Problem Description
answer
\(\mathrm{Subtask 1}\)
The problem is not seeking a maximum decrease sequences found \ (n-\ Le 500 \) , direct \ (O (n ^ 2) \) violence to DP.
\(\mathrm{Subtask 2}\)
Set \ (opt_i \) representative of the interval \ ([. 1, I] \) , and to \ (I \) is the end of the longest sequence is not reduced.
Consider split point, the \ (I \) is split into \ (I \) and \ (n-I + \) .
If \ (. 1 opt_i = \) , from the source node to \ (I \) connected edges.
If \ (n-opt_i = \) , from \ (i + n \) to sink even edges.
Collage right above two are \ (INF \) .
Next, from the (I \) \ to \ (i + n \) connected side, the right side of \ (1 \) , representing each number can only be used once.
Next enumeration \ (I, J \) , and \ (I <J \) , \ (\ FORALL a_j \ Le a_i \) and \ (. 1 + opt_i opt_j = \) , in \ (j + n, i \) is connected between the side, the right side is \ (1 \) .
Running \ (\ mathrm {Dinic} \ ) can.
\(\mathrm{Subtask 3}\)
Only need to lift the \ (1 \) and (n \) \ traffic restrictions can be.
\(\mathrm{Code}\)
#include<bits/stdc++.h>
using namespace std;
template <typename Tp>
void read(Tp &x){
x=0;char ch=1;int fh;
while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
if(ch=='-'){
fh=-1;ch=getchar();
}
else fh=1;
while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+ch-'0';
ch=getchar();
}
x*=fh;
}
const int maxn=507;
int n,a[maxn],len=1;
int opt[maxn],ans;
int S,T;
int Head[20000],v[200000],w[200000],tot=1;
int d[20000],Next[200000];
bool bfs(){
memset(d,0,sizeof(d));
queue<int>q;q.push(S);d[S]=1;
while(!q.empty()){
int x=q.front();q.pop();
for(int i=Head[x];i;i=Next[i]){
if(d[v[i]]||!w[i]) continue;
q.push(v[i]);d[v[i]]=d[x]+1;
if(v[i]==T) return true;
}
}
return false;
}
int dfs(int x,int flow){
if(x==T) return flow;
int rest=flow;
for(int i=Head[x];i&&rest;i=Next[i]){
if(d[v[i]]!=d[x]+1||!w[i]) continue;
int k=dfs(v[i],min(rest,w[i]));
if(!k) d[v[i]]=0;
else{
w[i]-=k,w[i xor 1]+=k;
rest-=k;
}
}
return flow-rest;
}
void add(int x,int y,int z){v[++tot]=y,Next[tot]=Head[x],Head[x]=tot,w[tot]=z;}
int main(){
read(n);
for(int i=1;i<=n;i++){
read(a[i]);opt[i]=1;
}
for(int i=1;i<=n;i++){
for(int j=1;j<i;j++){
if(a[i]>=a[j]){
opt[i]=max(opt[i],opt[j]+1);
len=max(opt[i],len);
}
}
}
printf("%d\n",len);S=n*2+1,T=S+1;
for(int i=1;i<=n;i++){
if(opt[i]==1) add(S,i,1),add(i,S,0);
}
for(int i=1;i<=n;i++){
if(opt[i]==len) add(i+n,T,1),add(T,i+n,0);
}
for(int i=1;i<=n;i++) add(i,i+n,1),add(i+n,i,0);
for(int i=1;i<=n;i++){
for(int j=1;j<i;j++){
if(a[i]>=a[j]&&opt[i]==opt[j]+1){
add(j+n,i,1);add(i,j+n,0);
}
}
}
while(bfs()){
int t;
while(t=dfs(S,0x3f3f3f3f)) ans+=t;
}
printf("%d\n",ans);
add(1,1+n,0x3f3f3f3f);add(n+1,1,0);
add(S,1,0x3f3f3f3f);add(1,S,0);
if(opt[n]==len) add(n,n+n,0x3f3f3f3f),add(2*n,n,0),add(n*2,T,0x3f3f3f3f),add(T,n*2,0);
while(bfs()){
ans+=dfs(S,0x3f3f3f3f);
}
printf("%d\n",ans);
return 0;
}