll res=x;dwn(i,63,0) res=max(res,res^p[i]);return res;} 37 void dfs(int x) 38 { 39 vis[x]=1;ren if(!vis[to[i]]) dis[to[i]]=dis[x]^val[i],dfs(to[i]); 40 else ins(dis[x]^dis[to[i]]^val[i]); 41 } 42 int main() 43 { 44 n=read(),m=read();int a,b;ll c; 45 rep(i,1,m) a=read(),b=read(),c=read(),add(a,b,c),add(b,a,c); 46 dfs(1);printf("%lld\n",query(dis[n])); 47 }