线段树维护(最大区间和，最大子段和，最长连续上升子序列）

本文主要介绍用线段树来维护(最大区间和，最大子段和，最长连续上升子序列）的问题。

HDU 1540 Tunnel Warfare（最长连续区间+单点修改）

洛谷 P2894 [USACO08FEB]酒店Hotel（最长连续区间+区间修改）

吉首大学2019年程序设计竞赛-白山茶与红玫瑰（最长连续区间+区间修改）

SPOJ - GSS1 Can you answer these queries I（最大子段和）

HDU3308 LCIS（区间最长连续上升子序列）

HDU 1540 Tunnel Warfare（最长连续区间+单点修改）

题意：有三种操作：

操作一：某个村庄被毁灭。

操作二：给出一个村庄的坐标，求包含包含这个村庄的的最长未被村庄的长度。

操作三：最后一个被摧毁的村庄被修复。

题解：首先我们可以想到用线段树来维护最长连续区间长度，我们现在可以用1代表改点没有被破坏，用0表示改点被破坏，然后用一个栈或数组来装上次破坏的点。然后就是线段树维护1的最长长度了，详解请看代码。

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>const int maxn = 5e4 + 5;
const int mod = 10007;
const int inf = 1e9;
const long long onf = 1e18;
#define me(a, b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI 3.14159265358979323846
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
int ls[maxn << 2], rs[maxn << 2], sum[maxn << 2];
///sum表示区间最长连续序列，ls表示从左端点开始最长连续前缀，rs表示从r开始最长后缀
int n, m;void push_up(int l, int r, int rt) {int mid = (l + r) >> 1;ls[rt] = ls[rt << 1];if (ls[rt << 1] == mid - l + 1)///要是左区间最长长度等于左区间长度，那么区间最长前缀长度要加上右儿子区间的最长前缀ls[rt] += ls[rt << 1 | 1];rs[rt] = rs[rt << 1 | 1];if (rs[rt << 1 | 1] == r - mid)///与上面情况一样rs[rt] += rs[rt << 1];sum[rt] = max(ls[rt], rs[rt]);sum[rt] = max(sum[rt], max(sum[rt << 1], sum[rt << 1 | 1]));sum[rt] = max(sum[rt], rs[rt << 1] + ls[rt << 1 | 1]);///区间的最长连续长度为，左区间，右区间最长连续长度，和左区间的最长后缀+右区间的最长前缀的三者最大。（最后一种情况表示最长连续区间在中间）
}void build(int l, int r, int rt) {if (l == r) {ls[rt] = rs[rt] = sum[rt] = 1;///初始化所有点都没有被破坏return;}int mid = (l + r) >> 1;build(lson);build(rson);push_up(l, r, rt);
}void push_date(int pos, int x, int l, int r, int rt) {sum[rt] = ls[rt] = rs[rt] = 0;if (l == r) {sum[rt] = ls[rt] = rs[rt] = x;return;}int mid = (l + r) >> 1;if (pos <= mid)push_date(pos, x, lson);elsepush_date(pos, x, rson);push_up(l, r, rt);
}int query(int pos, int l, int r, int rt) {if (l == r)return sum[rt];int mid = (l + r) >> 1;if (pos <= mid) {///点在左儿子区间里if (pos >= mid - rs[rt << 1] + 1)return ls[rt << 1 | 1] + rs[rt << 1];///要是点正好在左儿子的最长后缀中，肯定最长区间和就是左儿子的最长后缀长度+右儿子的最长前缀（左儿子的最长后缀与右儿子的最长前缀是连着的）elsereturn query(pos, lson);///要是没有落在最长后缀中，就不用考虑是否加上右儿子的最长前缀，直接往下找就行了} else {if (pos <= mid + ls[rt << 1 | 1])///与上面情况类似return ls[rt << 1 | 1] + rs[rt << 1];elsereturn query(pos, rson);}
}int main() {while (scanf("%d%d", &n, &m) != EOF) {build(1, n, 1);stack<int> q;while (m--) {char str[2];int x;scanf("%s", str);if (str[0] == 'D') {scanf("%d", &x);push_date(x, 0, 1, n, 1);q.push(x);} else if (str[0] == 'Q') {scanf("%d", &x);printf("%d\n", query(x, 1, n, 1));} else {if (q.size()) {push_date(q.top(), 1, 1, n, 1);q.pop();}}}}return 0;
}

