算法笔记--单调队列优化dp

时间:2022-08-20 08:41:43

单调队列:队列中元素单调递增或递减,可以用双端队列实现(deque),队列的前面和后面都可以入队出队。

单调队列优化dp:

问题引入:

dp[i] = min( a[j] ) ,i-m < j <= i

普通的做法是O(nlogn),但是当n很大是,这个复杂度就不行了,考虑用单调队列优化来达到O(n)。

单调队列优化dp时维护的一般都是两个值{ id(下标),value(值)},且它们都保持单调。

对于这个问题,我们维护一个两个值都单调递增的序列。

查询:队首不断删除,直到队首下标大于等于i - m + 1,队首就是答案。

插入:因为要保证下标单调递增,所以从队尾加入元素a[i],因为又要保证值单调递增,所以我们不断删除队尾大于a[i]的元素,直到队尾小于a[i]或者队列为空,然后在队尾添加a[i]。

为什么我们能直接删除队尾大于a[i]的元素呢?

因为队尾删除的那些元素下标比a[i]小且值比a[i]大,如果这些元素可以是答案,那么a[i]肯定比他们好,所以这些值不会对答案产生贡献,所以直接删除就好了。

最后,每个元素最多入队一次,出队一次,复杂度为O(n)。

PS:只维护下标也可以,因为知道了下标就知道了值(如果你保存下来的话),不过如果n太大(以至于数组存不下)只能两个都维护了

例1:POJ 2823

代码:

#include<iostream>
#include<cstdio>
#include<queue>
#include<deque>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head const int N = 1e6 + ;
int ans1[N], ans2[N];
int main() {
int n, k, x;
scanf("%d %d", &n, &k);
deque<pii>mn, mx;
for (int i = ; i <= n; i++) {
scanf("%d", &x);
while(!mn.empty() && mn.back().se >= x) mn.pop_back();
mn.pb(mp(i, x));
while(!mn.empty() && mn.front().fi < i-k+) mn.pop_front();
ans1[i] = mn.front().se;
while(!mx.empty() && mx.back().se <= x) mx.pop_back();
mx.pb(mp(i, x));
while(!mx.empty() && mx.front().fi < i-k+) mx.pop_front();
ans2[i] = mx.front().se;
}
for (int i = k; i <= n; i++) printf("%d%c", ans1[i], " \n"[i==n]);
for (int i = k; i <= n; i++) printf("%d%c", ans2[i], " \n"[i==n]);
return ;
}

例2:POJ - 3017

思路:multiset + 单调队列优化

单调队列维护的是下标递增的序列,但以这些下标为下标的数列是单调递减的,对于当前的i和队列中的一个下标tmp,他在队列中的上一下标为_tmp,则区间[_tmp+1,i]中的最大值为a[tmp],即转移方程为

dp[i] = min(dp[_tmp] + a[tmp]),tmp属于单调队列,对于单调队列中的第一个元素tmp,_tmp为第一个使得sum[i] - sum[_tmp] >= m的位置

#include<iostream>
#include<cstdio>
#include<queue>
#include<deque>
#include<set>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head const int N = 1e5 + ;
LL sum[N], dp[N];
int a[N];
deque<LL>q;
multiset<LL>s;
int main() {
int n;
LL m;
bool f = false;
scanf("%d %lld", &n, &m);
for (int i = ; i <= n; i++) {
scanf("%d", &a[i]);
if(a[i] > m) f = true;
sum[i] = sum[i-] + a[i];
}
if(f) return *puts("-1");
int p = ;
for (int i = ; i <= n; i++) {
while(sum[i] - sum[p] > m) p++;
while(!q.empty() && a[q.back()] <= a[i]) {
int tmp = q.back();
q.pop_back();
if(!q.empty()) s.erase(dp[q.back()] + a[tmp]);
}
if(!q.empty()) s.insert(dp[q.back()] + a[i]);
q.push_back(i);
while(!q.empty() && sum[i] - sum[q.front()-] > m) {
int tmp = q.front();
q.pop_front();
if(!q.empty()) s.erase(dp[tmp] + a[q.front()]);
}
dp[i] = dp[p] + a[q.front()];
if(s.size() != )dp[i] = min(dp[i], *s.begin());
//cout << dp[i] << endl;
}
printf("%lld\n", dp[n]);
return ;
}

例3: POJ 2373

思路:

dp[0] = 0

dp[i] = min {dp[j] + 1} 2*a <= i-j <= 2*b

喷水半径R为整数,L为偶数,所以洒水机在整数点上

#include<iostream>
#include<cstdio>
#include<queue>
#include<deque>
#include<set>
#include<cstring>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head const int N = 1e6 + ;
int dp[N], cnt[N], pre[N], suf[N];
deque<int>q;
int main() {
int n, L, a, b, l, r;
while( ~scanf("%d %d", &n, &L)) {
mem(cnt, );
scanf("%d %d", &a, &b);
a *= ;
b *= ;
for (int i = ; i < n; i++) {
scanf("%d %d", &l, &r);
cnt[l+]++;
cnt[r]--;
}
for (int i = ; i <= L; i++) cnt[i] += cnt[i-];
q.clear();
q.push_back();
for (int i = ; i <= L; i += ) {
if(cnt[i]) continue;
while(!q.empty() && (i-q.front()) > b) {
q.pop_front();
}
if(q.empty() || i - q.front() < a) continue;
dp[i] = dp[q.front()] + ; while(!q.empty() && dp[q.back()] > dp[i]) q.pop_back();
q.push_back(i);
}
if(dp[L])printf("%d\n", dp[L]);
else printf("-1\n");
}
return ;
}

例4: 5429 多重背包

思路: 单调队列优化多重背包

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head const int N = 7e3 + ;
int dp[N];
int v[N], w[N], c[N];
int main() {
int n, m;
scanf("%d %d", &n, &m);
for (int i = ; i <= n; i++) scanf("%d %d %d", &v[i], &w[i], &c[i]);
for (int i = ; i <= n; i++) {
for (int p = ; p < v[i]; p++) {
int cnt = c[i];
deque<pii> q;
q.push_back({dp[p] + cnt*w[i], });
for (int tot = ; p + tot*v[i] <= m; tot ++) {
cnt--;
while(!q.empty() && tot - q.front().se > c[i]) q.pop_front();
while(!q.empty() && q.back().fi <= dp[p+tot*v[i]] + cnt*w[i]) q.pop_back();
q.push_back({dp[p+tot*v[i]] + cnt*w[i], tot});
dp[p+tot*v[i]] = q.front().fi + (tot - c[i]) * w[i];
}
}
}
printf("%d\n", dp[m]);
return ;
}