Educational Codeforces Round 27 G. Shortest Path Problem?

对每一个可以走成环的地方记录一个可以改变的费用

想要异或的值最小那么一定是尽量跟小于等于自己的异或

只需要保存这个环跟原来已记录的费用异或出的最小费用（为0表示无变化）

代码学习了qls的(

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.5f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const int mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  100000+10;const int maxm = 1000000+10;int in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}vector<int>base;vector<pii>G[maxn];int t[maxn];bool vis[maxn];void add(int x){    int sz = base.size();    for(int i=0;i<sz;++i)    {        x = min(x,x^base[i]);    }    if(x)base.pb(x);}void dfs(int cur,int now){    t[cur] = now, vis[cur]=1;    int sz = G[cur].size();    for(int i=0;i<sz;++i)    {        pii p = G[cur][i];        int to = p.X;        int w = p.Y;        if(vis[to])add(now^w^t[to]);        else dfs(to,now^w);    }}int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL//    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);    int n,m;    sdd(n,m);    while(m--)    {        int u,v,w;        sddd(u,v,w);        G[u].pb(mp(v,w));        G[v].pb(mp(u,w));    }    dfs(1,0);    int sz = base.size();    for(int i=0;i<sz;++i)t[n] = min(t[n],t[n]^base[i]);    int ans = t[n];    ansn();    return 0;}