CSA：Count Arrays（dp）

最新推荐文章于 2021-11-01 15:18:17 发布

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本文探讨了一个有趣的问题：给定长度为N的二进制数组和Q个区间，如何计算满足每个区间内至少包含一个0的数组数量。文章通过动态规划的方法给出了详细的解决方案，并附带了完整的代码实现。

Count Arrays

Time limit: 1000 ms
Memory limit: 256 MB

You are given $Q$ segments. For every $\le i \le Q$ , segment number $i$ is $[l_i,r_i]$ .

A binary array of length $N$ is considered good if and only if for each $\le i \le Q$ there is at least one position between $l_i$ and $r_i$ (inclusive) in this array that contains $0$ .

Find the number of good arrays.

Standard input

The first line contains two integers $N$ and $Q$ .

The next $Q$ lines contain two integers each, $l_i$ and $r_i$ , representing the segments.

Standard output

Print the answer modulo $10^9+7$ on the first line.

Constraints and notes

$\le N, Q \le 10^5$
$\le l_i \le r_i \le N$ for each $\le i \le Q$

Input	Output	Explanation
3 2 1 2 2 3	5	Good arrays are $000$ , $001$ , $010$ , $100$ , $101$
5 1 1 5	31	Out of 32 possible arrays only one ( $11111$ ) is bad.

题意：一个N长度的01串和Q个区间，要求每个区间都至少有1个0，问符合的串共有几种？

思路：dp[i]表示前i个位置且s[i]=0的合法方案数(i后面可以看成全为1)，我们维护一个最大左界L，方程转移到位置j时dp[j]=sigma(dp[L]~dp[j-1])。

# include <bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
const int maxn = 1e5+30;
int dp[maxn], s[maxn], pre[maxn];
int main()
{
    int n, q, l, r;
    scanf("%d%d",&n,&q);
    while(q--)
    {
        scanf("%d%d",&l,&r);
        pre[r] = max(pre[r], l);
    }
    dp[0] = s[0] = 1;
    int bl=0;
    for(int i=1; i<=n; ++i)
    {
        dp[i] = s[i-1];
        if(bl) dp[i] = (dp[i]-s[bl-1]+mod)%mod;
        bl = max(bl, pre[i]);
        s[i] = (s[i-1]+dp[i])%mod;
    }
    printf("%d\n",bl?(s[n]-s[bl-1]+mod)%mod:s[n]);
    return 0;
}