csp-s模拟3

A. 奇观

观察到 \(c\) 和 \(f\) 互不影响，所以分开算就行，枚举相连的边太多了，会 \(T\),所以我们把总情况找出来，减去删去的边的

方案数即可，记 \(f_{u,x}\) 表示 \(u\) 节点往后跟 \(x\) 个长度的方案数，有 \(f_{u,x}=\sum_{x->y} \limits f_{y,x-1}\)

边界 \(f_{u,1}=\) 其度数,这是 \(c\) 的，\(f\) 就是长度为 \(3\) 的链后面随便取俩点

点击查看代码

#include<bits/stdc++.h>
#define int long long
const int mod=998244353;
const int maxn=1e5+10;
using namespace std;
int n,m,sum,ans,temp,in[maxn];
int f[maxn][4];
vector<int>q[maxn];
unordered_map<int,int>a[maxn]; signed main()
{freopen("a.in","r",stdin);freopen("a.out","w",stdout);ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n>>m;fill(in+1,in+1+n,n-1);for(int i=1,x,y;i<=m;i++){cin>>x>>y;in[x]--,in[y]--;q[x].push_back(y);q[y].push_back(x); }for(int i=1;i<=n;i++){f[i][1]=in[i];temp=(temp+f[i][1])%mod;}for(int i=1;i<=n;i++){f[i][2]=temp-f[i][1];for(int j=0;j<q[i].size();j++){f[i][2]-=f[q[i][j]][1];}ans=(ans+f[i][2])%mod;}temp=0; for(int i=1;i<=n;i++){f[i][3]=ans-f[i][2];for(int j=0;j<q[i].size();j++){f[i][3]-=f[q[i][j]][2];}temp=(temp+f[i][3])%mod;}
//	cout<<temp;sum=1ll*temp*temp%mod; ans=0;for(int i=1;i<=n;i++){ans=(ans+1ll*f[i][2]*in[i]%mod*in[i]%mod)%mod;}
//	cout<<ans;cout<<1ll*sum*ans%mod;return 0;
}
/*
4 1
1 2
*/

B. 铁路

新开一个数组来对应新建节点的位置，把位置指到合并的点中深度最小的点即可,这里合并时用并查集维护每个联通块

压缩路径也到深度最小的点，这样就对应起来了

点击查看代码

#include<bits/stdc++.h>
const int maxn=5e5+10;
using namespace std;
int head[maxn],nxt[maxn<<1],to[maxn<<1],tot;
int n,m,fa[maxn],last[maxn],pos[maxn<<1],size;
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
void add(int x,int y)
{to[++tot]=y;nxt[tot]=head[x];head[x]=tot;
}
void addm(int x,int y)
{add(x,y),add(y,x);
}
int pre[maxn],deep[maxn],st[20][maxn],cnt;
struct Lca
{int get(int x,int y){return deep[x]<deep[y]?x:y;}void lsx(int x,int fa){pre[x]=++cnt,last[x]=fa,st[0][cnt]=fa;deep[x]=deep[fa]+1;for(int i=head[x];i;i=nxt[i]) if(to[i]!=fa) lsx(to[i],x);}void init(){lsx(1,0);for(int i=1;(1<<i)<=n;i++)for(int j=1;j<=n-(1<<i)+1;j++)st[i][j]=get(st[i-1][j],st[i-1][j+(1<<i-1)]);}int lca(int x,int y){if(x==y) return x;if((x=pre[x])>(y=pre[y])) swap(x,y);int l=__lg(y-x);return get(st[l][x+1],st[l][y-(1<<l)+1]);}
}e;void merge(int x,int y,int i)
{int p=e.lca(x,y);int a=find(x),b=find(y),c=find(p);
//	cout<<a<<" "<<b<<" "<<p<<endl; while(find(a)!=find(c)){a=fa[a]=find(last[a]);size--;}while(find(b)!=find(c)){
//		cout<<b<<" "<<last[b]<<"!"<<endl; b=fa[b]=find(last[b]);	size--;}pos[n+i]=c;
}int main()
{freopen("a.in","r",stdin);freopen("a.out","w",stdout);ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n>>m;size=n;for(int i=1;i<=n;i++)fa[i]=i,pos[i]=i;for(int i=1;i<n;i++){int x,y;cin>>x>>y;addm(x,y);}e.init();for(int i=1;i<=m;i++){int x,y;cin>>x>>y;x=pos[x];y=pos[y];
//		cout<<x<<" "<<y<<"!"<<endl; merge(x,y,i);cout<<size<<'\n';	}return 0;
}
/*
10 5
2 1
9 7
4 10
1 8
3 4
1 7
3 2
6 1
1 51 6
8 4
5 12
13 13
9 14*/

