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].
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
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.
首先考虑怎么求一个01串有多少种不同的子序列
dp[i][0]表示考虑到第i位时以0结尾的不同的子序列个数
dp[i][1]表示考虑到第i位时以1结尾的不同的子序列个数
若第i+1位为1，则有：
以01为结尾的子序列个数为dp[i][0]
以11为结尾的子序列个数为dp[i][1]
只有一个1的子序列个数为1
以0为结尾的子序列个数为dp[i][0]
以上4种情况统计了考虑到i+1处时所有的子序列
于是有 dp[i+1][1]=dp[i][0]+dp[i][1]+1
dp[i+1][0]=dp[i][0]
(1 1 0
（dp[i][0],dp[i][1],1）* 0 1 0 =(dp[i+1][0],dp[i+1][1],1)
0 1 1)
记为A矩阵。
若i+1位为0同理有：
dp[i+1][1]=dp[i][0]
dp[i+1][0]=dp[i][0]+dp[i][1]+1
对应矩阵为（记为B矩阵）
1 0 0
1 1 0
1 0 1
其次考虑优化的问题。将以上的两种转移视为矩阵，用线段树维护矩阵的乘积即可。
对于将所有0换成1,1换成0的操作而言，等价于将所有A矩阵换成B，B换成A,而A和B通过交换1,2行及1,2列可互相转换，或者说，乘以初等矩阵Fs,t ,该矩阵的逆为自身。（记其为F）
0 1 0
1 0 0
0 0 1
于是有FAFFAF……FBF……=FAAB……F，或者说，将大量0换成1,1换成0的时候只需要在乘积最外面进行一次交换1,2行与1,2列的操作。
然而。。仍然TLE..估计是被卡了常数。心塞。
#include<bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i=a;i<=b;++i) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 int N,Q; const long long int mo=1e9+7; char s[5010]; int lazy[5050<<2]; struct Matrix{ int n,m; long long a[3][3]; Matrix (){clear();} void clear(){ n=m=3; memset(a,0,sizeof(a)); } Matrix operator *(const Matrix &b) const{ Matrix tmp; for (int i=0;i<n;++i) for (int j=0;j<b.m;++j) for (int k=0;k<m;++k) tmp.a[i][j]=(tmp.a[i][j]+a[i][k]*b.a[k][j])%mo; return tmp; } }; Matrix A0,A1,E; Matrix cnt[5050<<2]; inline void init() { A0.a[0][0]=1,A0.a[0][1]=0,A0.a[0][2]=0; A0.a[1][0]=1,A0.a[1][1]=1,A0.a[1][2]=0; A0.a[2][0]=1,A0.a[2][1]=0,A0.a[2][2]=1; A1.a[0][0]=1,A1.a[0][1]=1,A1.a[0][2]=0; A1.a[1][0]=0,A1.a[1][1]=1,A1.a[1][2]=0; A1.a[2][0]=0,A1.a[2][1]=1,A1.a[2][2]=1; E.a[0][0]=1,E.a[0][1]=0,E.a[0][2]=0; E.a[1][0]=0,E.a[1][1]=1,E.a[1][2]=0; E.a[2][0]=0,E.a[2][1]=0,E.a[2][2]=1; } inline void Pushup(int rt) { cnt[rt]=cnt[rt<<1]*cnt[rt<<1|1]; } inline void build(int l,int r,int rt) { if(l==r) { if(s[l-1]-'0'==0) { cnt[rt]=A0; } else cnt[rt]=A1; return; } int m=(l+r)>>1; build(lson); build(rson); Pushup(rt); } inline void change(Matrix &X) { swap(X.a[0][0],X.a[1][1]); swap(X.a[0][1],X.a[1][0]); swap(X.a[2][0],X.a[2][1]); } inline void pushdown(int rt) { if(lazy[rt]) { change(cnt[rt<<1]); change(cnt[rt<<1|1]); lazy[rt<<1]^=1; lazy[rt<<1|1]^=1; lazy[rt]=0; } } inline void update(int a,int b,int l,int r,int rt) { if(l>=a&&r<=b) { change(cnt[rt]); lazy[rt]^=1; return; } pushdown(rt); int m=(l+r)>>1; if(a<=m) update(a,b,lson); if(b>m) update(a,b,rson); Pushup(rt); } inline void Input() { scanf("%d%d",&N,&Q); scanf("%s",s); } inline Matrix query(int a,int b,int l,int r,int rt) { if(l>=a&&r<=b) return cnt[rt]; pushdown(rt); Matrix t1=E,t2=E; int m=(r+l)>>1; if(a<=m) t1=query(a,b,lson); if(b>m) t2=query(a,b,rson); return t1*t2; } int main() { //freopen("in.txt","r",stdin); int T,type,l,r; scanf("%d",&T); init(); rep(t,1,T) { Input(); build(1,N,1); rep(i,1,Q) { scanf("%d%d%d",&type,&l,&r); if(type==1) { update(l,r,1,N,1); // rep(j,1,2*N) printf("i=%d dp%d=%lld ",i,j,(cnt[j].a[2][0]+cnt[j].a[2][1])%mo); } else { Matrix tmp; tmp=query(l,r,1,N,1); printf("%lld ",(tmp.a[2][0]+tmp.a[2][1])%mo); } } } return 0; }
代码如下：