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 + 5; int ans1[N], ans2[N]; int main() { int n, k, x; scanf("%d %d", &n, &k); deque<pii>mn, mx; for (int i = 1; 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+1) 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+1) mx.pop_front(); ans2[i] = mx.front().se; } for (int i = k; i <= n; i++) printf("%d%c", ans1[i], " "[i==n]); for (int i = k; i <= n; i++) printf("%d%c", ans2[i], " "[i==n]); return 0; }
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 + 5; 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 = 1; i <= n; i++) { scanf("%d", &a[i]); if(a[i] > m) f = true; sum[i] = sum[i-1] + a[i]; } if(f) return 0*puts("-1"); int p = 0; for (int i = 1; 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()-1] > 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() != 0)dp[i] = min(dp[i], *s.begin()); //cout << dp[i] << endl; } printf("%lld ", dp[n]); return 0; }
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 + 5; 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, 0); scanf("%d %d", &a, &b); a *= 2; b *= 2; for (int i = 0; i < n; i++) { scanf("%d %d", &l, &r); cnt[l+1]++; cnt[r]--; } for (int i = 1; i <= L; i++) cnt[i] += cnt[i-1]; q.clear(); q.push_back(0); for (int i = 2; i <= L; i += 2) { 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()] + 1; while(!q.empty() && dp[q.back()] > dp[i]) q.pop_back(); q.push_back(i); } if(dp[L])printf("%d ", dp[L]); else printf("-1 "); } return 0; }
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 + 5; int dp[N]; int v[N], w[N], c[N]; int main() { int n, m; scanf("%d %d", &n, &m); for (int i = 1; i <= n; i++) scanf("%d %d %d", &v[i], &w[i], &c[i]); for (int i = 1; i <= n; i++) { for (int p = 0; p < v[i]; p++) { int cnt = c[i]; deque<pii> q; q.push_back({dp[p] + cnt*w[i], 0}); for (int tot = 1; 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 ", dp[m]); return 0; }