T1：
原题是
考虑对于每个点维护表示所有入点的
然后用暴力维护一下每个入点的
以及全局最大最小值
每次暴力修改即可
要维护东西和影响的有点多
复杂度
#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
static char ibuf[RLEN],*ib,*ob;
(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
char ch=gc();
int res=0,f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define ll long long
inline ll readl(){
char ch=gc();
ll res=0;bool f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=100005;
multiset<ll> st[N],ansmx,ansmn;
int n,q,a[N];
ll b[N],f[N],e[N],d[N];
inline ll Val(int x){
return f[x]+b[x]-e[x]*d[x]+e[x];
}
inline void reset(int x){
e[x]=b[x]/d[x];
}
inline ll mx(int x){
return st[x].size()?*(st[x].rbegin()):0;
}
inline ll mn(int x){
return st[x].size()?*(st[x].bg()):0;
}
inline ll querymx(int x){
return mx(x)+e[x];
}
inline ll querymn(int x){
return mn(x)+e[x];
}
inline void delet(int x){
if(!st[x].size())return;
ansmx.erase(ansmx.find(querymx(x)));
ansmn.erase(ansmn.find(querymn(x)));
}
inline void ins(int x){
if(!st[x].size())return;
ansmx.insert(querymx(x));
ansmn.insert(querymn(x));
}
int main(){
n=read(),q=read();
for(int i=1;i<=n;i++)b[i]=readl();
for(int i=1;i<=n;i++)a[i]=read(),d[i]++,d[a[i]]++;
for(int i=1;i<=n;i++)d[i]++,reset(i);
for(int i=1;i<=n;i++)f[a[i]]+=e[i];
for(int i=1;i<=n;i++)st[a[i]].insert(Val(i));
for(int i=1;i<=n;i++)ins(i);
while(q--){//cerr<<q<<'
';
int op=read();
if(op==1){
int u=read(),x=a[u],y=read();
delet(x),delet(a[x]);
st[x].erase(st[x].find(Val(u)));
st[a[x]].erase(st[a[x]].find(Val(x)));
delet(a[a[x]]);
st[a[a[x]]].erase(st[a[a[x]]].find(Val(a[x])));
// cerr<<"Pos 1
";
f[a[x]]-=e[x];
f[x]-=e[u],d[x]--;
reset(x);
f[a[x]]+=e[x];
// cerr<<"Pos 2
";
st[a[a[x]]].insert(Val(a[x]));
ins(a[a[x]]);
st[a[x]].insert(Val(x));
ins(a[x]);
ins(x);
// cerr<<"Pos 3
";
a[u]=y;
// cerr<<"Pos 4
";
delet(y),delet(a[y]);
st[y].insert(Val(u));
st[a[y]].erase(st[a[y]].find(Val(y)));
delet(a[a[y]]);
st[a[a[y]]].erase(st[a[a[y]]].find(Val(a[y])));
// cerr<<"Pos 5
";
f[a[y]]-=e[y];
f[y]+=e[u],d[y]++;
reset(y);
f[a[y]]+=e[y];
st[a[a[y]]].insert(Val(a[y]));
ins(a[a[y]]);
st[a[y]].insert(Val(y));
ins(a[y]);
ins(y);
}
else if(op==2){
int i=read();
cout<<Val(i)+e[a[i]]<<'
';
}
else{
cout<<(*(ansmn.bg()))<<" "<<(*(ansmx.rbegin()))<<'
';
}
}
}
T2：
原题是
显然要的是压列的状态
但是由于直接压有后效性所以只能压对角线
压一下每个位置是否被覆盖，向下还是向右可以做到
考虑直接压相邻两行对角线的状态是
但实际有用总共只有个位置
轮廓线状压即可
复杂度
#include<bits/stdc++.h>
using namespace std;
const int RLEN=(1<<20)|5;
inline char gc(){
static char ibuf[RLEN],*ib,*ob;
(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
return (ib==ob)?EOF:*ib++;
}
#define gc getchar
inline int read(){
char ch=gc();
int res=0,f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define pb push_back
#define re register
#define fi first
#define se second
#define pii pair<int,int>
#define cs const
#define bg begin
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
