codevs 4927 线段树练习5 题解

一、题目：

codevs原题

二、思路：

这将是我NOIP2018前的最后一篇题解，从现在到NOIP我将进入总复习阶段。祝NOIP rp++。

那么这道题显然会有两个标记，一个add标记，一个set标记。那么怎样才能使这两个标记互不影响呢？

我们规定一个顺序，当set标记打过来时，我们首先将add标记清零，再打set标记。

改了改代码风格，具体请见博客置顶文章。

三、代码：

/*
 * @Author: 岸芷汀兰
 * @Date: 2018-10-15 22:11:37
 * @LastEditors: 岸芷汀兰
 * @LastEditTime: 2018-10-15 23:13:52
 * @Description: 4927 of codevs
 */

#include<iostream>
#include<cstdio>
#include<cstring>    
#include<algorithm>    
#include<string>

#define LL long long
#define mem(s,v) memset(s,v,sizeof(s))

using namespace std;
template<class Type>
inline Type read(void) {
	Type x = 0, f = 1; char ch = getchar();
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return f * x;
}

const int maxn = 100005;

int n, m;

LL a[maxn];

struct Segment_Tree {
#define lson (o << 1)
#define rson (o << 1 | 1)
	LL set[maxn << 2], add[maxn << 2], sum[maxn << 2], maxv[maxn << 2], minv[maxn << 2], len[maxn << 2];
	inline void pushup(int o) {
		sum[o] = sum[lson] + sum[rson];
		maxv[o] = max(maxv[lson], maxv[rson]);
		minv[o] = min(minv[lson], minv[rson]);
	}
	inline void build(int o, int l, int r) {
		len[o] = r - l + 1;
		add[o] = sum[o] = maxv[o] = minv[o] = 0;
		set[o] = -1;
		if (l == r) { sum[o] = a[l]; maxv[o] = a[l]; minv[o] = a[l]; return; }
		int mid = (l + r) >> 1;
		build(lson, l, mid);
		build(rson, mid + 1, r);
		pushup(o);
	}
	inline void Add(int o,int v){
		add[o]+=v;
		sum[o]+=len[o]*v;
		minv[o]+=v;maxv[o]+=v;
		return;
	}
	inline void Set(int o,int v){
		set[o]=v;add[o]=0;
		sum[o]=len[o]*v;
		maxv[o]=minv[o]=v;
		return;
	}
	inline void pushdown(int o){
		if(set[o]!=-1){
			Set(lson,set[o]);Set(rson,set[o]);
			set[o]=-1;
		}
		if(add[o]){
			Add(lson,add[o]);Add(rson,add[o]);
			add[o]=0;
		}
		return;
	}
	inline void Update(int o,int l,int r,int ql,int qr,int k,LL v){
		if(ql<=l&&r<=qr){
			if(k==1){//add
				Add(o,v);return;
			}
			else{//set
				Set(o,v);return;
			}
		}
		pushdown(o);
		int mid=(l+r)>>1;
		if(ql<=mid)Update(lson,l,mid,ql,qr,k,v);
		if(qr>mid)Update(rson,mid+1,r,ql,qr,k,v);
		pushup(o);
	}
	inline LL query(int o,int l,int r,int ql,int qr,int k){
		if(ql<=l&&r<=qr){
			switch(k){
				case 1://sum
					return sum[o];
				case 2://max
					return maxv[o];
				case 3://min
					return minv[o];
			}
		}
		pushdown(o);
		int mid=(l+r)>>1;LL ans=0;
		switch(k){
			case 1://sum
				if(ql<=mid)ans+=query(lson,l,mid,ql,qr,k);
				if(qr>mid)ans+=query(rson,mid+1,r,ql,qr,k);
				return ans;
			case 2://max
				ans=-0x3f3f3f3f3f3f3f3f;
				if(ql<=mid)ans=max(ans,query(lson,l,mid,ql,qr,k));
				if(qr>mid)ans=max(ans,query(rson,mid+1,r,ql,qr,k));
				return ans;
			case 3://min
				ans=0x3f3f3f3f3f3f3f3f;
				if(ql<=mid)ans=min(ans,query(lson,l,mid,ql,qr,k));
				if(qr>mid)ans=min(ans,query(rson,mid+1,r,ql,qr,k));
				return ans;
		}
	}
}T;

int main() {
	n = read<int>(); m = read<int>();
	for (register int i = 1; i <= n; ++i) {
		a[i] = read<LL>();
	}
	T.build(1, 1, n);
	for (register int i = 1; i <= m; ++i) {
		string s; cin >> s;
		if (s == "add") {
			int l = read<int>(), r = read<int>(); LL c = read<LL>();
			T.Update(1, 1, n, l, r, 1, c);
		}
		else if (s == "set") {
			int l = read<int>(), r = read<int>(); LL c = read<LL>();
			T.Update(1, 1, n, l, r, 2, c);
		}
		else if (s == "sum") {
			int l = read<int>(), r = read<int>();
			printf("%lld
", T.query(1, 1, n, l, r, 1));
		}
		else if (s == "max") {
			int l = read<int>(), r = read<int>();
			printf("%lld
", T.query(1, 1, n, l, r, 2));
		}
		else if (s == "min") {
			int l = read<int>(), r = read<int>();
			printf("%lld
", T.query(1, 1, n, l, r, 3));
		}
	}
	return 0;
}