当一张图每条边最多属于一个环的时候，这个图叫做仙人掌
圆方树用来解决仙人掌上的问题
1.仙人掌上最大独立集
仙人掌上做dp可以分成树边和环边两种边
所以相当于实现树形和环形dp
void tarjan(ll x,ll y)
{
dfn[x]=low[x]=++cnt;
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (v==y) continue;
if (!dfn[v])
{
fa[v]=x;
tarjan(v,x);
low[x]=min(low[x],low[v]);
} else low[x]=min(low[x],dfn[v]);
if (low[v]>dfn[x]) {树形dp;}
}
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (fa[v]!=x&&dfn[v]>dfn[x]) 环形dp;
}
}
圆方树dp就套上面的板子就可以了
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#include <immllrin.h>
//#include <emmllrin.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (ll i=h;i<=t;i++)
#define dep(i,t,h) for (ll i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rll register ll
inline ll rd(){
ll x=0;char c=getchar();bool f=0;
while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
return f?-x:x;
}
char ss[1<<24],*A=ss,*B=ss;
IL char gc()
{
return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T>void maxa(T &x,T y)
{
if (y>x) x=y;
}
template<class T>void mina(T &x,T y)
{
if (y<x) x=y;
}
template<class T>void read(T &x)
{
ll f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
while(c=gc(),c>47&&c<58) x=x*10+(c^48); x*=f;
}
struct cp {
ll x,y;
cp operator +(cp B)
{
return (cp){x+B.x,y+B.y};
}
cp operator -(cp B)
{
return (cp){x-B.x,y-B.y};
}
ll operator *(cp B)
{
return x*B.y-y*B.x;
}
ll half() { return y < 0 || (y == 0 && x < 0); }
};
struct re{
ll a,b,c;
};
const ll N=3e5;
ll n,m,q,f1,f2;
ll dfn[N],low[N],cnt,fa[N],f[N][2];
ll head[N],l;
re e[N*2];
void arr(ll x,ll y)
{
e[++l].a=head[x];
e[l].b=y;
head[x]=l;
}
void solve(ll x,ll v)
{
ll f0=0,f1=-1e9,t0,t1;
for (ll i=v;i!=x;i=fa[i])
{
t0=f[i][0]+max(f0,f1);
t1=f[i][1]+f0;
f0=t0; f1=t1;
}
f[x][0]+=max(f0,f1);
f0=-1e9,f1=0;
for (ll i=v;i!=x;i=fa[i])
{
t0=f[i][0]+max(f0,f1);
t1=f[i][1]+f0;
f0=t0; f1=t1;
}
f[x][1]+=f0;
}
void tarjan(ll x,ll y)
{
dfn[x]=low[x]=++cnt;
// cerr<<x<<endl;
f[x][0]=0; f[x][1]=1;
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (v==y) continue;
if (!dfn[v])
{
fa[v]=x;
tarjan(v,x);
low[x]=min(low[x],low[v]);
} else low[x]=min(low[x],dfn[v]);
if (low[v]>dfn[x]) {f[x][0]+=max(f[v][1],f[v][0]),f[x][1]+=f[v][0];}
}
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (fa[v]!=x&&dfn[v]>dfn[x]) solve(x,v);
}
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
rep(i,1,m)
{
ll u,v,w;
cin>>u>>v;
arr(u,v); arr(v,u);
}
tarjan(1,0);
cout<<max(f[1][0],f[1][1]);
return 0;
}View Code
2.Winter Festival
题目描述
Peter最喜欢的季节是冬天。为了庆祝冬天的活动，Peter准备装饰他所在的城市。
城市有许多区域，有一些道路连接着某些区域对。Peter想给每条道路装饰成数字0,1或2之一。
一组方案合法当且仅当满足以下两个条件：
- 对于任意两条不同的边(x, y), (x, z)，它们的边权和mod 3不等于1。
- 对于任意一个简单环，里面的所有边的边权和必须是奇数。
请找到一组合法方案，使得所有边的边权之和最小。(n,m<=100000)
题解：
比较显然对于边权和mod 3不等于1这个条件 等价于不能同时出现0,1且不能出现两个2
于是如果这个是树我们可以记录f[x][i][j][k]表示当前在x点0/1/2有无出现
现在变成仙人掌多一个环形dp
代码好像写的比较麻烦
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#include <immllrin.h>
//#include <emmllrin.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (ll i=h;i<=t;i++)
#define dep(i,t,h) for (ll i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rll register ll
#define me1(g) memset(g,1,sizeof(g))
