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[THUSCH2017] 大魔法师 卡题记录

发布时间 2023-08-30 09:55:59作者: koukilee

题目：fzqoj - luogu

前情提示： 此题极度卡常!!!，否则你就会像我这个蒟蒻一样卡题\(3h\)：

死亡记录

前置知识：

   1.线段树的区间修改，不会的可以点这 - 基础 ： 进阶

   2.基本的矩阵乘法：Fibonacci

题解部分

对于题目给出的6种操作，我们可以用线段树与矩阵乘法来维护

思路

维护一个四元组，并转化成矩阵: [\(A\) \(i\) \(B\)\(i\)  \(C\)\(i\) \(V\)] ，每一个操作也可以看成一个矩阵

对于操作1：

\[\begin{equation} \left[ \begin{array}{cccc} 1& 0 &0 & 0\\ 1& 1 &0&0\\ 0 &0 &1 &0\\ 0&0 &0&1 \end{array} \right ] \end{equation} \]

  以此类推，对于操作2：

\[\begin{equation}\left[ \begin{array}{cccc} 1& 0 &0 & 0\\ 0& 1 &0&0\\ 0 &1 &1 &0\\ 0&0 &0&1\end{array}\right ]\end{equation} \]

  那么对于操作3：

\[\begin{equation}\left[ \begin{array}{cccc} 1& 0 &1 & 0\\ 0& 1 &0&0\\ 0 &0 &1 &0\\ 0&0 &0&1\end{array}\right ]\end{equation} \]

  对于操作4，就需要使用第4维了：

\[\begin{equation}\left[ \begin{array}{cccc} 1& 0 &0 & 0\\ 0& 1 &0&0\\ 0 &0 &1 &0\\ v&0 &0&1\end{array}\right ]\end{equation} \]

  操作5类似：

\[\begin{equation}\left[ \begin{array}{cccc} 1& 0 &0 & 0\\ 0& 1 &0&0\\ 0 &0 &1 &0\\ 0&0 &v&1\end{array}\right ]\end{equation} \]

  操作6也就出来了：

\[\begin{equation}\left[ \begin{array}{cccc} 1& 0 &0 & 0\\ 0& 1 &0&0\\ 0 &0 &0 &0\\ 0&0 &v&1\end{array}\right ]\end{equation} \]

做到这里，只需要写一颗线段树，维护区间矩阵乘/加即可。

30 Pts 代码如下

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

struct matrix{
	int x[4][4];
}g[1000010], tree[1000100], lazy[1000100];

ll n, m, opt, l, r, v, mod = 998244353;

inline ll read(){
	ll f = 1, sum = 0; char ch = getchar();
	while(ch < '0' || ch > '9') f = (ch == '-' ? -1 : 1), ch = getchar();
	while(ch >= '0' && ch <= '9') sum = (sum << 1) + (sum << 3) + ch - '0', ch = getchar();
	return f * sum;
}

matrix matrix_add(matrix a, matrix b){
	matrix ans;
	memset(ans.x, 0, sizeof(ans.x));
	for(ll i = 0; i <= 3; i++)
		ans.x[0][i] = (a.x[0][i] + b.x[0][i]) % mod;
	return ans;
}

matrix matrix_plus(matrix a, matrix b){
	matrix ans;
	memset(ans.x, 0, sizeof(ans.x));
	for(ll i = 0; i <= 3; i++){
		for(ll j = 0; j <= 3; j++){
			for(ll k = 0; k <= 3; k++){
				ans.x[i][j] = (ans.x[i][j] + a.x[i][k] * b.x[k][j] % mod) % mod;
			}
		}
	}
	return ans;
}

void pushdown(ll k){
	tree[k * 2] = matrix_plus(tree[k * 2], lazy[k]);
	tree[k * 2 + 1] = matrix_plus(tree[k * 2 + 1], lazy[k]);
	lazy[k * 2] = matrix_plus(lazy[k * 2], lazy[k]);
	lazy[k * 2 + 1] = matrix_plus(lazy[k * 2 + 1], lazy[k]);
	memset(lazy[k].x, 0, sizeof(lazy[k].x));
	lazy[k].x[0][0] = lazy[k].x[1][1] = lazy[k].x[2][2] = lazy[k].x[3][3] = 1;
} 

void build(ll k, ll l, ll r){
	memset(lazy[k].x, 0, sizeof(lazy[k].x));
	lazy[k].x[0][0] = lazy[k].x[1][1] = lazy[k].x[2][2] = lazy[k].x[3][3] = 1;
	if(l == r){
		tree[k] = g[l];
		return;
	}
	ll mid = (l + r) >> 1;
	build(k * 2, l, mid);
	build(k * 2 + 1, mid + 1, r);
	tree[k] = matrix_add(tree[k * 2], tree[k * 2 + 1]); 
}