洛谷 P2894 [USACO08FEB]酒店Hotel（最长连续区间+区间修改）

题意：有n代表有n个房间，编号为1-n，开始都为空房，有m行操作，以下每行先输入一个数 i ，表示一种操作：

若i为1，表示查询房间，再输入一个数x，表示在1-n 房间中找到长度为x的连续空房，输出连续x个房间中左端的房间号，尽量让这个房间号最小，若找不到长度为x的连续空房，输出0。

若i为2，表示退房，再输入两个数 x，y 代表让房间号 x-x+y-1 的房间为空。

题解：这是一个最长连续区间，区间修改的裸题，只是注意找连续区间最左端的房间号（可以好好理解下下面代码的Find_pos函数）。

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>const int maxn = 5e4 + 5;
const int mod = 10007;
const int inf = 1e9;
const long long onf = 1e18;
#define me(a, b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI 3.14159265358979323846
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
int ls[maxn << 2], rs[maxn << 2], sum[maxn << 2];
int lazy[maxn << 2];
int n, m;void push_up(int l, int r, int rt) {int mid = (l + r) >> 1;ls[rt] = ls[rt << 1];if (ls[rt << 1] == mid - l + 1)ls[rt] += ls[rt << 1 | 1];rs[rt] = rs[rt << 1 | 1];if (rs[rt << 1 | 1] == r - mid)rs[rt] += rs[rt << 1];sum[rt] = max(ls[rt], rs[rt]);sum[rt] = max(sum[rt], max(sum[rt << 1], sum[rt << 1 | 1]));sum[rt] = max(sum[rt], rs[rt << 1] + ls[rt << 1 | 1]);
}void push_down(int l, int r, int rt) {if (lazy[rt] != -1) {int mid = (l + r) >> 1;lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];sum[rt << 1] = ls[rt << 1] = rs[rt << 1] = (mid - l + 1) * lazy[rt];sum[rt << 1 | 1] = ls[rt << 1 | 1] = rs[rt << 1 | 1] = (r - mid) * lazy[rt];lazy[rt] = -1;}
}void build(int l, int r, int rt) {sum[rt] = ls[rt] = rs[rt] = 0;if (l == r) {sum[rt] = ls[rt] = rs[rt] = 1;return;}int mid = (l + r) >> 1;build(lson);build(rson);push_up(l, r, rt);
}void push_date(int L, int R, int opt, int l, int r, int rt) {if (L <= l && R >= r) {sum[rt] = ls[rt] = rs[rt] = (r - l + 1) * opt;lazy[rt] = opt;return;}push_down(l, r, rt);int mid = (l + r) >> 1;if (R <= mid)push_date(L, R, opt, lson);else if (L > mid)push_date(L, R, opt, rson);else {push_date(L, mid, opt, lson);push_date(mid + 1, R, opt, rson);}push_up(l, r, rt);
}int Find_pos(int len, int l, int r, int rt) {if(l==r)return l;push_down(l, r, rt);int mid = (l + r) >> 1;if(len<=sum[rt<<1])return Find_pos(len,lson);else if(rs[rt<<1]+ls[rt<<1|1]>=len)return mid-rs[rt<<1]+1;elsereturn Find_pos(len,rson);
}int main() {me(lazy, -1);scanf("%d%d", &n, &m);build(1, n, 1);while (m--) {int opt, l, len;scanf("%d", &opt);if (opt == 1) {scanf("%d", &len);if (len <= sum[1]) {int pos = Find_pos(len, 1, n, 1);printf("%d\n", pos);push_date(pos, pos+len-1, 0, 1, n, 1);} elseprintf("0\n");} else {scanf("%d%d", &l, &len);push_date(l, l + len - 1, 1, 1, n, 1);}}return 0;
}

吉首大学2019年程序设计竞赛-白山茶与红玫瑰（最长连续区间+区间修改）

题意：现在给出一段序列，只有0和1，现在有两种操作：

操作一：给出[l,r]，求出该区间内最长连续0区间的长度。

操作二：给出[l,r]，将该区间的0全部变成1,1全部变成0。

题解：现在建立两棵线段树，一个保存最长连续1区间长度，一个保存最长连续0区间长度，在执行操作一：就是常规求法就行了，在进行操作二时，将对应区间的对应数组的值全部交换，这样两者刚好反过来，就是我们要求的值。

