多校联合练习赛1 Problem1005 Deque LIS+LDS 再加一系列优化
Deque
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 731 Accepted Submission(s): 236
Problem Description
Today, the teacher gave Alice extra homework for the girl weren't attentive in his class. It's hard, and Alice is going to turn to you for help.
The teacher gave Alice a sequence of number(named A) and a deque. The sequence exactly contains N integers. A deque is such a queue, that one is able to push or pop the element at its front end or rear end. Alice was asked to take out the elements from the sequence in order(from A_1 to A_N), and decide to push it to the front or rear of the deque, or drop it directly. At any moment, Alice is allowed to pop the elements on the both ends of the deque. The only limit is, that the elements in the deque should be non-decreasing.
Alice's task is to find a way to push as many elements as possible into the deque. You, the greatest programmer, are required to reclaim the little girl from despair.
The teacher gave Alice a sequence of number(named A) and a deque. The sequence exactly contains N integers. A deque is such a queue, that one is able to push or pop the element at its front end or rear end. Alice was asked to take out the elements from the sequence in order(from A_1 to A_N), and decide to push it to the front or rear of the deque, or drop it directly. At any moment, Alice is allowed to pop the elements on the both ends of the deque. The only limit is, that the elements in the deque should be non-decreasing.
Alice's task is to find a way to push as many elements as possible into the deque. You, the greatest programmer, are required to reclaim the little girl from despair.
Input
The first line is an integer T(1≤T≤10) indicating the number of test cases.
For each case, the first line is the length of sequence N(1≤N≤100000).
The following line contains N integers A 1,A 2,…,A N.
For each case, the first line is the length of sequence N(1≤N≤100000).
The following line contains N integers A 1,A 2,…,A N.
Output
For each test case, output one integer indicating the maximum length of the deque.
Sample Input
3 7 1 2 3 4 5 6 7 5 4 3 2 1 5 5 5 4 1 2 3
Sample Output
7 5 3
Source
Recommend
liuyiding
我直接附上标准题解吧。。当时是看出这是最长下降跟最长上升序列的求和的，只是因为没相处nlogn算法，最终没有做出来。 今天终于AC掉了。 刚开始考虑的时候，还是没有考虑到
5
5 5 5 5 5 这种需要过滤掉重复的方法，导致WA了很多次，后来多开了一个数组保留才AC掉。
标准题解是：
Problem E. Deque考虑题目的一个简化版本:使双端队列单调上升。对于序列 A 和队列 Q,找到队列中最早出现的数字Ax,则Ax将 Q 分成的两个部分分别是原序列中以Ax开始的最长上升和最长下降序列,答案即为这两者之和的最大值。而对于本题,由于存在相同元素,所以只要找到以Ax为起点的最长不下降序列和最长不上升序列的和,然后减去两个里面出现Ax 次数的最小值即可。
我的代码是：
/*
* @author ipqhjjybj
* @date 20130723
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define MAXN 100005
#define clr(x,k) memset((x),(k),sizeof(x))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
int n;
int a[MAXN];
int d_p[MAXN]; //存储递减序列的数字
int d_f[MAXN];
int n_d_f[MAXN];//相同数字出现次数
int i_p[MAXN]; //存储递增序列的数字
int i_f[MAXN];
int n_i_f[MAXN];//相同数字出现次数
int find_dec(int l,int r,int x){
int i=l,j=r,mid;
while(i<=j){
mid=(i+j)>>1;
if(d_p[mid]<=x) i=mid+1;
else j=mid-1;
}return i;
}
int find_inc(int l,int r,int x){
int i = l,j=r,mid;
while(i<=j){
mid=(i+j)>>1;
if(i_p[mid]>=x) i=mid+1;
else j=mid-1;
}return i;
}
int main(){
//freopen("1005.in","r",stdin);
int z;
scanf("%d",&z);
while(z--){
clr(n_i_f,0),clr(n_d_f,0);
scanf("%d",&n);
for(int i = 1;i <= n;i++)
scanf("%d",&a[i]);
int ans=1;
int d_ans=1;
d_p[1]=a[n];
d_f[n]=1;
n_d_f[1]=1;
int i_ans=1;
i_p[1]=a[n];
i_f[n]=1;
n_i_f[1]=1;
for(int i=n-1;i>0;i--){
int t1=find_dec(1,d_ans,a[i]);
if(t1>d_ans)d_ans++;
d_p[t1]=a[i];
n_d_f[t1]=(t1>1&&d_p[t1-1]==a[i])?n_d_f[t1-1]+1:1;
d_f[i]=t1;
int t2=find_inc(1,i_ans,a[i]);
if(t2>i_ans)i_ans++;
i_p[t2]=a[i];
n_i_f[t2]=(t2>1&&i_p[t2-1]==a[i])?n_i_f[t2-1]+1:1;
i_f[i]=t2;
//printf("m_t1=%d m_t2=%d\n",m_t1,m_t2);
int temp = t1+t2-min(n_d_f[t1],n_i_f[t2]);
ans = max(ans,temp);
}
printf("%d\n",ans);
}
return 0;
}
325MS过掉。。
附上标准程序
#include "iostream"
#include "cstring"
#include "cstdio"
#include "vector"
#include "algorithm"
#include "map"
using namespace std;
const int N = 100010;
int a[N];
int num_up[N],num_down[N];
int dp_up[N],dp_down[N];
int n;
void getdp(int dp[],int num[])
{
dp[n]=1;
vector<int> v;
v.push_back(a[n]);
vector<int>::iterator iter;
for(int i=n-1;i>=1;i--){
int sz=v.size();
if(a[i]>v[sz-1]){
v.push_back(a[i]);
dp[i]=sz+1;
num[i]=1;
}else if(a[i]==v[sz-1]){
iter=upper_bound(v.begin(),v.end(),a[i]);
dp[i]=iter-v.begin()+1;
v.push_back(a[i]);
pair<vector<int>::iterator,vector<int>::iterator> bounds;
bounds=equal_range(v.begin(),v.end(),a[i]);
num[i]=bounds.second-bounds.first;
}else{
iter=upper_bound(v.begin(),v.end(),a[i]);
dp[i]=iter-v.begin()+1;
*iter=a[i];
pair<vector<int>::iterator,vector<int>::iterator> bounds;
bounds=equal_range(v.begin(),v.end(),a[i]);
num[i]=bounds.second-bounds.first;
}
}
}
void debug(int a[])
{
for(int i=1;i<=n;i++){
printf("%d ",a[i]);
}
printf("\n");
}
int main(void)
{
int T;
scanf("%d",&T);
while(T--){
int ans=0;
scanf("%d",&n);
map<int,int> mp;
mp.clear();
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
getdp(dp_up,num_up);
for(int i=1;i<=n;i++){
a[i]=-a[i];
}
getdp(dp_down,num_down);
for(int i=1;i<=n;i++){
mp[a[i]]++;
ans=max(ans,dp_down[i]+dp_up[i]-min(num_up[i],num_down[i]));
}
printf("%d\n",ans);
}
return 0;
}
标准程序用了986MS。。。可能我的程序优化的还好点吧。。。
总而言之是水平不够。。当时没想出nlogn算法。。