【2020省选模拟】题解

T1：

把前序遍历左右看做括号序列，就是$0\le s\le m-2$，折线容斥即可

#include
using namespace std;
#define cs const
#define pb push_back
#define pii pair
#define fi first
#define se second
#define ll long long

namespace IO{

cs int rlen=1<<20|1;
char ibuf[rlen],*ib,*ob;
inline char gc(){
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,rlen,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=0;
	while(!isdigit(ch))f=ch=='-',ch=gc();
	while(isdigit(ch))res=(res*10)+(ch^48),ch=gc();
	return f?-res:res;
}

}
using IO::read;

cs int mod=998244353;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
inline int dec(int a,int b){return a<b?a-b+mod:a-b;}
inline int mul(int a,int b){return (ll)a*b%mod;}
inline void Add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
inline void Dec(int &a,int b){a=a<b?a-b+mod:a-b;}
inline void Mul(int &a,int b){a=(ll)a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))if(b&1)Mul(res,a);return res;}
inline int Inv(int x){return ksm(x,mod-2);}

char xx;
cs int N=2e7+5;
int fac[N],ifac[N];
inline void init_fac(){
	ifac[0]=fac[0]=1;
	for(int i=1;i<N;i++)fac[i]=mul(fac[i-1],i);
	ifac[N-1]=Inv(fac[N-1]);
	for(int i=N-2;i;i--)ifac[i]=mul(ifac[i+1],i+1);
}

inline int C(int n,int m){return (n<0||m<0||n<m)?0:mul(fac[n],mul(ifac[m],ifac[n-m]));}
inline int calc(int x,int y){return C(x+y,x);}
int n,m;
char yy;
inline void move1(int &x,int &y){swap(x,y),x++,y--;}
inline void move2(int &x,int &y){swap(x,y),x-=m-1,y+=m-1;}
int main(){
	init_fac();
	n=read(),m=read();
	int res=calc(n-1,n-1),x=n-1,y=n-1;
	while(x>=0&&y>=0){
		move1(x,y),Dec(res,calc(x,y));
		move2(x,y),Add(res,calc(x,y));
	}
	x=n-1,y=n-1;
	while(x>=0&&y>=0){
		move2(x,y),Dec(res,calc(x,y));
		move1(x,y),Add(res,calc(x,y));
	}cout<<res<<'\n';return 0;
}

T2：

把关键点提出来状压，每个答案枚举关键点计算
先floyd预处理每种状态的最短路可以方便转移

#include
using namespace std;
#define cs const
#define pb push_back
#define pii pair
#define fi first
#define se second
#define ll long long
#define bg begin

namespace IO{

cs int rlen=1<<22|1;
char ibuf[rlen],*ib,*ob;
inline char gc(){
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,rlen,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=0;
	while(!isdigit(ch))f=ch=='-',ch=gc();
	while(isdigit(ch))res=(res*10)+(ch^48),ch=gc();
	return f?-res:res;
}

}
using IO::read;
template<typename tp>inline void chemx(tp &a,tp b){(a<b)?(a=b):0;}
template<typename tp>inline void chemn(tp &a,tp b){(a>b)?(a=b):0;}