注意：因为翻转两次就等于没有翻转，所以lazy数组，初值为0，1表示要翻转，然后每次修改都将对应lazy数组值异或1就行了。

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>const int maxn = 1e5 + 5;
const int mod = 10007;
const int inf = 1e9;
const long long onf = 1e18;
#define me(a, b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
#define lson L,mid,rt<<1
#define rson mid+1,R,rt<<1|1
#define PI 3.14159265358979323846
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
int ls1[maxn << 2], rs1[maxn << 2], sum1[maxn << 2];
int ls2[maxn << 2], rs2[maxn << 2], sum2[maxn << 2];
int lazy[maxn << 2];
int n, m, x;void push_up(int l, int r, int rt) {int mid = (l + r) >> 1;ls1[rt] = ls1[rt << 1];if (ls1[rt << 1] == mid - l + 1)ls1[rt] += ls1[rt << 1 | 1];rs1[rt] = rs1[rt << 1 | 1];if (rs1[rt << 1 | 1] == r - mid)rs1[rt] += rs1[rt << 1];sum1[rt] = max(sum1[rt << 1], sum1[rt << 1 | 1]);sum1[rt] = max(sum1[rt], max(ls1[rt], rs1[rt]));sum1[rt] = max(sum1[rt], rs1[rt << 1] + ls1[rt << 1 | 1]);ls2[rt] = ls2[rt << 1];if (ls2[rt << 1] == mid - l + 1)ls2[rt] += ls2[rt << 1 | 1];rs2[rt] = rs2[rt << 1 | 1];if (rs2[rt << 1 | 1] == r - mid)rs2[rt] += rs2[rt << 1];sum2[rt] = max(sum2[rt << 1], sum2[rt << 1 | 1]);sum2[rt] = max(sum2[rt], max(ls2[rt], rs2[rt]));sum2[rt] = max(sum2[rt], rs2[rt << 1] + ls2[rt << 1 | 1]);
}void push_down(int rt) {if (lazy[rt]) {lazy[rt << 1] ^= 1;lazy[rt << 1 | 1] ^= 1;swap(ls1[rt << 1], ls2[rt << 1]);swap(rs1[rt << 1], rs2[rt << 1]);swap(sum1[rt << 1], sum2[rt << 1]);swap(ls1[rt << 1 | 1], ls2[rt << 1 | 1]);swap(rs1[rt << 1 | 1], rs2[rt << 1 | 1]);swap(sum1[rt << 1 | 1], sum2[rt << 1 | 1]);lazy[rt] = 0;}
}void build(int l, int r, int rt) {if (l == r) {scanf("%d", &x);ls1[rt] = rs1[rt] = sum1[rt] = x;if (x == 1)ls2[rt] = rs2[rt] = sum2[rt] = 0;elsels2[rt] = rs2[rt] = sum2[rt] = 1;return;}int mid = (l + r) >> 1;build(l, mid, rt << 1);build(mid + 1, r, rt << 1 | 1);push_up(l, r, rt);
}void push_date(int l, int r, int L, int R, int rt) {if (L <= l && R >= r) {lazy[rt] ^= 1;swap(ls1[rt], ls2[rt]);swap(rs1[rt], rs2[rt]);swap(sum1[rt], sum2[rt]);return;}push_down(rt);int mid = (l + r) >> 1;if (R <= mid)push_date(l, mid, L, R, rt << 1);else if (L > mid)push_date(mid + 1, r, L, R, rt << 1 | 1);else {push_date(l, mid, L, mid, rt << 1);push_date(mid + 1, r, mid + 1, R, rt << 1 | 1);}push_up(l, r, rt);
}int query(int l, int r, int L, int R, int rt) {if (L <= l && R >= r)return sum1[rt];push_down(rt);int mid = (l + r) >> 1;if (R <= mid) {return query(l, mid, L, R, rt << 1);} else if (L > mid) {return query(mid + 1, r, L, R, rt << 1 | 1);} else {int len1 = query(l, mid, L, mid, rt << 1);int len2 = query(mid + 1, r, mid + 1, R, rt << 1 | 1);int len3 = min(mid - L + 1, rs1[rt << 1]) + min(R - mid, ls1[rt << 1 | 1]);return max(len1, max(len2, len3));}return 0;
}int main() {scanf("%d", &n);build(1, n, 1);scanf("%d", &m);while (m--) {int opt, l, r;scanf("%d%d%d", &opt, &l, &r);if (opt == 1)push_date(1, n, l, r, 1);elseprintf("%d\n", query(1, n, l, r, 1));}return 0;
}

