【线段树】各种模板集合

这里放蒟蒻 lrj124 碰到的线段树区间操作集合

No.1 区间加

懒标记

/************************************************
*Author        :  lrj124
*Created Time  :  2019.04.04.14:17
*Mail          :  1584634848@qq.com
*Problem       :  seg
************************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100000 + 10;
int n,m;
struct seg { int l,r;  ll mark,sum; } tree[maxn<<2];
inline void pushup(int root) { tree[root].sum = tree[root<<1].sum+tree[root<<1|1].sum; }
inline void pushdown(int root) {
	if (tree[root].mark) {
		tree[root<<1].mark += tree[root].mark;
		tree[root<<1|1].mark += tree[root].mark;
		tree[root<<1].sum += tree[root].mark*(tree[root<<1].r-tree[root<<1].l+1);
		tree[root<<1|1].sum += tree[root].mark*(tree[root<<1|1].r-tree[root<<1|1].l+1);
		tree[root].mark = 0;
	}
}
inline void Build(int l,int r,int root) {
	tree[root].l = l;
	tree[root].r = r;
	if (l == r) {
		scanf("%lld",&tree[root].sum);
		return;
	}
	int mid = l+r>>1;
	Build(l,mid,root<<1);
	Build(mid+1,r,root<<1|1);
	pushup(root);
}
inline void update(int l,int r,int ql,int qr,int root,ll x) {
	if (qr < l || ql > r) return;
	if (ql <= l && r <= qr) {
		tree[root].mark += x;
		tree[root].sum += x*(r-l+1);
		return;
	}
	pushdown(root);
	int mid = l+r>>1;
	update(l,mid,ql,qr,root<<1,x);
	update(mid+1,r,ql,qr,root<<1|1,x);
	pushup(root);
}
inline ll query(int l,int r,int ql,int qr,int root) {
	if (qr < l || ql > r) return 0;
	if (ql <= l && r <= qr) return tree[root].sum;
	pushdown(root);
	int mid = l+r>>1;
	return query(l,mid,ql,qr,root<<1)+query(mid+1,r,ql,qr,root<<1|1);
}
int main() {
	//freopen("seg.in","r",stdin);
	//freopen("seg.out","w",stdout);
	scanf("%d%d",&n,&m);
	Build(1,n,1);
	while (m--) {
		int op,x,y;
		ll k;
		scanf("%d%d%d",&op,&x,&y);
		if (op == 1) {
			scanf("%lld",&k);
			update(1,n,x,y,1,k);
		} else printf("%lld\n",query(1,n,x,y,1));
	}
	return 0;
}