cs int N=405,M=(1<<16)|5,T=34;
bool isgt[N];
int dis[N][N],f[M][T],g[T][T],ans[N][N],inf;
int v[T],z[T],gt[N],ky[N],ban[N];
int n,m,K,q,lim;
int d[M][T][T];
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	freopen("my.out","w",stdout);
	#endif
	n=read(),m=read(),K=read(),lim=1<<K;
	memset(g,127/2,sizeof(g));
	memset(dis,127/2,sizeof(dis)),inf=dis[0][0];
	for(int i=1;i<=n;i++)dis[i][i]=0;
	for(int i=1;i<=m;i++){
		int u=read(),v=read(),w=read();
		chemn(dis[u][v],w),chemn(dis[v][u],w);
	}
	for(int i=0;i<K;i++){
		v[i]=read(),z[i]=read();
		isgt[v[i]]=1;
		gt[v[i]]|=1<<i,ky[z[i]]|=1<<i;
	}
	for(int k=1;k<=n;k++)if(!isgt[k])
	for(int i=1;i<=n;i++)if(i!=k)
	for(int j=1;j<=n;j++)if(j!=k&&j!=i)
	chemn(dis[i][j],dis[i][k]+dis[k][j]);
	for(int i=0;i<K;i++)
	for(int j=0;j<K;j++){
		d[0][i][j]=dis[z[i]][z[j]];
		d[0][i+K][j]=dis[v[i]][z[j]];
		d[0][i][j+K]=dis[z[i]][v[j]];
		d[0][i+K][j+K]=dis[v[i]][v[j]];
	}
	for(int s=1;s<lim;s++){
		int (*D)[T]=d[s],(*pre)[T]=d[s^(s&(-s))];
		for(int i=0;i<K+K;i++)
		for(int j=0;j<K+K;j++)
			D[i][j]=pre[i][j];
		int u=__builtin_ctz(s);
		if((gt[v[u]]&s)==gt[v[u]])
		for(int i=0;i<K+K;i++)
		for(int j=0;j<K+K;j++){
			int vl=D[i][u+K]+D[u+K][j];
			chemn(D[i][j],vl),chemn(D[j][i],vl);
		}
	}
	for(int t=0;t<K;t++){
		memset(f,127/2,sizeof(f));
		f[ky[z[t]]][t]=0;
		for(int s=1<<t;s<lim;s=(s+1)|(1<<t)){
			for(int i=0;i<K;i++)if(((s>>i)&1)&&f[s][i]<inf){
				int vl=f[s][i],*D=d[s][i];
				for(int j=0;j<K;j++)if(!((s>>j)&1)){
					if((gt[z[j]]&s)==gt[z[j]])chemn(f[s|ky[z[j]]][j],vl+D[j]);
				}
				for(int j=0;j<K;j++)
					if((gt[v[j]]&s)==gt[v[j]])chemn(g[t][j],vl+D[j+K]);
			}
		}
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		if(isgt[i]){ans[i][j]=-1;continue;}
		if(!isgt[j]){
			ans[i][j]=dis[i][j];
			for(int a=0;a<K;a++)
			for(int b=0;b<K;b++){
				if(!isgt[z[a]])chemn(ans[i][j],dis[i][z[a]]+g[a][b]+dis[v[b]][j]);
			}
		}
		else{
			ans[i][j]=inf;
			int b;
			for(b=0;b<K;b++)if(v[b]==j)break;
			for(int a=0;a<K;a++){
				if(!isgt[z[a]])chemn(ans[i][j],dis[i][z[a]]+g[a][b]);
			}
		}
		if(ans[i][j]>=inf)ans[i][j]=-1;
	}
	q=read();
	while(q--){
		int s=read(),t=read();
		cout<<ans[s][t]<<'\n';
	}return 0;
}

T3：

分治，map+哈希维护一下每种字符之间合法交换之后的字符以及个数即可

#include
using namespace std;
#define cs const
#define pb push_back
#define pii pair
#define fi first
#define se second
#define ll long long
#define bg begin

namespace IO{

cs int rlen=1<<22|1;
char ibuf[rlen],*ib,*ob;
inline char gc(){
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,rlen,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=0;
	while(!isdigit(ch))f=ch=='-',ch=gc();
	while(isdigit(ch))res=(res*10)+(ch^48),ch=gc();
	return f?-res:res;
}
inline int readstring(char *s){
	char ch=gc();int top=0;
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	s[top+1]='\0';return top;
}

}
using IO::read;
using IO::readstring;

template<typename tp>inline void chemx(tp &a,tp b){(a<b)?(a=b):0;}
template<typename tp>inline void chemn(tp &a,tp b){(a>b)?(a=b):0;}

cs int m1=1e9+7,m2=1e9+9;
inline int add1(int a,int b){return a+b>=m1?a+b-m1:a+b;}
inline int add2(int a,int b){return a+b>=m2?a+b-m2:a+b;}
inline int dec1(int a,int b){return a<b?a-b+m1:a-b;}
inline int dec2(int a,int b){return a<b?a-b+m2:a-b;}
inline int mul1(int a,int b){return (ll)a*b%m1;}
inline int mul2(int a,int b){return (ll)a*b%m2;}
cs pii bs=pii(5,13);
inline pii operator +(cs pii &a,cs pii &b){return pii(add1(a.fi,b.fi),add2(a.se,b.se));}
inline pii operator -(cs pii &a,cs pii &b){return pii(dec1(a.fi,b.fi),dec2(a.se,b.se));}
inline pii operator *(cs pii &a,cs pii &b){return pii(mul1(a.fi,b.fi),mul2(a.se,b.se));}

