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根据 n n n 奇偶性可以推断答案 合法解只需要在任何一棵生成树上构造即可
贪心肯定要在最大生成树上
然后从前往后看一条未选的边能不能选即可
#includebits/stdc.h
using namespace std;
#ifdef LOCAL#define …47.92.197.167:5283/contest/425/problem/3
根据 n n n 奇偶性可以推断答案 合法解只需要在任何一棵生成树上构造即可
贪心肯定要在最大生成树上
然后从前往后看一条未选的边能不能选即可
#includebits/stdc.h
using namespace std;
#ifdef LOCAL#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else#define debug(...) void(0)
#endif
#define int long long
inline int read(){int x0,f1;char chgetchar(); while(ch0||
ch9){if(ch-)f-1;chgetchar();}while(ch0ch9){
x(x1)(x3)(ch^48);chgetchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
#define fi first
#define se second
//srand(time(0));
#define N 900010
//#define M
//#define mo
int n, m, i, j, k;
int c[N], ans[N], u, v, mn[N], f[N], cho[N];
vectorpairint, int G[N], T; int fa(int x) {if(f[x]x) return x; return f[x]fa(f[x]);
}void dfs1(int x, int fa, int id) {for(auto t : G[x]) {int yt.fi, it.se; if(yfa) continue; debug(%d - %d (%d)\n, x, y, i); dfs1(y, x, i); }if(fa c[x]%20) c[x], c[fa], ans[id]1;
}void dfs2(int x, int fa) {for(auto t : G[x]) {int yt.fi, idt.se; if(yfa) continue; mn[y]id; dfs2(y, x); mn[x]min(mn[x], mn[y]); }
}int dfs3(int x, int fa, int id) {cho[x]1e9; if(id!1e9 ans[id]0) cho[x]id; if(idmn[x]) {if(ans[id]1) return 1e9; else cho[x]id; }for(auto t : G[x]) {int yt.fi, it.se; if(yfa) continue; dfs3(y, x, i); debug(# %lld - %lld\n, x, y); cho[x]min(cho[x], cho[y]);
// if(k!1e9) break;}if(idcho[x] ans[id]1) return cho[x]1e9;
// debug(%lld : %lld | %d || %d\n, x, flg, id, mn[x]);
// if(flg idm) ans[id]^1; return cho[x];
}void dfs4(int x, int fa, int id) {if(cho[x]1e9) return ; sort(G[x].begin(), G[x].end(), [] (pairint, intx, pairint, inty) { return cho[x.fi]cho[y.fi]; }); for(auto t : G[x]) {int yt.fi, it.se; if(yfa) continue; dfs4(y, x, i); break; }if(cho[x]!1e9 fa) ans[id]^1;
}signed main()
{
// freopen(lilac.in, r, stdin);
// freopen(lilac.out, w, stdout);#ifdef LOCALfreopen(in.txt, r, stdin);freopen(out.txt, w, stdout);#endif
// Tread();
// while(T--) {
//
// }nread(); mread(); for(i1; im; i) {uread()1; vread()1; debug(%d %d\n, u, v); T.pb({u, v}); }for(i1; in; i) f[i]i; for(im-1; i0; --i) {auto tT[i]; ut.fi; vt.se; if(fa(u)fa(v)) { ans[i1]1; c[u]; c[v]; continue; }f[fa(u)]fa(v); G[u].pb({v, i1}); G[v].pb({u, i1}); }for(i1; im; i) debug(%d, ans[i]); debug(\n); for(i1; in; i) debug(%d , c[i]); debug(\n); dfs1(1, 0, 0); if(n%20) { for(i1; im; i) printf(%d, ans[i]); return 0; }mn[1]1e9; dfs2(1, 0); for(i1; in; i) debug(%d , mn[i]); debug(\n); for(i1; im; i) debug(%d, ans[i]); debug(\n); dfs3(1, 0, 1e9); for(i1; im; i) debug(%d , cho[i]); debug(\n); dfs4(1, 0, 1e9); for(i1; im; i) printf(%d, ans[i]); return 0;
}