int mod;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int M=(1<<14)|5;
int f[2][M],g[2][M];
int n,m,lim,now;
class BearDestroys{
public :
inline int solve(){
lim=1<<(m+1);
now=0,f[0][0]=1;
for(int i=1;i<=n+m-1;i++){
now^=1;
memset(f[now],0,sizeof(f[now]));
memset(g[now],0,sizeof(g[now]));
for(int s=0;s<lim/2;s++)Add(f[now][s<<1],f[now^1][s]),Add(g[now][s<<1],g[now^1][s]);
for(int j=1;j<=m;j++){
if(j>i)continue;now^=1;
memset(f[now],0,sizeof(f[now]));
memset(g[now],0,sizeof(g[now]));
if(i-j+1>n){
for(int s=0;s<lim;s++)
Add(f[now][(s&(1<<j))?(s^(1<<j)):s],f[now^1][s]),
Add(g[now][(s&(1<<j))?(s^(1<<j)):s],g[now^1][s]);
continue;
}
for(int s=0;s<lim;s++)if(f[now^1][s]||g[now^1][s]){
if(s&(1<<j))Add(f[now][s^(1<<j)],mul(2,f[now^1][s])),Add(g[now][s^(1<<j)],mul(2,g[now^1][s]));
else{
if(i-j+1==n){
if(j==m)Add(f[now][s],mul(2,f[now^1][s])),Add(g[now][s],mul(2,g[now^1][s]));
else{
Add(f[now][s|(1<<j)],mul(f[now^1][s],2));
Add(g[now][s|(1<<j)],mul(add(g[now^1][s],f[now^1][s]),2));
}
continue;
}
if(j==m){
if(s&(1<<(j-1))){
Add(f[now][(s&(1<<j))?(s-(1<<j)):s],mul(2,f[now^1][s]));
Add(g[now][(s&(1<<j))?(s-(1<<j)):s],mul(2,g[now^1][s]));
}
else{
Add(f[now][((s&(1<<j))?(s-(1<<j)):s)|(1<<(j-1))],mul(2,f[now^1][s]));
Add(g[now][((s&(1<<j))?(s-(1<<j)):s)|(1<<(j-1))],mul(2,add(f[now^1][s],g[now^1][s])));
}
continue;
}
if(!(s&(1<<(j-1)))){
Add(f[now][s|(1<<(j-1))],f[now^1][s]);
Add(g[now][s|(1<<(j-1))],add(f[now^1][s],g[now^1][s]));
}
Add(f[now][s|(1<<j)],mul(f[now^1][s],1+(!!(s&(1<<(j-1))))));
Add(g[now][s|(1<<j)],mul(add(g[now^1][s],f[now^1][s]),1+(!!(s&(1<<(j-1))))));
}
}
}
}
int res=0;
for(int i=0;i<lim;i++)Add(res,g[now][i]);
return g[now][0];
}
inline int sumUp(int _n,int _m,int _mod){
m=_n,n=_m,mod=_mod;
return solve();
}
};
BearDestroys stargazer;
int main(){
#ifdef Stargazer
freopen("lx.cpp","r",stdin);
#endif
int n=read(),m=read(),mod=read();
cout<<stargazer.sumUp(n,m,mod);
}
T3：
广二原题…
虽然考场的时候早就忘了
乱推了一发
写了个暴力枚举子集
乱剪枝就跑过去了
数据真水
下来重新写了发复杂度正确的代码
#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
static char ibuf[RLEN],*ib,*ob;
(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
char ch=gc();
int res=0,f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
int mod;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int N=21,C=(1<<20)|5;
int n,c[N],lim,a[C];
int sum[C];
int cnt[C];
bool dfs(int sta,int num){
while(sta){
bool flag=1;
if(sum[sta]*num==sum[lim-1]*cnt[sta])return true;
for(int s=sta;s;s=sta&(s-1))
if(sum[s]*num==sum[lim-1]*(cnt[s]-1)){
flag=0,sta^=s;break;
}
if(flag)return false;
}
return true;
}
int main(){
n=read(),lim=1<<n;
for(int i=1;i<=n;i++)c[i]=read(),a[1<<(i-1)]=c[i];
if(n==1){puts("1");return 0;}
for(int i=1;i<lim;i++){
sum[i]=sum[i-(i&(-i))]+a[i&(-i)],cnt[i]=cnt[i-(i&(-i))]+1;
}
for(int i=(n+1)/2;i<=n-2;i++){
if(dfs(lim-1,i)){
cout<<i;return 0;
}
}
cout<<n-1;return 0;
}