#define mep(x,y) memcpy(x,y,sizeof(y))
inline ll rd(){
ll x=0;char c=getchar();bool f=0;
while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
return f?-x:x;
}
char ss[1<<24],*A=ss,*B=ss;
IL char gc()
{
return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T>void maxa(T &x,T y)
{
if (y>x) x=y;
}
template<class T>void mina(T &x,T y)
{
if (y<x) x=y;
}
template<class T>void read(T &x)
{
ll f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
while(c=gc(),c>47&&c<58) x=x*10+(c^48); x*=f;
}
struct cp {
ll x,y;
cp operator +(cp B)
{
return (cp){x+B.x,y+B.y};
}
cp operator -(cp B)
{
return (cp){x-B.x,y-B.y};
}
ll operator *(cp B)
{
return x*B.y-y*B.x;
}
ll half() { return y < 0 || (y == 0 && x < 0); }
};
struct re{
ll a,b,c;
};
const ll N=3e5;
ll n,m,q,f1,f2;
ll dfn[N],low[N],cnt,fa[N],f[N][2][2][2],cnt1[N];
ll head[N],l,b[N];
re e[N*2];
void arr(ll x,ll y)
{
e[++l].a=head[x];
e[l].b=y;
head[x]=l;
}
bool tt=0;
void cl(int x)
{
if (cnt1[x]==0) cnt1[x]++;
else tt=1;
}
void mina(int &x,int y)
{
if (y<x) x=y;
}
void solve(ll x,ll v,int k)
{
for (ll i=v;i!=x;i=fa[i])
{
cl(b[i]); cl(b[i]^1);
}
cl(k); cl(k^1);
ll g[2][2][2][2][3];
me1(g);
rep(i1,0,1)
rep(j1,0,1)
rep(k1,0,1)
rep(t,0,2)
{
bool t1=i1|(t==0);
bool t2=j1|(t==1);
int t3=k1+(t==2);
if (t1&t2||t3>=2) continue;
g[t1][t2][t3][t%2][t]=min(g[t1][t2][t3][t%2][t],f[v][i1][j1][k1]+t);
}
for (ll i=fa[v];i!=fa[x];i=fa[i])
{
ll h[2][2][2][2][3]; me1(h);
rep(i1,0,1)
rep(j1,0,1)
rep(k1,0,1)
rep(i2,0,1)
rep(j2,0,1)
rep(k2,0,1)
rep(t,0,2)
{
bool t1=i2|(t==0);
bool t2=j2|(t==1);
int t3=k2+(t==2);
if ((t1&t2)||t3>=2) continue;
t1=i1|(t==0);
t2=j1|(t==1);
t3=k1+(t==2);
if ((t1&t2)||t3>=2) continue;
rep(p1,0,1)
rep(p2,0,2)
mina(h[t1][t2][t3][(p1+t)%2][p2],f[i][i1][j1][k1]+g[i2][j2][k2][p1][p2]+t);
}
mep(g,h);
}
me1(f[x]);
rep(i1,0,1)
rep(j1,0,1)
rep(k1,0,1)
rep(p1,1,1)
rep(t,0,2)
{
bool t1=i1|(t==0);
bool t2=j1|(t==1);
int t3=k1+(t==2);
if (t1&t2||t3>=2) continue;
mina(f[x][t1][t2][t3],g[i1][j1][k1][p1][t]);
}
}
void tarjan(ll x,ll y)
{
if (tt) return;
dfn[x]=low[x]=++cnt;
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (v==y) continue;
if (!dfn[v])
{
fa[v]=x; b[v]=u;
tarjan(v,x);
low[x]=min(low[x],low[v]);
} else low[x]=min(low[x],dfn[v]);
if (low[v]>dfn[x])
{
ll g[2][2][2];
me1(g);
rep(i1,0,1)
rep(j1,0,1)
rep(k1,0,1)
rep(i2,0,1)
rep(j2,0,1)
rep(k2,0,1)
rep(t,0,2)
{
bool t1=i2|(t==0);
bool t2=j2|(t==1);
int t3=k2+(t==2);
if ((t1&t2)||t3>=2) continue;
t1=i1|(t==0);
t2=j1|(t==1);
t3=k1+(t==2);
if ((t1&t2)||t3>=2) continue;
g[t1][t2][t3]=min(g[t1][t2][t3],f[x][i1][j1][k1]+f[v][i2][j2][k2]+t);
}
mep(f[x],g);
}
}
for (ll u=head[x];u;u=e[u].a) {
ll v=e[u].b;
if (fa[v]!=x&&dfn[v]>dfn[x]) solve(x,v,u);
}
}
int main()
{
ios::sync_with_stdio(false);
l=1;
cin>>n>>m;
rep(i,1,m)
{
ll u,v;
cin>>u>>v;
arr(u,v); arr(v,u);
}
ll ans=0;
rep(i,1,n)
if (!dfn[i])
{
tarjan(i,0);
vector<int> ve;
rep(i1,0,1)
rep(j1,0,1)
rep(k1,0,1)
ve.push_back(f[i][i1][j1][k1]);
sort(ve.begin(),ve.end());
ans+=ve[0];
}
if (tt||ans>2*m) cout<<-1<<endl;
else
{
cout<<ans<<endl;
}
return 0;
}View Code
3.仙人掌上最短路
这个题一定要建出圆方树
圆方树的建立方法是对于树边直接连，对于环边把环边都连到一个新点上