void ctree(ll k, ll l, ll r, ll x, ll y, ll vv, ll opt){
	if(x <= l && r <= y){
		matrix ne;
		memset(ne.x, 0, sizeof(ne.x));
		ne.x[0][0] = ne.x[1][1] = ne.x[2][2] = ne.x[3][3] = 1;
		if(opt == 1) ne.x[1][0] = 1;
		else if(opt == 2) ne.x[2][1] = 1;
		else if(opt == 3) ne.x[0][2] = 1;
		else if(opt == 4) ne.x[3][0] = vv;
		else if(opt == 5) ne.x[1][1] = vv;
		else if(opt == 6) ne.x[2][2] = 0, ne.x[3][2] = vv;
		tree[k] = matrix_plus(tree[k], ne);
		lazy[k] = matrix_plus(lazy[k], ne);
		return; 
	}
	pushdown(k);
	ll mid = (l + r) >> 1;
	if(x <= mid) ctree(k * 2, l, mid, x, y, vv, opt);
	if(y > mid) ctree(k * 2 + 1, mid + 1, r, x, y, vv, opt);
	tree[k] = matrix_add(tree[k * 2], tree[k * 2 + 1]); 
}

matrix query(ll k, ll l, ll r, ll x, ll y){
	if(x <= l && r <= y) return tree[k];
	pushdown(k);
	matrix nex;
	memset(nex.x, 0, sizeof(nex.x));
	ll mid = (l + r) >> 1;
	if(x <= mid) nex = matrix_add(nex, query(k * 2, l, mid, x, y));
	if(y > mid) nex = matrix_add(nex, query(k * 2 + 1, mid + 1, r, x, y));
	return nex; 
}

int main(){
	n = read();
	for(ll i = 1; i <= n; i++){
		for(ll j = 0; j <= 2; j++) g[i].x[0][j] = read();
		g[i].x[0][3] = 1;
	}
	build(1, 1, n);
	m = read();
	while(m--){
		opt = read(), l = read(), r = read();
		if(opt == 7){
			matrix ans = query(1, 1, n, l, r);
			for(ll i = 0; i <= 2; i++) printf("%lld ", ans.x[0][i]);
			putchar('\n');
		}
		else{
			if(opt >= 4) v = read();
			ctree(1, 1, n, l, r, v, opt);
		}
		v = 0;
	}
	return 0;
}

But

因为这份代码的常数极高，你会得到满屏的TLE

1

至此，拉开了卡常的序幕

First

我们可以发现，取模的常数是极高的，那我们可以将%模数，改成减去模数。

同时，memset对long long赋值时，所需要的常数是很大的，我们其实可以将long long改成int，在做乘法时强制转换成 long long即可

修改部分代码:

inline int add(int a){
	return a > mod ? a - mod : a;
}

荣获45分

Second

接下来，我们现将\(*2\)全部改成了\(<<1\)

很好，快了100毫秒

Third

通过BDFS得知，循环是很耗时的。那么我们可以进行如下优化：

如果两个相乘的数，有一个为0，那么其实就可以不用算了：

修改部分代码：

inline matrix matrix_plus(matrix a, matrix b){
	matrix ans;
	memset(ans.x, 0, sizeof(ans.x));
	for(register int i = 0; i <= 3; i++){
		for(register int k = 0; k <= 3; k++){
			if(!a.x[i][k]) continue;
			for(register int j = 0; j <= 3; j++){
				if(!b.x[k][j]) continue; /*相等直接跳过*/
				ans.x[i][j] = add(ans.x[i][j] + 1ll * a.x[i][k] * b.x[k][j] % mod);
			}
		}
	}
	return ans;
}

。。。。在卡题2h后，我A了...

