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; }
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; }
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; }