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+,M=1e6+,inf=;

inline LL read(){
    LL x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}

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

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

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

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

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

    E.arr[][] = ;E.arr[][] = ;E.arr[][] = ;
    E.arr[][] = ;E.arr[][] = ;

    F.arr[][] = ;F.arr[][] = ;F.arr[][] = ;
    F.arr[][] = ;F.arr[][] = ;

    again.arr[][] = ;

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

 

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)dp(i,0/1) 表示考虑到 s_is​i​​，以 0/10/1 结尾的串的数量。那么 dp(i, 0) =dp(i - 1, 0) + dp(i - 1, 1) + 1dp(i,0)=dp(i−1,0)+dp(i−1,1)+1. 11 也同理。
那么假设在某个区间之前，dp(i, 0/1) = (x, y)dp(i,0/1)=(x,y) 的话，过了这段区间，就会变成 (ax + by + c, dx + ey + f)(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。