定义老点叫做圆点新点叫做方点
圆点到圆点的距离就是原树距离
圆点到方点的距离就是圆点到环上dfs序最小点的环上最短距离
求两个点距离时，先求出lca=k
若k是圆点，距离就是lca[x]+lca[y]-2*lca[k]
若k是方点，距离就是lca[x]+lca[y]-lca[a]-lca[b]+dis(a,b)
其中a,b分别代表x,y到k差一步的点，dis(a,b)代表a，b在环上的距离
正确性比较容易证明
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#include <immllrin.h>
//#include <emmllrin.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (ll i=h;i<=t;i++)
#define dep(i,t,h) for (ll i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rll register ll
inline ll rd(){
ll x=0;char c=getchar();bool f=0;
while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
return f?-x:x;
}
char ss[1<<24],*A=ss,*B=ss;
IL char gc()
{
return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T>void maxa(T &x,T y)
{
if (y>x) x=y;
}
template<class T>void mina(T &x,T y)
{
if (y<x) x=y;
}
template<class T>void read(T &x)
{
ll f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
while(c=gc(),c>47&&c<58) x=x*10+(c^48); x*=f;
}
struct cp {
ll x,y;
cp operator +(cp B)
{
return (cp){x+B.x,y+B.y};
}
cp operator -(cp B)
{
return (cp){x-B.x,y-B.y};
}
ll operator *(cp B)
{
return x*B.y-y*B.x;
}
ll half() { return y < 0 || (y == 0 && x < 0); }
};
struct re{
ll a,b,c;
};
const ll N=3e5;
ll n,m,q,f1,f2;
ll dis[N],s[N],b[N];
ll id[N];
struct yy{
re e[N*2];
ll head[N],d[N];
ll bz[20][N],l;
void arr(ll x,ll y,ll z)
{
e[++l].a=head[x];
e[l].b=y;
e[l].c=z;
// if (x<y) cerr<<x<<" "<<y<<endl;
head[x]=l;
}
void dfs(ll x,ll y)
{
bz[0][x]=y;
rep(i,1,19) bz[i][x]=bz[i-1][bz[i-1][x]];
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (v==y) continue;
dis[v]=dis[x]+e[u].c;
d[v]=d[x]+1;
dfs(v,x);
}
}
ll lca(ll x,ll y)
{
if (d[x]<d[y]) swap(x,y);
dep(i,19,0) if (d[bz[i][x]]>=d[y]) x=bz[i][x];
if (x==y) return x;
dep(i,19,0) if (bz[i][x]!=bz[i][y]) x=bz[i][x],y=bz[i][y];
f1=x; f2=y;
return bz[0][x];
}
}G2;
struct xx{
ll dfn[N],low[N],cnt,fa[N],n;
ll head[N],l;
re e[N*2];
void arr(ll x,ll y,ll z)
{
e[++l].a=head[x];
e[l].b=y;
e[l].c=z;
head[x]=l;
}
void solve(ll x,ll v,ll w)
{
n++;
ll now=w,now2=0;
for (ll i=v;i!=fa[x];i=fa[i])
{
s[i]=now;
now+=b[i];
id[i]=now2++;
}
s[n]=s[x]; s[x]=0;
for (ll i=v;i!=fa[x];i=fa[i]) G2.arr(n,i,min(s[i],s[n]-s[i])),G2.arr(i,n,min(s[i],s[n]-s[i]));
}
void tarjan(ll x,ll y)
{
dfn[x]=low[x]=++cnt;
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (v==y) continue;
if (!dfn[v])
{
fa[v]=x; b[v]=e[u].c;
tarjan(v,x);
low[x]=min(low[x],low[v]);
} else low[x]=min(low[x],dfn[v]);
if (low[v]>dfn[x]) {G2.arr(x,v,e[u].c); G2.arr(v,x,e[u].c);}
}
for (ll u=head[x];u;u=e[u].a)
{
ll v=e[u].b;
if (fa[v]!=x&&dfn[v]>dfn[x]) solve(x,v,e[u].c);
}
}
}G1;
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m>>q;
rep(i,1,m)
{
ll u,v,w;
cin>>u>>v>>w;
G1.arr(u,v,w); G1.arr(v,u,w);
}
G1.n=n;
G1.tarjan(1,0);
G2.dfs(1,0);
rep(i,1,q)
{
ll u,v;
cin>>u>>v;
ll g=G2.lca(u,v);
if (g<=n) cout<<dis[u]+dis[v]-2*dis[g]<<endl;
else
{
ll ans=dis[u]+dis[v]-dis[f1]-dis[f2];
if (id[f1]<=id[f2]) ans+=min(s[f1]+s[g]-s[f2],s[f2]-s[f1]);
else ans+=min(s[f2]+s[g]-s[f1],s[f1]-s[f2]);;
cout<<ans<<endl;
}
}
return 0;
}View Code
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