THUSC 2017 大魔法师

一个序列，每个物品有三个权值 $A,B,C$

要求维护：

1.区间 $A_i+=B_i$

2.区间 $B_i+=C_i$

3.区间 $C_i+=A_i$

4.区间 $A_i+=v$

5.区间 $B_i \times = v$

6.区间 $C_i = v$

7.询问区间 $A,B,C$ 各自的和

 

线段树，每个点维护 $A,B,C,区间长度$

每次修改相当于区间乘一个转移矩阵

时间复杂度 $O(16nlogn)$

垫底

#include <bits/stdc++.h>
#define LL long long
using namespace std;
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
inline int read() {
    int x = 0, f = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar())if (ch == '-')f = -f;
    for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0';
    return x * f;
}
const int mod = 998244353, maxn = 2.5e5 + 10;
#define ls (x << 1)
#define rs ((x << 1) | 1)
int n, q, A[maxn], B[maxn], C[maxn];
struct Matrix {
    int a[4][4];
    Matrix() {memset(a, 0, sizeof(a));}
    Matrix operator * (const Matrix &b) const {
        Matrix c;
        rep(i, 0, 3) rep(j, 0, 3) rep(k, 0, 3)
            (c.a[i][j] += (1LL * a[i][k] * b.a[k][j] % mod)) %= mod;
        return c;
    }
    Matrix operator + (const Matrix &b) const {
        Matrix c;
        rep(i, 0, 3) rep(j, 0, 3) c.a[i][j] = (a[i][j] + b.a[i][j]) % mod;
        return c;
    }
}tag[maxn << 2];
int seg[maxn << 2][4];
void mul(int *f, Matrix gg) {
    int tmp[4] = {0, 0, 0, 0};
    rep(i, 0, 3) rep(j, 0, 3) (tmp[j] += (1LL * f[i] * gg.a[i][j] % mod)) %= mod;
    rep(i, 0, 3) f[i] = tmp[i];
}
inline int clear(Matrix x) {
    if (!(x.a[0][0] == 1 && x.a[1][1] == 1 && x.a[2][2] == 1 && x.a[3][3] == 1))return false;
    if (x.a[0][1] || x.a[0][2] || x.a[0][3] || x.a[1][0] || x.a[1][2] || x.a[1][3])return false;
    if (x.a[2][0] || x.a[2][1] || x.a[2][3] || x.a[3][0] || x.a[3][1] || x.a[3][2]) return false;
    return true;
}
inline void pushup(int x) {
    rep(i, 0, 3) seg[x][i] = (seg[ls][i] + seg[rs][i]) % mod;
}
inline void pushdown(int x) {
    if(clear(tag[x])) return;
    tag[ls] = tag[ls] * tag[x], tag[rs] = tag[rs] * tag[x];
    mul(seg[ls], tag[x]), mul(seg[rs], tag[x]);
    rep(i, 0, 3) rep(j, 0, 3) tag[x].a[i][j] = (i == j);
}
inline void build(int x, int l, int r) {
    if(l == r) {
        seg[x][0] = A[l]; seg[x][1] = B[l]; seg[x][2] = C[l]; seg[x][3] = 1;
        return;
    }
    int mid = (l + r) >> 1;
    build(ls, l, mid); build(rs, mid + 1, r);
    pushup(x);
}
Matrix cur; int res[4];
inline void update(int x, int l, int r, int L, int R) {
    if(L <= l && r <= R) {
        mul(seg[x], cur);
        tag[x] = tag[x] * cur;
        return;
    }
    pushdown(x);
    int mid = (l + r) >> 1;
    if(L <= mid) update(ls, l, mid, L, R);
    if(R > mid) update(rs, mid + 1, r, L, R);
    pushup(x);
}
inline void query(int x, int l, int r, int L, int R) {
    if(L <= l && r <= R) {
        rep(i, 0, 3) (res[i] += seg[x][i]) %= mod;
        return;
    }
    pushdown(x);
    int mid = (l + r) >> 1;
    if(L <= mid) query(ls, l, mid, L, R);
    if(R > mid) query(rs, mid + 1, r, L, R);
}
int main() {
    n = read();
    rep(i, 1, (n<<2)) rep(j, 0, 3) rep(k, 0, 3) tag[i].a[j][k] = (j == k);
    rep(i, 1, n) A[i] = read(), B[i] = read(), C[i] = read();
    build(1, 1, n);
    q = read();
    while(q--) {
        int opt = read(), l = read(), r = read();
        rep(i, 0, 3) rep(j, 0, 3) cur.a[i][j] = (i == j);
        if(opt == 1) cur.a[1][0]++;
        else if(opt == 2) cur.a[2][1]++;
        else if(opt == 3) cur.a[0][2]++;
        else if(opt == 4) (cur.a[3][0] += read()) %= mod;
        else if(opt == 5) cur.a[1][1] = read();
        else if(opt == 6) cur.a[2][2] = 0, cur.a[3][2] = read();
        if(opt != 7) update(1, 1, n, l, r);
        if(opt == 7) {
            rep(i, 0, 3) res[i] = 0;
            query(1, 1, n, l, r);
            printf("%d %d %d\n", res[0], res[1], res[2]);
            continue;
        }
    }
}

View Code

 

标签：ch,rs,int,mid,魔法师,区间,2017,THUSC,getchar
来源： https://www.cnblogs.com/Kong-Ruo/p/10491587.html