HDU 6155 Subsequence Count 线段树维护矩阵

Subsequence Count

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)

Problem Description

Given a binary string S[1,...,N] (i.e. a sequence of 0's and 1's), and Q queries on the string.

There are two types of queries:

1. Flipping the bits (i.e., changing all 1 to 0 and 0 to 1) between l and r (inclusive).
2. Counting the number of distinct subsequences in the substring S[l,...,r].

 

Input

The first line contains an integer T, denoting the number of the test cases.

For each test, the first line contains two integers N and Q.

The second line contains the string S.

Then Q lines follow, each with three integers type, l and r, denoting the queries.

1≤T≤5

1≤N,Q≤105

S[i]∈{0,1},∀1≤i≤N

type∈{1,2}

1≤l≤r≤N

 

Output

For each query of type 2, output the answer mod (109+7) in one line.

 

Sample Input

2 4 4 1010 2 1 4 2 2 4 1 2 3 2 1 4 4 4 0000 1 1 2 1 2 3 1 3 4 2 1 4

 

Sample Output

11 6 8 10

 

题解：

　　设定dp[i][0/1] 到第i个字符以0/1结尾的子序列方案

　　若s[i] = =1 : dp[i][1] = dp[i-1][0] + dp[i-1][1] + 1;

　　　　　　　　dp[i][0] = dp[i-1][0];

       若是s[i] == 0: dp[i][0] =  dp[i-1][0] + dp[i-1][1] + 1;

　　　　　　　　dp[i][1] = dp[i-1][1];

　　写成矩阵，用线段树维护一段连续矩阵乘积，有点卡常数

#include<bits/stdc++.h>
using namespace std;
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N=5e5+20,M=1e6+10,inf=2147483647;

inline LL read(){
    LL 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 x*f;
}

const LL mod = 1e9+7;
char s[N];

struct Matix {
    LL arr[3][3];
}E,F,again,EE;
inline Matix mul(Matix a,Matix b) {
    Matix ans;
    memset(ans.arr,0,sizeof(ans.arr));
    for(int i = 0; i < 3; i++) {
        for(int j = 0; j < 3; j++) {
            for(int k = 0; k < 3; k++)
                ans.arr[i][j] += a.arr[i][k] * b.arr[k][j],ans.arr[i][j] %= mod;
        }
    }
    return ans;
}


Matix v[N * 4],now,facE[N],facF[N];
int lazy[N * 4],fi[N * 4],se[N * 4];


void change(int i) {
    swap(v[i].arr[0][0],v[i].arr[1][0]);
    swap(v[i].arr[0][1],v[i].arr[1][1]);
    swap(v[i].arr[0][2],v[i].arr[1][2]);
    swap(v[i].arr[0][0],v[i].arr[0][1]);
    swap(v[i].arr[1][0],v[i].arr[1][1]);
    swap(v[i].arr[2][0],v[i].arr[2][1]);
}
void push_down(int i,int ll,int rr) {
    if(!lazy[i]) return;
    lazy[ls] ^= 1;
    lazy[rs] ^= 1;
    change(ls);change(rs);
    lazy[i] ^= 1;
}
inline void push_up(int i,int ll,int rr) {
    v[i] = mul(v[ls],v[rs]);
}
void build(int i,int ll,int rr) {
    lazy[i] = 0;
    if(ll == rr) {
        if(s[ll] == '1') v[i] = E,fi[i] = 1,se[i] = 0;
        else v[i] = F,fi[i] = 0,se[i] = 1;
        return ;
    }
    build(ls,ll,mid);
    build(rs,mid+1,rr);
    push_up(i,ll,rr);
}
inline void update(int i,int ll,int rr,int x,int y) {
    push_down(i,ll,rr);
    if(ll == x && rr == y) {
        lazy[i] ^= 1;
        change(i);
        return ;
    }
    if(y <= mid) update(ls,ll,mid,x,y);
    else if(x > mid) update(rs,mid+1,rr,x,y);
    else update(ls,ll,mid,x,mid),update(rs,mid+1,rr,mid+1,y);
    push_up(i,ll,rr);
}
inline Matix ask(int i,int ll,int rr,int x,int y) {
    push_down(i,ll,rr);
    if(ll == x && rr == y) {
        return v[i];
    }
    if(y <= mid) return ask(ls,ll,mid,x,y);
    else if(x > mid) return ask(rs,mid+1,rr,x,y);
    else return mul(ask(ls,ll,mid,x,mid),ask(rs,mid+1,rr,mid+1,y));
    push_up(i,ll,rr);
}

int main() {
    EE.arr[0][0] = 1,EE.arr[1][1] = 1,EE.arr[2][2] = 1;

    E.arr[0][0] = 1;E.arr[0][1] = 1;E.arr[0][2] = 1;
    E.arr[1][1] = 1;E.arr[2][2] = 1;

    F.arr[0][0] = 1;F.arr[1][0] = 1;F.arr[1][1] = 1;
    F.arr[1][2] = 1;F.arr[2][2] = 1;

    again.arr[0][2] = 1;

    int T;
    T = read();
    while(T--) {
        int n,Q;
        n = read();
        Q = read();
        scanf("%s",s+1);
        build(1,1,n);
        while(Q--) {
            int op,l,r;
            op = read();
            l = read();
            r = read();
            if(op == 1)
                update(1,1,n,l,r);
            else {
                now = mul(again,ask(1,1,n,l,r));
                printf("%lld
",(now.arr[0][0]+now.arr[0][1])%mod);
            }
        }
    }
    return 0;
}

 

先考虑怎么算 s_1, s_2, ldots, s_ns​1​​,s​2​​,…,s​n​​ 的答案。设 $dp(i,0/1)$ 表示考虑到 s_is​i​​，以 $0/1$ 结尾的串的数量。那么 $dp(i,0)=dp(i-1,0)+dp(i-1,1)+1$. $1$ 也同理。
那么假设在某个区间之前，$dp(i,0/1)=(x,y)$ 的话，过了这段区间，就会变成 $(ax+by+c,dx+ey+f)$ 的形式，只要用线段树维护这个线性变化就好了。