CSP 后多校四
这场考试确实很让我想吐槽，个人认为是一套很垃圾的 \(NOIP\) 模拟赛.
\(T1\) 和 \(T3\) 傻了吧唧地给了两道水题，一个半小时 \(A\) 完了，\(T2\) 一个大模拟题面给得不清楚直接弃了，\(T4\) 特意整一个斯特林数凸显这场考试有多么地 \(NOIP\).
总的来说这场考试个人感觉没啥意义，做得没有什么收获就很难受.
A. 特殊字符串
签到题，随便写.
A_code
#include<bits/stdc++.h>
using namespace std;
namespace BSS{
#define ll long long
#define lf double
#define ull unsigned ll
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define lbt(x) ((x)&(-(x)))
#define Fill(x,y) memset(x,y,sizeof(x))
#define Copy(x,y) memcpy(x,y,sizeof(x))
#define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
auto read=[]()->ll{
ll w=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar();
return cit?w:(-w);
};
} using namespace BSS;
const ll N=1e5+21;
char ch[N];
ll m,n,ans;
ll dp[30],val[N];
ll ic[30][30];
signed main(){
File(shiki);
n=read(),scanf("%s",ch+1);
for(ll i=1;i<=n;i++) val[i]=ch[i]-'a'+1;
m=read(); ll u,v; Fill(dp,-0x3f);
for(ll i=1;i<=m;i++){
scanf("%s",ch+1),u=ch[1]-'a'+1;
scanf("%s",ch+1),v=ch[1]-'a'+1;
ic[u][v]+=read();
}
for(ll i=1;i<=n;i++){
u=val[i]; if(ic[u][u]) dp[u]+=ic[u][u];
for(ll j=1;j<=26;j++){
if(u==j) continue;
dp[u]=max(dp[u],dp[j]+ic[j][u]);
}
dp[u]=max(dp[u],0ll);
}
for(ll i=1;i<=26;i++) ans=max(ans,dp[i]);
printf("%lld\n",ans),exit(0);
}
B. 宝可梦
大模拟，找准顺序就行了.
B_code
#include<bits/stdc++.h>
using namespace std;
namespace BSS{
#define ll long long
#define lf double
#define ull unsigned ll
#define pb push_back
#define kap make_pair
#define lb lower_bound
#define ub upper_bound
#define lbt(x) ((x)&(-(x)))
#define Fill(x,y) memset(x,y,sizeof(x))
#define Copy(x,y) memcpy(x,y,sizeof(x))
#define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
auto read=[]()->ll{
ll w=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar();
return cit?w:(-w);
};
} using namespace BSS;
const ll N=4e6+21,inf=1e15;
char ch[N];
ll m,n,sx,sy,ex,ey,nx,ny,cnt,ans,ops;
ll rk[N];
ll dx[4]={0,-1,0,1},dy[4]={1,0,-1,0};
vector<ll> mt[N];
map<char,int> dir;
auto getid=[](ll x,ll y,ll d)->ll{ return (x-1)*m*4+(y-1)*4+d; };
void solve(ll x,ll y,ll d){
for(ll di=(d+3)%4;;(++di)%=4){
nx=x+dx[di],ny=y+dy[di];
if(nx<1 or nx>n or ny<1 or ny>m) continue;
if(!mt[nx][ny]) continue;
if(!rk[getid(x,y,di)]) rk[getid(x,y,di)]=++cnt,solve(nx,ny,di);
break;
}
}
auto Work=[]()->void{
ll d,x,y,sx,sy,ex,ey,id,ans=inf;
sx=read(),sy=read(),ex=read(),ey=read(),scanf("%s",ch+1),d=dir[ch[1]];
if(sx==ex and sy==ey) { puts("0"); return ; }
for(ll i=0;i<4;i++){
id=rk[getid(ex,ey,i)];
if(!id) continue;
// cout<<id<<' '<<rk[getid(sx,sy,d)]<<endl;
if(id>=rk[getid(sx,sy,d)]) ans=min(ans,id-rk[getid(sx,sy,d)]);
else ans=min(ans,id-rk[getid(sx,sy,d)]+cnt);
}
printf("%lld\n",ans);
};
signed main(){
File(pokemon);
n=read(),m=read(); ll flag=1;
for(ll i=1;i<=n;i++){
mt[i].pb(0),scanf("%s",ch+1);
for(ll j=1;j<=m;j++) mt[i].pb(ch[j]=='.' ? 1 : 0);
}
dir['R']=0,dir['U']=1,dir['L']=2,dir['D']=3;
for(ll i=1;flag and i<=n;i++){
for(ll j=1;flag and j<=m;j++)
if(mt[i][j]) solve(i,j,3),flag=0;
}
for(int Ts=read();Ts;Ts--) Work();
exit(0);
}
C. 矩阵
签到题，随便写.