No.2 区间加，区间乘

加个乘法标记

#include <cstdio>
const int maxn = 100000 + 10;
struct Seg { long long l,r,sum,add,mul; } tree[maxn*4];
long long p;
long long n,m;
inline void pushup(long long root) { tree[root].sum = tree[root<<1].sum+tree[root<<1|1].sum; tree[root].sum %= p; }
inline void BuildTree(long long l,long long r,long long root) {
	tree[root].l = l;
	tree[root].r = r;
	tree[root].mul = 1;
	if (l == r) {
		scanf("%lld",&tree[root].sum);
		tree[root].sum %= p;
		return;
	}
	long long mid = l+r>>1;
	BuildTree(l,mid,root<<1);
	BuildTree(mid+1,r,root<<1|1);
	pushup(root);
}
inline void pushdown(long long root) {
	if (tree[root].mul != 1) {
		tree[root<<1].mul = tree[root<<1].mul*tree[root].mul%p;
		tree[root<<1|1].mul = tree[root<<1|1].mul*tree[root].mul%p;
		tree[root<<1].add = tree[root<<1].add*tree[root].mul%p;
		tree[root<<1|1].add = tree[root<<1|1].add*tree[root].mul%p;
		tree[root<<1].sum = tree[root<<1].sum*tree[root].mul%p;
		tree[root<<1|1].sum = tree[root<<1|1].sum*tree[root].mul%p;
		tree[root].mul = 1;
	}
	if (tree[root].add != 0) {
		tree[root<<1].add = (tree[root<<1].add+tree[root].add)%p;
		tree[root<<1|1].add = (tree[root<<1|1].add+tree[root].add)%p;
		tree[root<<1].sum = (tree[root<<1].sum+tree[root].add*(tree[root<<1].r-tree[root<<1].l+1))%p;
		tree[root<<1|1].sum = (tree[root<<1|1].sum+tree[root].add*(tree[root<<1|1].r-tree[root<<1|1].l+1))%p;
		tree[root].add = 0;
	}
}
inline void UpdateAdd(long long ql,long long qr,long long l,long long r,long long root,long long x) {
	if (ql > r || qr < l) return;
	if (ql <= l && qr >= r) {
		tree[root].add = (tree[root].add+x)%p;
		tree[root].sum = (tree[root].sum+x*(r-l+1))%p;
		return;
	}
	pushdown(root);
	long long mid = l+r>>1;
	UpdateAdd(ql,qr,l,mid,root<<1,x);
	UpdateAdd(ql,qr,mid+1,r,root<<1|1,x);
	pushup(root);
}
inline void UpdateMul(long long ql,long long qr,long long l,long long r,long long root,long long x) {
	if (ql > r || qr < l) return;
	if (ql <= l && qr >= r) {
		tree[root].add = tree[root].add*x%p;
		tree[root].mul = tree[root].mul*x%p;
		tree[root].sum = tree[root].sum*x%p;
		return;
	}
	pushdown(root);
	long long mid = l+r>>1;
	UpdateMul(ql,qr,l,mid,root<<1,x);
	UpdateMul(ql,qr,mid+1,r,root<<1|1,x);
	pushup(root);
}
inline long long Query(long long ql,long long qr,long long l,long long r,long long root) {
	if (ql > r || qr < l) return 0;
	if (ql <= l && qr >= r) return tree[root].sum;
	pushdown(root);
	long long mid = l+r>>1;
	return (Query(ql,qr,l,mid,root<<1)+Query(ql,qr,mid+1,r,root<<1|1))%p;
}
int main() {
	scanf("%lld%lld%lld",&n,&m,&p);
	BuildTree(1,n,1);
	while (m--) {
		long long val;
		long long op,l,r;
		scanf("%lld%lld%lld",&op,&l,&r);
		if (op == 1) {
			scanf("%lld",&val);
			UpdateMul(l,r,1,n,1,val);
		} else if (op == 2) {
			scanf("%lld",&val);
			UpdateAdd(l,r,1,n,1,val);
		} else printf("%lld\n",Query(l,r,1,n,1));
	}
	return 0;
}

No.3 把区间中等于 \(x\) 的数改成 \(y\)

对每个数开一个 tag，表示每个数会映射到哪个数。

/************************************************
*Author        :  lrj124
*Created Time  :  2019.10.04.14:07
*Mail          :  1584634848@qq.com
*Problem       :  cf911g
************************************************/
#include <cstdio>
const int maxn = 200000 + 10;
int n,q,a[maxn],tag[maxn<<2][101];
inline void pushdown(int root) {
	for (int i = 1;i <= 100;i++) {
		tag[root<<1][i] = tag[root][tag[root<<1][i]];
		tag[root<<1|1][i] = tag[root][tag[root<<1|1][i]];
	}
	for (int i = 1;i <= 100;i++) tag[root][i] = i;
}
inline void update(int l,int r,int ul,int ur,int num,int root,int x) {
	if (l > ur || r < ul) return;
	if (ul <= l && r <= ur) {
		for (int i = 1;i <= 100;i++)
			if (tag[root][i] == num) tag[root][i] = x;
		return;
	}
	pushdown(root);
	int mid = l+r>>1;
	update(l,mid,ul,ur,num,root<<1,x);
	update(mid+1,r,ul,ur,num,root<<1|1,x);
}
inline void print(int l,int r,int root) {
	if (l == r) {
		printf("%d ",tag[root][a[l]]);
		return;
	}
	pushdown(root);
	int mid = l+r>>1;
	print(l,mid,root<<1);
	print(mid+1,r,root<<1|1);
}
int main() {
// 	freopen("cf911g.in","r",stdin);
// 	freopen("cf911g.out","w",stdout);
	scanf("%d",&n);
	for (int i = 1;i <= n;i++) scanf("%d",&a[i]);
	for (int i = 1;i <= n<<2;i++) for (int j = 1;j <= 100;j++) tag[i][j] = j;
	for (scanf("%d",&q);q--;) {
		int l,r,x,y;
		scanf("%d%d%d%d",&l,&r,&x,&y);
		update(1,n,l,r,x,1,y);
	}
	print(1,n,1);
	return 0;
}