SPOJ - GSS1 Can you answer these queries I（最大子段和）

题意：给出一段序列，给出m次询问，让你给出该区间内最大连续子段和。

题解：跟最长连续子序列差不多，这是一道模板题，直接上代码。

#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>const int mod = 998244353;
const int maxn = 5e4 + 5;
const int inf = 1e9;
const long long onf = 1e18;
#define me(a, b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
#define lson rt<<1
#define rson rt<<1|1
#define PI 3.14159265358979323846
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
struct node {int val, lmax, rmax, max;
} tree[maxn << 2];void push_up(int rt) {tree[rt].max = max(max(tree[lson].max, tree[rson].max), tree[lson].rmax + tree[rson].lmax);tree[rt].lmax = max(tree[lson].lmax, tree[lson].val + tree[rson].lmax);tree[rt].rmax = max(tree[rson].rmax, tree[rson].val + tree[lson].rmax);tree[rt].val = tree[lson].val + tree[rson].val;
}void build(int l, int r, int rt) {if (l == r) {scanf("%d", &tree[rt].val);tree[rt].lmax = tree[rt].rmax = tree[rt].max = tree[rt].val;return;}int mid = (l + r) >> 1;build(Lson);build(Rson);push_up(rt);
}node query(int ql, int qr, int l, int r, int rt) {if (ql <= l && qr >= r)return tree[rt];int mid = (l + r) >> 1;if (qr <= mid)return query(ql, qr, Lson);else if (ql > mid)return query(ql, qr, Rson);else {node L = query(ql, mid, Lson);node R = query(mid + 1, qr, Rson);node ans;ans.max = max(max(L.max, R.max), L.rmax + R.lmax);ans.lmax = max(L.lmax, L.val + R.lmax);ans.rmax = max(R.rmax, R.val + L.rmax);return ans;}
}int main() {int n, m;scanf("%d", &n);build(1, n, 1);scanf("%d", &m);while (m--) {int l, r;scanf("%d%d", &l, &r);printf("%d\n", query(l, r, 1, n, 1).max);}return 0;
}

HDU3308 LCIS（区间最长连续上升子序列）

题意：求区间最长连续上升子序列。

PS：因为要更改节点的值，所以不能用dp做。这里就可以用线段树来维护。

这就是一个裸的线段树求最长上升子序列的题，细节看代码，主要是要注意更新左子树，右子树，和区间最长子序列长度。

#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>const int mod = 998244353;
const int maxn = 1e5 + 5;
const int inf = 1e9;
const long long onf = 1e18;
#define me(a, b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
#define lson rt<<1
#define rson rt<<1|1
#define PI 3.14159265358979323846
typedef long long ll;
typedef unsigned long long ull;
using namespace std;int a[maxn], sum[maxn << 2], ls[maxn << 2], rs[maxn << 2];void updata(int l, int r, int rt)///更新区间最长子序列长度
{int mid = (l + r) >> 1;ls[rt] = ls[lson], rs[rt] = rs[rson];sum[rt] = max(sum[lson], sum[rson]);if (a[mid] < a[mid + 1]) {///当满足这个条件，说明可以左儿子右儿子合并if (ls[rt] == mid - l + 1)///当左儿子区间全部都是上升子序列，又能区间合并，这时候左儿子区间加上右儿子的左区间ls[rt] = ls[lson] + ls[rson];if (rs[rt] == r - mid)///与上面类似rs[rt] = rs[rson] + rs[lson];sum[rt] = max(sum[rt], ls[rson] + rs[lson]);}
}void build(int l, int r, int rt)///建树
{if (l == r) {sum[rt] = ls[rt] = rs[rt] = 1;return;}int mid = (l + r) >> 1;build(Lson);build(Rson);updata(l, r, rt);
}void pushdata(int pos, int c, int l, int r, int rt)///更改节点值
{if (l == r) {a[l] = c;return;}int mid = (l + r) >> 1;if (pos <= mid)pushdata(pos, c, Lson);elsepushdata(pos, c, Rson);updata(l, r, rt);
}int query(int L, int R, int l, int r, int rt)///求出最长子序列长度
{if (l >= L && R >= r)return sum[rt];int ret = 0;int mid = (l + r) >> 1;if (mid >= L)ret = max(ret, query(L, R, Lson));if (mid < R)ret = max(ret, query(L, R, Rson));if (a[mid] < a[mid + 1]) {ret = max(ret, min(mid - L + 1, rs[lson]) + min(R - mid, ls[rson]));///[m+1,R]与[m+1,r]相交部分的最大前缀+[L,m]与[l,m]的最大后缀.}return ret;
}int main() {int t;cin >> t;while (t--) {int n, m;scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);build(1, n, 1);while (m--) {char s[2];int a, b;scanf("%s%d%d", s, &a, &b);if (s[0] == 'U')pushdata(a + 1, b, 1, n, 1);elseprintf("%d\n", query(a + 1, b + 1, 1, n, 1));}}return 0;
}