hdu 4578 Transformation （线段树 + 延迟标记）

最新推荐文章于 2024-08-19 01:03:13 发布

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#线段树 #hdu 4578

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本文介绍了解决HDU4578问题的方法，该问题涉及区间更新和查询，通过使用线段树和懒惰传播实现高效操作。文章详细解释了三种操作的优先级及其如何影响线段树节点的状态更新。

题目链接：hdu 4578

思路：这道题目有个关键点，就是三个操作的优先级，优先级如下：1改变>2乘>3加，可以这么理解，先定义三个lazy0，lazy1，lazy2的意义，分别表示将要加的数，将要乘上的数，将要变成的数c，那么一旦你执行操作1（改变），那么lazy2=c, lazy1要置为1，lazy0要置于0，同样，你一旦操作2乘c，那么你lazy1=lazy1*c，lazy0=lazy0*c。最后，你如果操作3加，那么你只要将lazy0=lazy0 + c. 所以可以看出优先级是改变>乘>加。因此pushdown的时候，因先对优先级高的先操作，因为它会影响后面的操作。

代码：

#include<cstdio>
#define ll long long 
const int mod = 10007;
const int maxn = 1e5+5;
#define mid (L + R)/2
#define ls rt*2
#define rs rt*2+1
#define lson ls,L,mid
#define rson rs,mid+1,R
ll sum[maxn<<2], sum1[maxn<<2], sum2[maxn<<2];
ll lazy0[maxn<<2], lazy1[maxn<<2], lazy2[maxn<<2];//加，乘，变 
void build(int rt, int L, int R)
{
	lazy0[rt]=lazy2[rt]=sum[rt]=sum1[rt]=sum2[rt]= 0;
	lazy1[rt] = 1;
	if(L == R) return ;
	build(lson);
	build(rson);
}
void pushDown(int rt, int  L, int R)
{
	if(lazy2[rt]){//变 
		sum2[ls] = 1ll*(((lazy2[rt]*lazy2[rt])%mod*lazy2[rt])%mod*(mid-L+1))%mod;
		sum2[rs] = 1ll*(((lazy2[rt]*lazy2[rt])%mod*lazy2[rt])%mod*(R-mid))%mod;
		sum1[ls] = 1ll*((lazy2[rt]*lazy2[rt])%mod*(mid-L+1))%mod;
		sum1[rs] = 1ll*((lazy2[rt]*lazy2[rt])%mod*(R-mid))%mod;
		sum[ls] = 1ll*lazy2[rt]*(mid-L+1)%mod; 
		sum[rs] = 1ll*lazy2[rt]*(R-mid)%mod;
		lazy2[ls] = lazy2[rs] = lazy2[rt];
		lazy1[ls] = lazy1[rs] = 1;
		lazy0[ls] = lazy0[rs] = 0;
		lazy2[rt] = 0;
	}
	if(lazy1[rt] != 1){//乘 
		sum2[ls] = 1ll*((sum2[ls]*lazy1[rt])%mod*(lazy1[rt]*lazy1[rt])%mod)%mod;
		sum2[rs] = 1ll*((sum2[rs]*lazy1[rt])%mod*(lazy1[rt]*lazy1[rt])%mod)%mod;
		sum1[ls] = 1ll*(sum1[ls]*lazy1[rt])%mod*lazy1[rt]%mod;
		sum1[rs] = 1ll*(sum1[rs]*lazy1[rt])%mod*lazy1[rt]%mod;
		sum[ls] = 1ll*sum[ls]*lazy1[rt]%mod;
		sum[rs] = 1ll*sum[rs]*lazy1[rt]%mod;
		lazy1[ls] = 1ll*lazy1[ls]*lazy1[rt]%mod;
		lazy1[rs] = 1ll*lazy1[rs]*lazy1[rt]%mod;
		if(lazy0[ls])
			lazy0[ls] = 1ll*lazy0[ls]*lazy1[rt]%mod;
		if(lazy0[rs])
			lazy0[rs] = 1ll*lazy0[rs]*lazy1[rt]%mod;
		lazy1[rt] = 1;
	}
	if(lazy0[rt]){//加 sum2[rt] + 1ll*(R-L+1)*val*val*val + 3ll*val*(sum[rt]*val + 1ll*sum1[rt]);
		sum2[ls] = sum2[ls] + 1ll*((mid-L+1)*(lazy0[rt]*lazy0[rt])%mod*lazy0[rt])%mod + 3ll*(lazy0[rt]*((sum[ls]*lazy0[rt])%mod + 1ll*sum1[ls])%mod)%mod;
		sum2[ls] %= mod;
		sum2[rs] = sum2[rs] + 1ll*((R-mid)*(lazy0[rt]*lazy0[rt])%mod*lazy0[rt])%mod + 3ll*lazy0[rt]*(sum[rs]*lazy0[rt] + 1ll*sum1[rs]);
