【BZOJ】【P3831】【Poi2014】【Little Bird】【题解】【单调队列+DP】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=3831

dp显然

f[i]=min(f[j]+(d[j]<=d[i])) j∈[i-k,i-1]

单调队列优化

最优解有可能是f[j]或f[k]+1之类的,先比较f,再比较d,它们是一样的(好像讲的不清楚= =、)

Code:

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
deque<int>dq;
int d[maxn],n,f[maxn];
int getint(){
int res=0;char c=getchar();
while(!isdigit(c))c=getchar();
while(isdigit(c))res=res*10+c-'0',c=getchar();
return res;
}
bool bigger(int x,int y){return f[x]==f[y]?d[x]<d[y]:f[x]>f[y];}
int main(){
n=getint();
for(int i=1;i<=n;i++)d[i]=getint();
int q=getint();
while(q--){
int k=getint();dq.clear();
for(int i=1;i<=n;i++){
while(dq.size()&&dq.front()+k<i)dq.pop_front();
if(i-1)f[i]=f[dq.front()]+(d[dq.front()]<=d[i]);
while(dq.size()&&bigger(dq.back(),i))dq.pop_back();
dq.push_back(i);
}printf("%d\n",f
);
}
return 0;
}