No.4 区间每个数开方

就是暴力递归到每个叶子节点，若当前区间 Max 小于 2 就 return

/************************************************
*Author        :  lrj124
*Created Time  :  2019.09.28.14:24
*Mail          :  1584634848@qq.com
*Problem       :  luogu4145
************************************************/
#include <algorithm>
#include <cstdio>
#include <cmath>
using ll = long long;
const int maxn = 100000 + 10;
ll sum[maxn<<2],maxv[maxn<<2];
inline void pushup(int root) {
	sum[root] = sum[root<<1]+sum[root<<1|1];
	maxv[root] = std :: max(maxv[root<<1],maxv[root<<1|1]);
}
inline void build(int l,int r,int root) {
	if (l == r) {
		scanf("%lld",&sum[root]);
		maxv[root] = sum[root];
		return;
	}
	int mid = l+r>>1;
	build(l,mid,root<<1);
	build(mid+1,r,root<<1|1);
	pushup(root);
}
inline void update(int l,int r,int ul,int ur,int root) {
	if (l > ur || r < ul || maxv[root] < 2) return;
	if (l == r) {
		maxv[root] = sum[root] = sqrt(sum[root]);
		return;
	}
	int mid = l+r>>1;
	update(l,mid,ul,ur,root<<1);
	update(mid+1,r,ul,ur,root<<1|1);
	pushup(root);
}
inline ll query(int l,int r,int ql,int qr,int root) {
	if (l > qr || r < ql) return 0;
	if (ql <= l && r <= qr) return sum[root];
	int mid = l+r>>1;
	return query(l,mid,ql,qr,root<<1)+query(mid+1,r,ql,qr,root<<1|1);
}
int main() {
	for (int n,q,x,y,z;scanf("%d",&n) ^ EOF;puts("")) {
		build(1,n,1);
		for (scanf("%d",&q);q--;) {
			scanf("%d%d%d",&x,&y,&z);
			if (y > z) std :: swap(y,z);
			if (x) printf("%lld\n",query(1,n,y,z,1));
			else update(1,n,y,z,1);
		}
	}
	return 0;
}

