T1 2A
考察快读的写法。
$code:$
T1
#include<bits/stdc++.h>
#define scanf SCANF=scanf
using namespace std;
namespace IO{
inline int read(){
char ch=getchar(); int x=0,f=1;
while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9'){ x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
return x*f;
}
inline void write(int x,char sp){
char ch[20]; int len=0;
if(x<0){ putchar('-'); x=~x+1; }
do{ ch[len++]=x%10+(1<<4)+(1<<5); x/=10; }while(x);
for(int i=len-1;~i;--i) putchar(ch[i]); putchar(sp);
}
inline int max(int x,int y){ return x<y?y:x; }
inline int min(int x,int y){ return x<y?x:y; }
inline void swap(int &x,int &y){ x^=y^=x^=y; }
inline void chmax(int &x,int y){ x=x<y?y:x; }
inline void chmin(int &x,int y){ x=x<y?x:y; }
} using namespace IO;
const int NN=50;
int SCANF;
int tmp,len,cnt,num[10];
bool flag;
char s[NN];
signed main(){
FILE *R=freopen("ip.in","r",stdin);
FILE *W=freopen("ip.out","w",stdout);
scanf("%s",s+1); len=strlen(s+1); tmp=1;
for(int i=1;i<=len;i++){
if(s[i]=='0'&&s[i+1]>='0'&&s[i+1]<='9') flag=1;
if(s[i]<'0'||s[i]>'9'){
++cnt;
if(s[i]!='.'||cnt>3) flag=1;
}
}
while(num[0]<5){
int x=0;
while(tmp<=len&&(s[tmp]<'0'||s[tmp]>'9')) ++tmp;
while(tmp<=len&&s[tmp]>='0'&&s[tmp]<='9'){ x=(x<<1)+(x<<3)+(s[tmp]^48); ++tmp; }
if(x>255) x=255, flag=1;
num[++num[0]]=x;
}
puts(flag?"NO":"YES");
for(int i=1;i<5;i++) write(num[i],i==4?'\n':'.');
return 0;
}T2 2B
不会打暴力，于是切了。
显然的贪心：先消$AP$后消$PP$。
又发现对于每个$A$，只要它右边有足够的$P$，总能被消。因此扫一边模拟即可。不知道为啥极限数据去了$3$个零。（$1e7 \to 1e4$
我写的比较麻烦，实际上从左向右扫，遇到$P$就减$A$的数量就行。
T2%%WTZ超简洁写法
#include<bits/stdc++.h>
#define int long long
#define inf 0x3f3f3f3f3f3f3f3f
#define chkmax(x,y) (x)=(x>y)?(x):(y)
#define chkmin(x,y) (x)=(x<y)?(x):(y)
using namespace std;
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
return x*f;
}
inline void write(int x){
static int sta[42];int top=0;
do{sta[++top]=x%10,x/=10;}while(x);
while(top) putchar(sta[top--]+'0');
return void();
}
int n,suma,sump;
char str[10010];
signed main(){
freopen("apstr.in","r",stdin);
freopen("apstr.out","w",stdout);
scanf("%s",str+1);n=strlen(str+1);
for(int i=1;i<=n;i++){
if(str[i]=='A'){suma++;}
else{
if(suma){suma--;}
else{sump++;sump%=2;}
}
}
printf("%lld\n",suma+sump);
return 0;
}T2我的不知道为啥很麻烦的写法
#include<bits/stdc++.h>
#define scanf SCANF=scanf
using namespace std;
namespace IO{
inline int read(){
char ch=getchar(); int x=0,f=1;
while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9'){ x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
return x*f;
}
inline void write(int x,char sp){
char ch[20]; int len=0;
if(x<0){ putchar('-'); x=~x+1; }
do{ ch[len++]=x%10+(1<<4)+(1<<5); x/=10; }while(x);
for(int i=len-1;~i;--i) putchar(ch[i]); putchar(sp);
}
inline int max(int x,int y){ return x<y?y:x; }
inline int min(int x,int y){ return x<y?x:y; }
inline void swap(int &x,int &y){ x^=y^=x^=y; }
inline void chmax(int &x,int y){ x=x<y?y:x; }
inline void chmin(int &x,int y){ x=x<y?x:y; }