C_code
#include<bits/stdc++.h>
using namespace std;
namespace BSS{
#define ll long long
#define lf double
#define ull unsigned ll
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define lbt(x) ((x)&(-(x)))
#define Fill(x,y) memset(x,y,sizeof(x))
#define Copy(x,y) memcpy(x,y,sizeof(x))
#define File(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
auto read=[]()->ll{
ll w=0; bool cit=1; char ch;
while(!isdigit(ch=getchar())) if(ch=='-') cit=0;
while(isdigit(ch)) w=(w<<3)+(w<<1)+(ch^48),ch=getchar();
return cit?w:(-w);
};
} using namespace BSS;
const ll N=1e6+21,inf=1e15;
ll m,n,ts,ans,cnt,tot;
ll in[N],bin[N],dis[N],vis[N];
vector<ll> to[N],mat[N];
struct I { ll x,y,w; } p[N];
struct II { ll u,v,w; } e[N<<4],o[N<<4];
struct III{
ll id,d;
III(){}
III(ll id,ll d) : id(id),d(d) {}
friend bool operator < (III i,III j){
return i.d>j.d;
}
};
priority_queue<III> que;
auto getid=[](ll x,ll y)->ll{ return (x-1)*m+y; };
auto add=[](ll u,ll v,ll w)->void{
e[++ts].u=u,e[ts].v=v,e[ts].w=w;
};
auto bfs=[]()->void{
ll u;
for(ll i=1;i<=tot;i++){
dis[u=bin[i]]=-inf,vis[u]=0;
if(!in[u]) que.push(III(u,0)),dis[u]=1;
}
while(que.size()){
u=que.top().id,que.pop();
if(vis[u]) continue; vis[u]=1;
for(auto v : to[u]){
if(dis[v]<dis[u]+1){
que.push(III(v,dis[v]=dis[u]+1));
}
}
}
for(ll i=1;i<=tot;i++){
ans=max(ans,dis[bin[i]]);
to[bin[i]].clear(),vis[bin[i]]=0;
}
};
signed main(){
File(matrix);
n=read(),m=read(); ll u,v;
for(ll i=1;i<=n;i++){
mat[i].pb(0);
for(ll j=1;j<=m;j++) mat[i].pb(read());
}
for(ll i=1;i<=n;i++){
for(ll j=1;j<=m;j++){
p[++cnt].x=i,p[cnt].y=j,p[cnt].w=mat[i][j];
if(i>1 and mat[i][j]%mat[i-1][j]==0){
if(mat[i][j]==mat[i-1][j]) puts("-1"),exit(0);
if(mat[i][j]/mat[i-1][j]>200) ans=max(ans,2ll);
else add(getid(i-1,j),cnt,mat[i][j]/mat[i-1][j]);
}
if(j>1 and mat[i][j]%mat[i][j-1]==0){
if(mat[i][j]==mat[i][j-1]) puts("-1"),exit(0);
if(mat[i][j]/mat[i][j-1]>200) ans=max(ans,2ll);
else add(getid(i,j-1),cnt,mat[i][j]/mat[i][j-1]);
}
if(i<n and mat[i][j]%mat[i+1][j]==0){
if(mat[i][j]==mat[i+1][j]) puts("-1"),exit(0);
if(mat[i][j]/mat[i+1][j]>200) ans=max(ans,2ll);
else add(getid(i+1,j),cnt,mat[i][j]/mat[i+1][j]);
}
if(j<m and mat[i][j]%mat[i][j+1]==0){
if(mat[i][j]==mat[i][j+1]) puts("-1"),exit(0);
if(mat[i][j]/mat[i][j+1]>200) ans=max(ans,2ll);
else add(getid(i,j+1),cnt,mat[i][j]/mat[i][j+1]);
}
}
}
sort(e+1,e+1+ts,[](II i,II j){ return i.w<j.w; });
for(ll i=1,j=1;i<=200;i++){
while(e[j].w==i){
u=e[j].u,v=e[j].v,to[u].pb(v),in[v]++;
bin[++tot]=u,bin[++tot]=v,j++;
}
bfs();
}
printf("%lld\n",max(ans,1ll)),exit(0);
}
D. 乘法
斯特林数，先鸽.
