[DP || 贪心 链表] BZOJ 4576 [Usaco2016 Open]262144 & BZOJ 4580 [Usaco2016 Open]248

有意思的一道题

官网上的做法是f[i][j]表示从j出发能够合并出i的末端 不存在则为0

然后就可以转移了

这有什么好呢 短

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc());
}

const int N=300005;

int n,a
;
int f[105]
;

int main(){
int ans=0;
freopen("two.in","r",stdin);
freopen("two.out","w",stdout);
read(n);
for (int i=1;i<=n;i++)
read(a[i]);
for (int i=1;i<=70;i++)
for (int j=1;j<=n;j++){
f[i][j]=a[j]==i?j:(f[i-1][j]?f[i-1][f[i-1][j]+1]:0);
if (f[i][j])
ans=max(ans,i);
}
printf("%d\n",ans);
return 0;
}

或者可以直接贪心 用链表维护

把能连续偶数个合并的直接合并 

要是是奇数个的话 把链断开 合并后两边都分到新的这个块

虽说复杂度官网上说是nlogn 但实际表现很好

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
typedef pair<int,int> abcd;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=300005;
const int M=500005;

int n,ans,a
;
int pre[M],nxt[M];
abcd A[M]; int cnt;
int del[M];
vector<int> Vec[105];

inline void Solve(int x){
if (del[x]) return;
int Pre=pre[x],Nxt=nxt[x];
if (A[x].second==1){
if (Pre) nxt[Pre]=0;
if (Nxt) pre[Nxt]=0;
}else if (~A[x].second&1){
A[x]=abcd(A[x].first+1,A[x].second>>1);
ans=max(ans,A[x].first);
while (Pre && A[Pre].first==A[x].first) del[Pre]=1,A[x].second+=A[Pre].second,Pre=pre[Pre];
while (Nxt && A[Nxt].first==A[x].first) del[Nxt]=1,A[x].second+=A[Nxt].second,Nxt=nxt[Nxt];
pre[x]=Pre; if (Pre) nxt[Pre]=x;
nxt[x]=Nxt; if (Nxt) pre[Nxt]=x;
Vec[A[x].first].push_back(x);
}else{
A[x]=A[++cnt]=abcd(A[x].first+1,A[x].second>>1);
ans=max(ans,A[x].first);
while (Pre && A[Pre].first==A[x].first) del[Pre]=1,A[x].second+=A[Pre].second,Pre=pre[Pre];
while (Nxt && A[Nxt].first==A[cnt].first) del[Nxt]=1,A[cnt].second+=A[Nxt].second,Nxt=nxt[Nxt];
pre[x]=Pre; nxt[x]=0;
if (Pre) nxt[Pre]=x;
nxt[cnt]=Nxt; pre[cnt]=0;
if (Nxt) pre[Nxt]=x;
Vec[A[x].first].push_back(x);
Vec[A[x].first].push_back(cnt);
}
}

int main(){
freopen("two.in","r",stdin);
freopen("two.out","w",stdout);
read(n);
for (int i=1;i<=n;i++){
read(a[i]); ans=max(ans,a[i]);
if (a[i]!=a[i-1]) A[++cnt]=abcd(a[i],1); else A[cnt].second++;
}
for (int i=1;i<=cnt;i++)
Vec[A[i].first].push_back(i),pre[i]=i-1,nxt[i]=i+1;
pre[1]=0; nxt[cnt]=0;
for (int i=1;i<=70;i++)
for (int j=0;j<(signed)Vec[i].size();j++)
Solve(Vec[i][j]);
printf("%d\n",ans);
return 0;
}

还有用单调栈维护贪心的：http://blog.csdn.net/aarongzk/article/details/51712701