} using namespace IO;
const int NN=1e5+5;
int SCANF;
int n,pos,ans,sump,suma,totp,tota,rest,s[NN];
char str[NN];
signed main(){
FILE *R=freopen("apstr.in","r",stdin);
FILE *W=freopen("apstr.out","w",stdout);
scanf("%s",str+1); n=strlen(str+1); s[0]=1;
for(int i=1;i<=n;i++) s[i]=(str[i]=='P');
for(pos=n;!s[pos];pos--) ++ans;//cout<<ans<<endl;
for(int i=1;i<=pos;i++) tota+=s[i]^1, totp+=s[i];
while(pos){
sump+=s[pos]; suma+=s[pos]^1;
if(s[pos-1]&&!s[pos]){//cout<<pos-1<<' '<<suma<<' '<<sump<<endl;
rest+=max(0,suma-sump);
sump-=min(sump,suma); suma=0;
}
pos--;
}//cout<<tota<<' '<<totp<<' '<<rest<<endl;
ans+=rest;
ans+=totp-tota+rest&1;
write(ans,'\n');
return 0;
}T2拿来对拍但依然能A的n2写法
#include<bits/stdc++.h>
#define scanf SCANF=scanf
using namespace std;
namespace IO{
inline int read(){
char ch=getchar(); int x=0,f=1;
while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9'){ x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
return x*f;
}
inline void write(int x,char sp){
char ch[20]; int len=0;
if(x<0){ putchar('-'); x=~x+1; }
do{ ch[len++]=x%10+(1<<4)+(1<<5); x/=10; }while(x);
for(int i=len-1;~i;--i) putchar(ch[i]); putchar(sp);
}
inline int max(int x,int y){ return x<y?y:x; }
inline int min(int x,int y){ return x<y?x:y; }
inline void swap(int &x,int &y){ x^=y^=x^=y; }
inline void chmax(int &x,int y){ x=x<y?y:x; }
inline void chmin(int &x,int y){ x=x<y?x:y; }
} using namespace IO;
const int NN=1e5+5;
int SCANF;
int n,pos,ans,sump,suma,totp,tota,rest,s[NN];
char str[NN];
signed main(){
freopen("in","r",stdin);
freopen("out2","w",stdout);
scanf("%s",str+1); n=strlen(str+1); s[0]=1;
for(int i=1;i<=n;i++) s[i]=(str[i]=='P');
while(n){
bool flag=0;
for(int i=1;i<=n;i++) if(s[i+1]==1&&!s[i]){
for(int j=i;j<=n-2;j++) s[j]=s[j+2];
s[n]=s[n-1]=-1;
flag=1; n-=2; break;
}
if(!flag) break;
}
// for(int i=1;i<=n;i++)cout<<s[i]<<' ';cout<<endl;
int tmp=n;
for(int i=1;i<=tmp;i+=2) if(s[i]==1&&s[i+1]==1) n-=2;
write(n,'\n');
return 0;
}T3 2C
前两个限制很好搞，第三个限制用$bitset$记每个点的祖先，两两枚举声明的类，如果二者祖先有交且都不是对方的祖先则不合法。
也可$O(nlog)$排序后扫一遍。
$code:$
T3
#include<bits/stdc++.h>
#define bst bitset<NN>
using namespace std;
namespace IO{
inline int read(){
char ch=getchar(); int x=0,f=1;
while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9'){ x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
return x*f;
}
inline void write(int x,char sp){
char ch[20]; int len=0;
if(x<0){ putchar('-'); x=~x+1; }
do{ ch[len++]=x%10+(1<<4)+(1<<5); x/=10; }while(x);
for(int i=len-1;~i;--i) putchar(ch[i]); putchar(sp);
}
inline int max(int x,int y){ return x<y?y:x; }
inline int min(int x,int y){ return x<y?x:y; }
inline void swap(int &x,int &y){ x^=y^=x^=y; }
inline void chmax(int &x,int y){ x=x<y?y:x; }
inline void chmin(int &x,int y){ x=x<y?x:y; }
} using namespace IO;
const int NN=1010;
map<string,int>has;
int n,tot,cnt;
bst anc[NN],t;
bool flag;
string nul,tmp[NN];
signed main(){
FILE *R=freopen("class.in","r",stdin);