C. 光纤

破防了，这东西谁愿意改谁改，我改不了一点。。。

找凸包的每一条边的高，找最小的那个就是答案

点击查看代码

#include<bits/stdc++.h>
#define int __int128
const int maxn=1e6+10;
using namespace std;
typedef pair<int,int>pr;
int n,top,st[maxn];
pr p[maxn];
bool used[maxn];template<typename T> inline T read() {T X = 0; bool flag = 1; char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-') flag = 0; ch = getchar();}while (ch >= '0' && ch <= '9') {X = (X << 1) + (X << 3) + ch - '0'; ch = getchar();}if (flag) return X;return ~ (X - 1);
}template<typename T> inline void write(T X) {if (X < 0) {putchar('-'); X = ~ (X - 1);}int s[50], top = 0;while (X) {s[++top] = X % 10; X /= 10;}if (!top) s[++top] = 0;while (top) putchar(s[top--] + '0');return;
}pr operator - (pr a,pr b)
{return {a.first-b.first,a.second-b.second};
}int cross(pr a,pr b,pr c)
{pr u,v;u=b-a,v=c-a;return u.first*v.second-v.first*u.second;
}pr s1(pr a,pr b,pr c)
{pr u,v;u=b-a,v=c-a;int x=(u.first*v.second-v.first*u.second);return {x*x,u.first*u.first+u.second*u.second};
}void pre()
{sort(p+1,p+n+1);for(int i=1;i<=n;i++){while(top>=2&&cross(p[st[top-1]],p[st[top]],p[i])>0){used[st[top]]=0;top--; }st[++top]=i;used[i]=1;}used[1]=0;for(int i=n-1;i>=1;i--){if(used[i])continue;while(top>=2&&cross(p[st[top-1]],p[st[top]],p[i])>0) top--;st[++top]=i;}
}pr yue(pr x)
{int a=x.first,b=x.second,c=__gcd(a,b);return {a/c,b/c};
}bool cmp(pr a,pr b)
{pr c={a.first*b.second,a.second*b.second};pr d={b.first*a.second,b.second*a.second};return c.first>d.first;
}pr min(pr a,pr b)
{if(!b.first) return a; return cmp(a,b)?b:a;
}void solve()
{pr a=p[st[1]],b=p[st[2]],c=p[st[3]];pr ans=s1(a,b,c);int k=3;for(int i=4;i<top;i++){pr d=s1(a,b,p[st[i]]);if(d.first>ans.first){ans=d;k=i;}}
//	write(ans.first); int x=2,y=3;pr l={0,1};for(int i=k,cnt=1;cnt<top-1;cnt++,x=(x+1)%top,y=(y+1)%top){if(!x)x++;if(!y)y++;while(1){a=p[st[x]],b=p[st[y]],c=p[st[i]];pr d=s1(a,b,c);d=yue(d),ans=yue(ans);if(cmp(d,l))l=d;else {i--;if(!i)i=top-1;break;}i=(i+1)%top;if(!i)i++;if(i+1==x||i+1==y) break; }
//		write(l.first); ans=min(ans,l);}
//	cout<<ans.first<<" "<<ans.second<<endl; ans={ans.first,ans.second*4};ans=yue(ans); write(ans.first);cout<<"/";write(ans.second);
}signed main()
{freopen("ex_A2.in","r",stdin);
//	freopen("a.out","w",stdout);
//	ios::sync_with_stdio(0);
//	cin.tie(0),cout.tie(0);n=read<int>();for(int i=1;i<=n;i++){int x,y;x=read<int>(),y=read<int>();p[i]={x,y};}pre(); solve();return 0;
}
/*
5
0 0
2 0
1 3
1 2
1 1
*/