最后，还有点小优化，就直接看最终AC代码吧：

我觉得我的代码还是非常简洁易懂的吧

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

struct matrix{
	int x[4][4];
}g[1000010], tree[1000100], lazy[1000100];

int n, m, opt, l, r, v, mod = 998244353;

inline int add(int a){
	return a > mod ? a - mod : a;
}

inline int read(){
	int f = 1, sum = 0; char ch = getchar();
	while(ch < '0' || ch > '9') f = (ch == '-' ? -1 : 1), ch = getchar();
	while(ch >= '0' && ch <= '9') sum = (sum << 1) + (sum << 3) + (ch ^ 48), ch = getchar();
	return f * sum;
}

inline matrix matrix_add(matrix a, matrix b){
	matrix ans;
	memset(ans.x, 0, sizeof(ans.x));
	ans.x[0][0] = (a.x[0][0] + b.x[0][0]) % mod;
	ans.x[0][1] = (a.x[0][1] + b.x[0][1]) % mod;
	ans.x[0][2] = (a.x[0][2] + b.x[0][2]) % mod;
	ans.x[0][3] = (a.x[0][3] + b.x[0][3]) % mod;
	return ans;
}

inline matrix matrix_plus(matrix a, matrix b){
	matrix ans;
	memset(ans.x, 0, sizeof(ans.x));
	for(register int i = 0; i <= 3; i++){
		for(register int k = 0; k <= 3; k++){
			if(!a.x[i][k]) continue;
			for(register int j = 0; j <= 3; j++){
				if(!b.x[k][j]) continue;
				ans.x[i][j] = add(ans.x[i][j] + 1ll * a.x[i][k] * b.x[k][j] % mod);
			}
		}
	}
	return ans;
}

inline void pushdown(int k){
	tree[k << 1] = matrix_plus(tree[k << 1], lazy[k]);
	tree[(k << 1) | 1] = matrix_plus(tree[(k << 1) | 1], lazy[k]);
	lazy[k << 1] = matrix_plus(lazy[k << 1], lazy[k]);
	lazy[(k << 1) | 1] = matrix_plus(lazy[(k << 1) | 1], lazy[k]);
	memset(lazy[k].x, 0, sizeof(lazy[k].x));
	lazy[k].x[0][0] = lazy[k].x[1][1] = lazy[k].x[2][2] = lazy[k].x[3][3] = 1;
} 

inline void build(int k, int l, int r){
	memset(lazy[k].x, 0, sizeof(lazy[k].x));
	lazy[k].x[0][0] = lazy[k].x[1][1] = lazy[k].x[2][2] = lazy[k].x[3][3] = 1;
	if(l == r){
		tree[k] = g[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(k << 1, l, mid);
	build((k << 1) | 1, mid + 1, r);
	tree[k] = matrix_add(tree[k << 1], tree[(k << 1) | 1]); 
}

inline void ctree(int k, int l, int r, int x, int y, int vv, int opt){
	if(x <= l && r <= y){
		matrix ne;
		memset(ne.x, 0, sizeof(ne.x));
		ne.x[0][0] = ne.x[1][1] = ne.x[2][2] = ne.x[3][3] = 1;
		if(opt == 1) ne.x[1][0] = 1;
		else if(opt == 2) ne.x[2][1] = 1;
		else if(opt == 3) ne.x[0][2] = 1;
		else if(opt == 4) ne.x[3][0] = vv;
		else if(opt == 5) ne.x[1][1] = vv;
		else if(opt == 6) ne.x[2][2] = 0, ne.x[3][2] = vv;
		tree[k] = matrix_plus(tree[k], ne);
		lazy[k] = matrix_plus(lazy[k], ne);
		return; 
	}
	pushdown(k);
	int mid = (l + r) >> 1;
	if(x <= mid) ctree(k << 1, l, mid, x, y, vv, opt);
	if(y > mid) ctree((k << 1) | 1, mid + 1, r, x, y, vv, opt);
	tree[k] = matrix_add(tree[k << 1], tree[(k << 1) | 1]); 
}

inline matrix query(int k, int l, int r, int x, int y){
	if(x <= l && r <= y) return tree[k];
	pushdown(k);
	matrix nex;
	memset(nex.x, 0, sizeof(nex.x));
	int mid = (l + r) >> 1;
	if(x <= mid) nex = matrix_add(nex, query(k << 1, l, mid, x, y));
	if(y > mid) nex = matrix_add(nex, query((k << 1) | 1, mid + 1, r, x, y));
	return nex; 
}

int main(){
	n = read();
	for(register int i = 1; i <= n; i++){
		g[i].x[0][0] = read(), g[i].x[0][1] = read(), g[i].x[0][2] = read();
		g[i].x[0][3] = 1;
	}
	build(1, 1, n);
	m = read();
	while(m--){
		opt = read(), l = read(), r = read();
		if(opt == 7){
			matrix ans = query(1, 1, n, l, r);
			printf("%d %d %d\n", ans.x[0][0], ans.x[0][1], ans.x[0][2]);
		}
		else{
			if(opt >= 4) v = read();
			ctree(1, 1, n, l, r, v, opt);
		}
	}
	return 0;
}

都看到这了，那就。。。

谢谢！！