lsum 、lmax 、rsum 、 rmax 都是代表着从最左边开始或者从最右边开始
第一类:
仅仅要求求解的区间要求是连续的
我们考虑设置三个变量 lsum , rsum , sum 分别记录从左边开始的连续区间大小,从右边开始的连续区间的大小,整个区间的连续区间的大小的最大值
然后我们采取线段树去维护这三个变量就可以了
因为我们要求的是连续的区间,所以我们 pushup 操作的时候也需要确保我们的区间是连续的
sum 维护的是整个范围内最大的连续区间 : 左右两个区间最大的或者是在中间的某一段
这一类连续区间的求解问题的典型例题:
P2894 [USACO08FEB]Hotel G
题目链接:https://www.luogu.com.cn/problem/P2894
#include <algorithm> #include <string> #include <cstring> #include <vector> #include <map> #include <stack> #include <set> #include <queue> #include <cmath> #include <cstdio> #include <iomanip> #include <ctime> #include <bitset> #include <cmath> #include <sstream> #include <iostream> #include <unordered_map> #define ll long long #define ull unsigned long long #define ls nod<<1 #define rs (nod<<1)+1 #define pii pair<int,int> #define mp make_pair #define pb push_back #define INF 0x3f3f3f3f3f3f3f3f #define max(a, b) (a>b?a:b) #define min(a, b) (a<b?a:b) const double eps = 1e-10; const int maxn = 5e4 + 10; const ll MOD = 99999999999999; int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; } using namespace std; int sum[maxn << 2],lazy[maxn << 2]; int lsum[maxn << 2],rsum[maxn << 2]; void build(int l,int r,int nod) { sum[nod] = lsum[nod] = rsum[nod] = (r-l+1); lazy[nod] = -1; if (l == r) return ; int mid = (l + r) >> 1; build(l,mid,ls); build(mid+1,r,rs); } void pushup(int l,int r,int nod) { int mid = (l + r) >> 1; sum[nod] = max(max(sum[ls],sum[rs]),rsum[ls] + lsum[rs]); lsum[nod] = lsum[ls]; rsum[nod] = rsum[rs]; if (lsum[ls] == (mid - l + 1)) lsum[nod] = lsum[ls] + lsum[rs]; if (rsum[rs] == (r - mid)) rsum[nod] = rsum[rs] + rsum[ls]; } void pushdown(int l,int r,int nod) { int mid = (l + r) >> 1; sum[ls] = lsum[ls] = rsum[ls] = (lazy[nod]) * (mid - l + 1); sum[rs] = lsum[rs] = rsum[rs] = (lazy[nod]) * (r - mid); lazy[ls] =lazy[rs] = lazy[nod]; lazy[nod] = -1; } void modify(int l,int r,int x,int y,int z,int nod) { if (x <= l && y >= r) { sum[nod] = lsum[nod] = rsum[nod] = (r-l+1)*z; lazy[nod] = z; return ; } if (lazy[nod] != -1) { pushdown(l,r,nod); } int mid = (l + r) >> 1; if (x <= mid) modify(l,mid,x,y,z,ls); if (y > mid) modify(mid+1,r,x,y,z,rs); pushup(l,r,nod); } int query(int l,int r,int siz,int nod) { int mid = (l + r) >> 1; if (l == r && siz == 1) return l; if (lazy[nod] != -1) pushdown(l,r,nod); if (sum[nod] >= siz) { if (sum[ls] >= siz) { return query(l,mid,siz,ls); } else if (rsum[ls] + lsum[rs] >= siz) { return mid + 1 - rsum[ls]; } else return query(mid+1,r,siz,rs); } return 0; } int main() { ios::sync_with_stdio(false); int n,m; cin >> n >> m; build(1,n,1); while (m--) { int op,x,y; cin >> op; if (op == 1) { cin >> x; int l = query(1,n,x,1); cout << l << endl; if (!l) continue; modify(1,n,l,l+x-1,0,1); } else { cin >> x >> y; modify(1,n,x,x+y-1,1,1); } } return 0; }
第二类:
要求求解的连续区间是最大的
因为要求的连续区间是最大的,所以我们需要维护四个变量 sum , lmax , rmax , ans 分别代表区间和,左边连续区间的最大值,右边连续区间的最大值,整个区间的连续区间的最大值
因为我们要求的是连续的区间,所以我们 pushup 操作的时候也需要确保我们的区间是连续的
lmax 维护: 左区间的lmax和整个左区间 + 右区间的lmax 中的最大值
这类问题的典型例题:
P4513 小白逛公园
题目链接:https://www.luogu.com.cn/problem/P4513
#include <algorithm> #include <string> #include <cstring> #include <vector> #include <map> #include <stack> #include <set> #include <queue> #include <cmath> #include <cstdio> #include <iomanip> #include <ctime> #include <bitset> #include <cmath> #include <sstream> #include <iostream> #include <unordered_map> #define ll long long #define ull unsigned long long #define ls nod<<1 #define rs (nod<<1)+1 #define pii pair<int,int> #define mp make_pair #define pb push_back #define INF 0x3f3f3f3f #define max(a, b) (a>b?a:b) #define min(a, b) (a<b?a:b) const double eps = 1e-10; const int maxn = 5e5 + 10; const ll MOD = 99999999999999; int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; } using namespace std; // #define int ll int a[maxn]; struct segment_tree { int l,r; int sum,lmax,rmax,ans; }tree[maxn << 2]; void pushup(int nod) { tree[nod].sum = tree[ls].sum + tree[rs].sum; tree[nod].lmax = max(tree[ls].lmax,tree[ls].sum + tree[rs].lmax); tree[nod].rmax = max(tree[rs].rmax,tree[rs].sum + tree[ls].rmax); tree[nod].ans = max(max(tree[ls].ans,tree[rs].ans),tree[ls].rmax + tree[rs].lmax); } void build(int l,int r,int nod) { tree[nod].l = l,tree[nod].r = r; if (l == r) { tree[nod].ans = tree[nod].lmax = tree[nod].rmax = tree[nod].sum = a[l]; return ; } int mid = (l + r) >> 1; build(l,mid,ls); build(mid+1,r,rs); pushup(nod); } void modify(int k,int z,int nod) { int l = tree[nod].l,r = tree[nod].r; if (l == r) { tree[nod].ans = tree[nod].lmax = tree[nod].rmax = tree[nod].sum = z; return ; } int mid = (l + r) >> 1; if (k <= mid) modify(k,z,ls); if (k > mid) modify(k,z,rs); pushup(nod); } segment_tree query(int x,int y,int nod) { int l = tree[nod].l,r = tree[nod].r; if (x <= l && y >= r) return tree[nod]; int mid = (l + r) >> 1; if (y <= mid) return query(x,y,ls); else if (x > mid) return query(x,y,rs); else { segment_tree le = query(x,y,ls),ri = query(x,y,rs),end; end.sum = le.sum + ri.sum; end.lmax = max(le.lmax,le.sum + ri.lmax); end.rmax = max(ri.rmax,ri.sum + le.rmax); end.ans = max(max(le.ans,ri.ans),le.rmax + ri.lmax); return end; } } int main() { ios::sync_with_stdio(false); int n,m; cin >> n >> m; for (int i = 1;i <= n;i++) cin >> a[i]; build(1,n,1); while (m--) { int op,x,y; cin >> op >> x >> y; if (op == 1 && x > y) swap(x,y); if (op == 1) cout << query(x,y,1).ans << endl; else modify(x,y,1); } return 0; }