No.5 区间取 Min,Max

把 pushup 改一改就行，

区间和 \(x\) 取 Max 时，区间最大和区间最小都和 \(x\) 取 Max

区间和 \(x\) 取 Min 时，区间最大和区间最小都和 \(x\) 取 Min

/************************************************
*Author        :  lrj124
*Created Time  :  2019.10.03.21:38
*Mail          :  1584634848@qq.com
*Problem       :  luogu4560
************************************************/
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn = 2000000 + 10;
int n,q,maxv[maxn<<2],minv[maxn<<2];
inline void pushup_max(int root,int x) {
	minv[root] = max(minv[root],x);
	maxv[root] = max(maxv[root],x);
}
inline void pushup_min(int root,int x) {
	minv[root] = min(minv[root],x);
	maxv[root] = min(maxv[root],x);
}
inline void pushdown(int root) {
	pushup_max(root<<1,maxv[root]);
	pushup_max(root<<1|1,maxv[root]);
	pushup_min(root<<1,minv[root]);
	pushup_min(root<<1|1,minv[root]);
	maxv[root] = 0; minv[root] = 999999;
}
inline void take_max(int l,int r,int ul,int ur,int root,int x) {
	if (l > ur || r < ul) return;
	if (ul <= l && r <= ur) {
		pushup_max(root,x);
		return;
	}
	pushdown(root);
	int mid = l+r>>1;
	take_max(l,mid,ul,ur,root<<1,x);
	take_max(mid+1,r,ul,ur,root<<1|1,x);
}
inline void take_min(int l,int r,int ul,int ur,int root,int x) {
	if (l > ur || r < ul) return;
	if (ul <= l && r <= ur) {
		pushup_min(root,x);
		return;
	}
	pushdown(root);
	int mid = l+r>>1;
	take_min(l,mid,ul,ur,root<<1,x);
	take_min(mid+1,r,ul,ur,root<<1|1,x);
}
inline void print(int l,int r,int root) {
	if (l == r) {
		printf("%d\n",maxv[root]);
		return;
	}
	pushdown(root);
	int mid = l+r>>1;
	print(l,mid,root<<1);
	print(mid+1,r,root<<1|1);
}
int main() {
	for (scanf("%d%d",&n,&q);q--;) {
		int op,l,r,w;
		scanf("%d%d%d%d",&op,&l,&r,&w);
		if (op == 1) take_max(1,n,l+1,r+1,1,w);
		else take_min(1,n,l+1,r+1,1,w);
	}
	print(1,n,1);
	return 0;
}

No.6 单点修改，区间询问最大连续子段和

只需要维护四个数：

max,lmax,rmax,sum

分别表示区间最大连续子段和，区间左缀最大连续和，区间右缀最大连续和，区间和

lmax = max(lmax,rson.sum+rson.lmax)

rmax = max(rmax,lson.sum+lson.rmax)

max = max(lson.max,rson.max,lson.rmax+rson.lmax)

代码

/************************************************
*Author        :  lrj124
*Created Time  :  2019.09.15.08:38
*Mail          :  1584634848@qq.com
*Problem       :  spoj1716
************************************************/
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn = 50000 + 10;
struct seg { int l,r,sum,max; } tree[maxn<<2];
int n,q;
inline void pushup(int root) {
	tree[root].sum = tree[root<<1].sum+tree[root<<1|1].sum;
	tree[root].l = max(tree[root<<1].l,tree[root<<1|1].l+tree[root<<1].sum);
	tree[root].r = max(tree[root<<1|1].r,tree[root<<1].r+tree[root<<1|1].sum);
	tree[root].max = max(tree[root<<1].r+tree[root<<1|1].l,max(tree[root<<1].max,tree[root<<1|1].max));
}
inline void build(int l,int r,int root) {
	if (l == r) {
		int x; scanf("%d",&x);
		tree[root] = { x,x,x,x };
		return;
	}
	int mid = l+r>>1;
	build(l,mid,root<<1);
	build(mid+1,r,root<<1|1);
	pushup(root);
}
inline void update(int l,int r,int num,int root,int x) {
	if (l > num || r < num) return;
	if (l == r) {
		tree[root] = { x,x,x,x };
		return;
	}
	int mid = l+r>>1;
	update(l,mid,num,root<<1,x);
	update(mid+1,r,num,root<<1|1,x);
	pushup(root);
}
inline seg query(int l,int r,int ql,int qr,int root) {
	if (ql <= l && r <= qr) return tree[root];
	int mid = l+r>>1;
	if (mid >= qr) return query(l,mid,ql,qr,root<<1);
	if (ql > mid) return query(mid+1,r,ql,qr,root<<1|1);
	seg lson = query(l,mid,ql,qr,root<<1),rson = query(mid+1,r,ql,qr,root<<1|1),ans;
	ans = { max(lson.l,rson.l+lson.sum),max(rson.r,lson.r+rson.sum),rson.sum+lson.sum,max(lson.r+rson.l,max(lson.max,rson.max)) };
	return ans;
}
int main() {
// 	freopen("spoj1716.in","r",stdin);
// 	freopen("spoj1716.out","w",stdout);
	scanf("%d",&n);
	build(1,n,1);
	for (scanf("%d",&q);q--;) {
		int x,y,z; scanf("%d%d%d",&x,&y,&z);
		if (x == 0) update(1,n,y,1,z);
		else printf("%d\n",query(1,n,y,z,1).max);
	}
	return 0;
}