2020杭电多校联合训练（第四场） E.Equal Sentences （dp）

题面

Problem Description
Sometimes, changing the order of the words in a sentence doesn't influence understanding. For example, if we change "what time is it", into "what time it is"; or change "orz zhang three ak world final", into "zhang orz three world ak final", the meaning of the whole sentence doesn't change a lot, and most people can also understand the changed sentences well.

Formally, we define a sentence as a sequence of words. Two sentences S and T are almost-equal if the two conditions holds:

1. The multiset of the words in S is the same as the multiset of the words in T.
2. For a word α, its ith occurrence in S and its ith occurrence in T have indexes differing no more than 1. (The kth word in the sentence has index k.) This holds for all α and i, as long as the word α appears at least i times in both sentences.

Please notice that "almost-equal" is not a equivalence relation, unlike its name. That is, if sentences A and B are almost-equal, B and C are almost-equal, it is possible that A and C are not almost-equal.

Zhang3 has a sentence S consisting of n words. She wants to know how many different sentences there are, which are almost-equal to S, including S itself. Two sentences are considered different, if and only if there is a number i such that the ith word in the two sentences are different. As the answer can be very large, please help her calculate the answer modulo 109+7.

Input
The first line of the input gives the number of test cases, T(1≤T≤100). T test cases follow.

For each test case, the first line contains an integer n(1≤n≤105), the number of words in the sentence.

The second line contains the sentence S consisting of n words separated by spaces. Each word consists of no more than 10 lowercase English letters.

The sum of n in all test cases doesn't exceed 2×105.

Output
For each test case, print a line with an integer, representing the answer, modulo 109+7.

Sample Input
2
6
he he zhou is watching you
13
yi yi si wu yi si yi jiu yi jiu ba yao ling

Sample Output
8
233

思路

基础dp，我们枚举是否交换这个单词和前面一个单词，那么状态就从dp[i-2]和dp[i-1] 转移了过来，边界注意0和1的下标值都为1；

代码实现

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=0;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
const int maxn=1e5+10;
const int mod=1e9+7;
int dp[maxn];
string ss[maxn];
int n;

int main () {
   int t;
   cin>>t;
   while (t--) {
       cin>>n;
       rep (i,1,n) cin>>ss[i];
       MT (dp,0);
       dp[0]=dp[1]=1;
       rep (i,2,n) {
           if (ss[i]!=ss[i-1]) dp[i]=(dp[i]+dp[i-2])%mod;
           dp[i]+=dp[i-1];
           dp[i]%=mod; 
       }
       cout<<dp[n]<<endl;
   }

   return 0;
}