HDU6155(dp + 线段树区间更新 + 矩阵性质)

Subsequence Count

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 304    Accepted Submission(s): 126

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

24 410102 1 42 2 41 2 32 1 44 400001 1 21 2 31 3 42 1 4

 

Sample Output

116810
题意：给你一个01串，两种操作，一种查询操作，查询l, r中有多少不同的子串（不一定要连续），第二种时修改操作，把区间l, r中的0,1翻转，0变成1， 1变成0.
解题思路：先考虑一个问题，给你一个串我们怎样求这个串的子串的个数。我们容易想到用dp。
dp[i][0]表示考虑到第i位以0结尾的子串个数，dp[i][1]表示考虑到第i位以1结尾的子串的个数，则易得以下状态转移方程：
 当s[i] = '0'时
dp[i][0] = dp[i - 1][0] + dp[i - 1][1] + 1;
dp[i][1] = dp[i - 1][1];
当s[i] = '1'时
dp[i][0] = dp[i - 1][0];
dp[i][1] = dp[i - 1][0] + dp[i - 1][1] + 1;
然后我们可以写出递推式的系数矩阵：
当s[i] = '0'时
dp[i][0]    1 1 1   dp[i - 1][0]
dp[i][1] =  0 1 0 * dp[i - 1][1]
1           0 0 1   1
当s[i] = '1'
dp[i][0]    1 1 1   dp[i - 1][0]
dp[i][1] =  0 1 0 * dp[i - 1][1]
1           0 0 1   1
然后我们用线段树维护区间矩阵乘积就行。
当遇到修改操作时，我们可以发现0变成1,1变成0，对应的区间矩阵结果就变成然后的矩阵第一列和第二列交换，然后第一行和第二行交换后的矩阵，自己可以用手演算一下，这是个规律（不知道那些大佬是怎样看出来的）。
#include <bits/stdc++.h>using namespace std;const int maxn = 1e5 + 10;typedef long long LL;const LL mod = 1e9 + 7;int n, q;char s[maxn];struct Matrix{    LL v[3][3];    Matrix operator *(const Matrix &res){        Matrix ans;        for(int i = 0; i <= 2; i++)        {            for(int j = 0; j <= 2; j++)            {                ans.v[i][j] = 0;                for(int k = 0; k <= 2; k++)                {                    ans.v[i][j] += (v[i][k] * res.v[k][j]) % mod;                    ans.v[i][j] %= mod;                }            }        }        return ans;    }    void rev(){//矩阵第一列与第二列交换，然后第一行与第二行交换        swap(v[0][0], v[1][1]);        swap(v[1][0], v[0][1]);        swap(v[0][2], v[1][2]);    }};Matrix x1 = {1, 1, 1, 0, 1, 0, 0, 0, 1};//s[i]为0时的递推矩阵Matrix x2 = {1, 0, 0, 1, 1, 1, 0, 0, 1};//s[i]为1时的递推矩阵Matrix s1 = {1, 0, 0, 0, 0, 0, 1, 0, 0};//s[i]为0时的初始矩阵Matrix s2 = {0, 0, 0, 1, 0, 0, 1, 0, 0};//s[i]为0时的初始矩阵struct node{    int l, r;    Matrix sum;//区间矩阵乘积    int add;//是否翻转，0表示不翻转，1表示翻转}Node[maxn<<2];void pushUp(int i){    int lson = i<<1;    int rson = lson|1;    Node[i].sum = Node[rson].sum * Node[lson].sum;}void pushDown(int i){    int lson = i<<1;    int rson = lson|1;    Node[lson].add ^= 1;    Node[rson].add ^= 1;    Node[lson].sum.rev();    Node[rson].sum.rev();    Node[i].add = 0;}void build(int i, int l, int r){    Node[i].l = l;    Node[i].r = r;    Node[i].add = 0;    if(l == r)    {        if(s[l] == '0') Node[i].sum = x1;        else Node[i].sum = x2;        return;    }    int f = i;    i <<= 1;    int mid = (l + r)>>1;    build(i, l, mid);    build(i|1, mid + 1, r);    pushUp(f);}void update(int i, int l, int r){    if(Node[i].l == l && Node[i].r == r)    {        Node[i].add ^= 1;        Node[i].sum.rev();        return;    }    if(Node[i].add) pushDown(i);    int f = i;    i <<= 1;    if(r <= Node[i].r) update(i, l, r);    else if(l >= Node[i|1].l) update(i|1, l, r);    else    {        update(i, l, Node[i].r);        update(i|1, Node[i|1].l, r);    }    pushUp(f);}Matrix queryMatrix(int i, int l, int r){    if(Node[i].l == l && Node[i].r == r)    {        return Node[i].sum;    }    if(Node[i].add) pushDown(i);    i <<= 1;    if(r <= Node[i].r) return queryMatrix(i, l, r);    else if(l >= Node[i|1].l) return queryMatrix(i|1, l, r);    else    {        Matrix t1 = queryMatrix(i, l, Node[i].r);        Matrix t2 = queryMatrix(i|1, Node[i|1].l, r);        return t2 * t1;    }}int queryId(int i, int loc){    if(Node[i].l == Node[i].r)    {        return Node[i].add;    }    if(Node[i].add) pushDown(i);    i <<= 1;    if(loc <= Node[i].r) return queryId(i, loc);    else if(loc >= Node[i|1].l) return queryId(i|1, loc);}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%d%d", &n, &q);        scanf("%s", s + 1);        build(1, 1, n);        int op, l, r;        for(int i = 1; i <= q; i++)        {            scanf("%d%d%d", &op, &l, &r);            if(op == 1)            {                update(1, l, r);            }            else            {                if(l == r)                {                    printf("1\n");                    continue;                }                int id = queryId(1, l);                Matrix re = queryMatrix(1, l + 1, r);                Matrix term;                if(id == 0)                {                    if(s[l] == '0') term = s1;                    else term = s2;                }                else                {                    if(s[l] == '0') term = s2;                    else term = s1;                }                term = re * term;                LL cnt = (term.v[0][0] + term.v[1][0]) % mod;                printf("%lld\n", cnt);            }        }    }    return 0;}

Source

2017中国大学生程序设计竞赛 - 网络选拔赛