FILE *W=freopen("class.out","w",stdout);
n=read();
while(n--){
cnt=0; flag=0;
cin>>tmp[0]>>nul;
while(tmp[cnt][0]!=';') cin>>tmp[++cnt]; --cnt;
if(has.find(tmp[0])!=has.end()) flag=1;
for(int i=1;i<=cnt;i++)
if(has.find(tmp[i])==has.end()){ flag=1; break; }
if(flag){ puts("greska"); continue; }
for(int i=1;i<cnt;i++)
for(int j=i+1;j<=cnt;j++){
int a=has[tmp[i]],b=has[tmp[j]];
t=anc[a]&anc[b];
if(t.none()) continue;
if(!anc[a][b]&&!anc[b][a]){ flag=1; break; }
}
puts(flag?"greska":"ok");
if(flag) continue;
has[tmp[0]]=++tot;
for(int i=1;i<=cnt;i++){
anc[tot]|=anc[has[tmp[i]]];
anc[tot].set(has[tmp[i]]);
}
}
return 0;
}
/*
10
shape : ;
rectangle : shape ;
circle : shape ;
circle : ;
square : shape rectangle ;
runnable : object ;
object : ;
runnable : object shape ;
thread : runnable ;
applet : square thread ;
*/T4 2D
枚举$k$，每次在子图中$DFS$记录答案，然后拓扑排序将度数为$k$的点删掉，继续枚举$k+1$。
看起来及其暴力，但由于$n,m$同级所以不大好卡，即使被卡了也可以优化枚举策略。
注意开始时要把度数为$0$的点先删掉。
$code:$
T4
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
namespace IO{
inline int read(){
char ch=getchar(); int x=0,f=1;
while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9'){ x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
return x*f;
}
inline void write(LL x,char sp){
char ch[20]; int len=0;
if(x<0){ putchar('-'); x=~x+1; }
do{ ch[len++]=x%10+(1<<4)+(1<<5); x/=10; }while(x);
for(int i=len-1;~i;--i) putchar(ch[i]); putchar(sp);
}
inline int max(int x,int y){ return x<y?y:x; }
inline int min(int x,int y){ return x<y?x:y; }
inline void swap(int &x,int &y){ x^=y^=x^=y; }
inline void chmax(int &x,int y){ x=x<y?y:x; }
inline void chmin(int &x,int y){ x=x<y?x:y; }
} using namespace IO;
const int NN=1e6+5;
int n,m,mx,idx,to[NN<<1],head[NN],nex[NN<<1],in[NN],deg[NN],res;
LL M,N,B,mm,nn,bb,now,ans;
int l,r,sum,q[NN];
bool ban[NN],vis[NN];
inline void add(int a,int b,int i){
to[++idx]=b; nex[idx]=head[a]; head[a]=idx; ++deg[a]; chmax(mx,deg[a]);
to[++idx]=a; nex[idx]=head[b]; head[b]=idx; ++deg[b]; chmax(mx,deg[b]);
}
void dfs(int s){
bb+=in[s]; ++nn; vis[s]=1;
for(int i=head[s];i;i=nex[i]){
int v=to[i];
if(ban[v]) continue;
++mm;
if(!vis[v]) dfs(v);
}
}
void topo(int lmt){
l=1; r=0;
for(int i=1;i<=n;i++)
if(!ban[i]&°[i]==lmt) ban[i]=1, q[++r]=i, --sum;
while(l<=r){
int x=q[l++];
for(int i=head[x];i;i=nex[i]) if(!ban[to[i]]){
--deg[to[i]];
if(deg[to[i]]==lmt)
ban[to[i]]=1, q[++r]=to[i], --sum;
}
}
}
signed main(){
FILE *R=freopen("kdgraph.in","r",stdin);
FILE *W=freopen("kdgraph.out","w",stdout);
n=sum=read(); m=read(); M=read(); N=read(); B=read(); ans=-1e18;
for(int a,b,i=1;i<=m;i++)
a=read(),b=read(), add(a,b,i);
for(int i=1;i<=n;i++){
in[i]=deg[i];
if(!in[i]) --sum, ban[i]=1;
}
for(int k=1;k<=mx&∑k++){
for(int i=1;i<=n;i++) vis[i]=0;
for(int i=1;i<=n;i++) if(!ban[i]&&!vis[i]){
mm=0; bb=0; nn=0;
dfs(i);
bb-=mm; mm>>=1;
now=mm*M-nn*N+bb*B;
if(now>=ans) res=k, ans=now;
}
topo(k);
}
write(res,' '); write(ans,'\n');
return 0;
}