		sum2[rs] %= mod;
		sum1[ls] = sum1[ls] + 1ll*lazy0[rt]*lazy0[rt]*(mid-L+1) + 2ll*sum[ls]*lazy0[rt];
		sum1[ls] %= mod;
		sum1[rs] = sum1[rs] + 1ll*lazy0[rt]*lazy0[rt]*(R-mid) + 2ll*sum[rs]*lazy0[rt];
		sum1[rs] %= mod;
		sum[ls] = sum[ls] + 1ll*lazy0[rt]*(mid-L+1);
		sum[ls] %= mod;
		sum[rs] = sum[rs] + 1ll*lazy0[rt]*(R-mid);
		sum[rt] %= mod;
		lazy0[ls] += lazy0[rt];
		lazy0[ls] %= mod;
		lazy0[rs] += lazy0[rt];
		lazy0[rs] %= mod; 
		lazy0[rt] = 0;
	}
}
void upDate(int rt, int L, int R, int l, int r, ll val, int op)
{
	if(l <= L && R <= r){//加 
		if(op == 1){
			sum2[rt] = sum2[rt] + 1ll*(R-L+1)*((val*val)%mod*val)%mod + 3ll*val*(((sum[rt]*val)%mod + 1ll*sum1[rt]))%mod;
			sum2[rt] %= mod;
			sum1[rt] = sum1[rt]+1ll*(R-L+1)*(val*val)%mod+2ll*(val*sum[rt])%mod;
			sum1[rt] %= mod;
			sum[rt] = sum[rt] + 1ll*val*(R - L + 1);
			sum[rt] %= mod;			
			lazy0[rt] += val;
			lazy0[rt] %= mod;
		}
		else if(op == 2){//乘 
			sum2[rt] = 1ll*sum2[rt]*((val*val)%mod*val)%mod;
			sum1[rt] = 1ll*(sum1[rt] * (val*val)%mod)%mod;
			sum[rt] = 1ll*val*sum[rt]%mod;
			if(lazy0[rt])
				lazy0[rt] = 1ll*lazy0[rt]*val%mod;
			lazy1[rt] = 1ll*lazy1[rt]*val%mod;
		}
		else if(op == 3){//变 
			sum2[rt] = 1ll*((val*val)%mod*val*(R-L+1))%mod;
			sum1[rt] = 1ll*val*val*(R-L+1)%mod;
			sum[rt] = 1ll*val*(R - L + 1)%mod;
			lazy0[rt] = 0;
			lazy1[rt] = 1;
			lazy2[rt] = val;
		}
		return ;
	}
	pushDown(rt, L, R);
	if(l <= mid) upDate(lson, l, r, val, op);
	if(r > mid) upDate(rson, l, r, val, op);
	sum[rt] = sum[ls] + sum[rs];
	sum1[rt] = sum1[ls] + sum1[rs];
	sum2[rt] = sum2[ls] + sum2[rs];
	sum[rt] %= mod;
	sum1[rt] %= mod;
	sum2[rt] %= mod;
}
ll query(int rt, int L, int R, int l, int r, int op)
{
	if(l <= L && R <= r){
		if(op == 1){
			return sum[rt];
		}
		else if(op == 2){
			return sum1[rt];
		}
		else if(op == 3){
			return sum2[rt];
		}
	}
	ll res = 0ll;
	pushDown(rt,L,R);
	if(l <= mid) res += query(lson,l,r,op);
	if(r > mid) res += query(rson,l,r,op);
	return res%mod;
}
int main()
{
	ll n, m;
	ll a, b, c, d;
	while(~scanf("%lld%lld", &n, &m)){
		if(n==0 && m== 0)break;
		build(1, 1, n);
		for(int i = 0; i < m; i++){
			scanf("%lld%lld%lld%lld", &a, &b, &c, &d);
			if(a == 1){
				upDate(1,1,n,b,c,d,1);
			}
			else if(a == 2){
				upDate(1,1,n,b,c,d,2);
			}
			else if(a == 3){
				upDate(1,1,n,b,c,d,3);
			}
			else if(a == 4){
				ll ans = query(1,1,n,b,c,d);
				printf("%lld\n", ans);
			}
		}
	}
	return 0;
}