cs int N=100005;
char str[N];
int n;ll ans;
pii pw[N];
struct Map{
	static cs int mod=10260817;
	int adj[mod],nxt[N<<1],vl[N<<1],stk[N<<1],top,cnt;
	pii to[N<<1];
	inline void clear(){
		while(top)adj[stk[top--]]=0;
		cnt=0;
	}
	void insert(pii x,int k){
		int p=((ll)x.fi*19260817+x.se)%mod;
		for(int e=adj[p];e;e=nxt[e])if(to[e]==x){vl[e]+=k;return;}
		stk[++top]=p;nxt[++cnt]=adj[p],adj[p]=cnt,to[cnt]=x,vl[cnt]=k;
	}
	int query(pii x){
		int p=((ll)x.fi*19260817+x.se)%mod;
		for(int e=adj[p];e;e=nxt[e])if(to[e]==x)return vl[e];
		return 0;
	}
}f[6];
inline int id(int a,int b){
	switch(a){
		case 1:return b==2?0:1;
		case 2:return b==1?2:3;
		case 3:return b==1?4:5;
	}assert(0);
}
struct hs{
	pii pre[N],suf[N];
	int n,ps,del,p[N],s[N];
	void clear(){n=del=0;}
	void push(char x){
		int c=x-'a'+1;
		if(kd==0)ps++;else ps--;
		if(n&&s[n]==c)p[++del]=c,n--;
		else s[++n]=c,pre[n]=pre[n-1]*bs+pii(c,c),suf[n]=suf[n-1]+pii(c,c)*pw[n-1],p[++del]=-1;
	}
	void back(){
		if(del){
			if(kd==0)ps--;else ps++;
			if(p[del]>0)s[++n]=p[del],pre[n]=pre[n-1]*bs+pii(p[del],p[del]),suf[n]=suf[n-1]+pii(p[del],p[del])*pw[n-1];
			else n--;del--;
		}
	}
	inline pii hasl(int l,int r){
		return pre[r]-pre[l-1]*pw[r-l+1];
	}
	int kd;
	void move(int p){
		if(kd==0){
			while(ps>p)back();
			while(ps<p)push(str[ps+1]);	
		}
		else{
			while(ps<p)back();
			while(ps>p)push(str[ps-1]);
		}
	}
}A,B,C;
inline pii calc(hs &a,char x,hs &b,int kd){
	a.push(x);	
	int l=1,r=min(a.n,b.n),res=0;
	while(l<=r){
		int mid=(l+r)>>1;
		if(a.hasl(a.n-mid+1,a.n)==b.hasl(b.n-mid+1,b.n))l=mid+1,res=mid;
		else r=mid-1;
	}
	pii now;
	if(kd==0)now=a.pre[a.n-res]*pw[b.n-res]+b.suf[b.n-res];
	else now=a.suf[a.n-res]+b.pre[b.n-res]*pw[a.n-res];
	a.back();return now;
}
void solve(int l,int r){
	if(l==r)return;
	int mid=(l+r)>>1;
	A.move(mid);C.clear();
	for(int i=mid;i>=l;i--){
		A.back();
		for(int j=1;j<=3;j++)if(j!=str[i]-'a'+1){
			f[id(str[i]-'a'+1,j)].insert(calc(A,'a'+j-1,C,0),1);
		}
		C.push(str[i]);
	}
	B.move(mid+1);C.clear();
	for(int i=mid+1;i<=r;i++){
		B.back();
		for(int j=1;j<=3;j++)if(j!=str[i]-'a'+1){	
			ans+=f[id(j,str[i]-'a'+1)].query(calc(C,'a'+j-1,B,1));
		}
		C.push(str[i]);
	}
	for(int i=0;i<6;i++)f[i].clear();
	solve(l,mid),solve(mid+1,r);
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	pw[0]=pii(1,1);
	for(int i=1;i<N;i++)pw[i]=pw[i-1]*bs;
	n=readstring(str);A.kd=0,B.kd=1,B.ps=n+1;
	for(int i=1;i<=n;i++)A.push(str[i]);
	for(int i=n;i>=1;i--)B.push(str[i]);
	solve(1,n);
	cout<<ans<<'\n';
	return 0;
}