题解 卡牌游戏/哪一天她能重回我身边
并不会做
首先是
- 一个处理形如「每个元素有两种值可以选」问题的常见转化方法：将这个元素看成连接这两个值的一条边
于是若将每张牌看成从底面数字连向顶面数字的一条边
问题就转化为要改变尽量少的边的方向使每个点的度数不超过1
于是分连通块，考虑这个连通块的贡献
若是棵树，可以树形DP
若是基环树，树的部分肯定是外向树，环上选修改次数小的方向即可
若边数大于点数一定无解
于是复杂度是 \(O(n)\) 的
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 200010
#define ll long long
#define fir first
#define sec second
#define make make_pair
#define pb push_back
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n;
int a[N], b[N], head[N], dsu[N], siz[N], ecnt[N], size;
int f[N], g[N], ring[N], top, cost;
ll ans1, ans2;
pair<int, int> e2[N];
vector<int> s[N];
bool vis[N], vis2[N], inring[N];
const ll mod=998244353;
struct edge{int to, next; bool back;}e[N<<1];
inline void add(int s, int t, bool typ) {e[++size].to=t; e[size].next=head[s]; e[size].back=typ; head[s]=size;}
inline int find(int p) {return dsu[p]==p?p:dsu[p]=find(dsu[p]);}
void dfs1(int u, int fa) {
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
dfs1(v, u);
f[u]+=f[v]+e[i].back;
}
}
void dfs2(int u, int fa) {
int tem=g[u]+f[u];
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
if (e[i].back) g[v]=tem-f[v]-1;
else g[v]=tem-f[v]+1;
dfs2(v, u);
}
}
int dfs3(int u, int in_edge) {
in_edge^=1;
vis[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) if (i!=in_edge) {
v = e[i].to;
if (vis[v]) {ring[++top]=u; inring[u]=1; cost+=e[i].back; return v==u?-1:v;}
int tem=dfs3(v, i);
if (~tem) {ring[++top]=u; inring[u]=1; cost+=e[i].back; return tem==u?-1:tem;}
}
return -1;
}
void dfs4(int u, int fa) {
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa || inring[v]) continue;
dfs4(v, u);
f[u]+=f[v]+e[i].back;
}
}
signed main()
{
freopen("card.in", "r", stdin);
freopen("card.out", "w", stdout);
int T=1;
while (T--) {
n=read(); size=1; ans1=0; ans2=1;
memset(head, -1, sizeof(head));
for (int i=1; i<=2*n; ++i) dsu[i]=i, siz[i]=1, ecnt[i]=0;
for (int i=1; i<=n; ++i) {
a[i]=read(), b[i]=read();
add(b[i], a[i], 0); add(a[i], b[i], 1);
e2[i]=make(b[i], a[i]);
int f1=find(a[i]), f2=find(b[i]);
if (f1==f2) {++ecnt[f1]; continue;}
dsu[f1]=f2; siz[f2]+=siz[f1]; ecnt[f2]+=ecnt[f1]+1;
}
// cout<<"bel: "; for (int i=1; i<=2*n; ++i) cout<<find(i)<<' '; cout<<endl;
for (int i=1; i<=2*n; ++i) s[find(i)].pb(i);
for (int i=1,bel; i<=2*n; ++i) if (!vis2[bel=find(i)] && siz[bel]>1) {
// cout<<"i: "<<i<<endl;
if (ecnt[bel]>siz[bel]) {puts("-1 -1"); goto jump;}
else if (ecnt[bel]==siz[bel]-1) {
dfs1(i, 0); dfs2(i, 0);
ll t1=INF, t2=1;
for (auto it:s[bel])
if (f[it]+g[it]<t1) t1=f[it]+g[it], t2=1;
else if (f[it]+g[it]==t1) ++t2;
// cout<<"t1: "<<t1<<endl;
ans1+=t1; ans2=ans2*t2%mod;
}
else {
top=0; cost=0;
ll t1=0, t2=1;
dfs3(i, 0);
for (int i=1; i<=top; ++i) {
dfs4(ring[i], 0);
t1+=f[ring[i]];
}
if (cost*2==top) t1+=cost, t2=2;
else t1+=min(cost, top-cost);
ans1+=t1; ans2=ans2*t2%mod;
}
vis2[bel]=1;
}
// cout<<"f: "; for (int i=1; i<=2*n; ++i) cout<<f[i]<<' '; cout<<endl;
// cout<<"g: "; for (int i=1; i<=2*n; ++i) cout<<g[i]<<' '; cout<<endl;
printf("%lld %lld\n", ans1, ans2);
jump: ;
}
return 0;
}
