START194 - Good Arrays (Not Cyclic)（线性dp）

思路讲解

这个的状态定义比较巧妙，定义的是a1和aN/2的差，a1+a2与aN/2+1+aN/2+2的差

FOR(i,1,N/2){
	FOR(d,-(N-1),N-1){
		FOR(j,0,N/2*N/2){
			if(j+d<0) continue;
			if(d<=0){
				dp[i][j]+=dp[i-1][j+d]*ct_[i][abs(d)];
				dp[i][j]%=mod;
			}else{
				dp[i][j]+=dp[i-1][j+d]*ct[i][d];
				dp[i][j]%=mod;
			}
			
		}
	}
}

AC代码

https://www.codechef.com/viewsolution/1172894021

源代码

// Problem: Good Arrays (Not Cyclic)
// Contest: CodeChef - START194B
// URL: https://www.codechef.com/START194B/problems/CNTGOODARRH
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(1e2)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)1e4+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;const double pi=acos(-1.0);

ll N,M,Q,X,K,T,lT;
char S[MAXN][MAXN];
//   dp[i][j] 第i行，diff=j。
ll dp[MAXN][MAXM];
// 辅助数组（count 简写）
// ct[i][j] i row, diff=j
ll ct[MAXN][MAXN];
// ct_[i][j] i row, diff=-j
ll ct_[MAXN][MAXN];
/*

*/
void init(){
	FOR(i,0,N+5){
		FOR(j,0,N+5){
			ct[i][j]=0;
			ct_[i][j]=0;
		}
	}
	FOR(i,0,N+5){
		FOR(j,0,N/2*N/2+5){
			dp[i][j]=0;
		}
	}
	dp[0][0]=1;
}
inline void solve(){
	cin>>N;
	init();
	vector<ll> cnt(N+5,0);
	FOR(i,1,N){
		FOR(j,1,N){
			cin>>S[i][j];
			if(S[i][j]=='1'){
				cnt[i]++;
			}
		}
	}
	ll ans=1;	// 统计总数目
	FOR(i,1,N){
		ans*=cnt[i];
		ans%=mod;
	}
	// 一般来说，求什么，dp里装的就是什么。
	// 那么这里面dp装的就是有多少个坏数组。
	// 那么我们一层一层推怎么样？
	ll n=N/2;
	FOR(i,1,N/2){	// 辅助数组计算
		FOR(d,0,N-1){
			FOR(j,1,N){
				if(j+d>N) break;
				if(S[i][j]=='1' && S[i+n][j+d]=='1'){
					ct[i][d]++;
					ct[i][d]%=mod;
				}
			}
		}
	}
	FOR(i,1,N/2){	// 辅助数组计算
		FOR(d,0,N-1){
			FOR(j,d+1,N){
				if(S[i][j]=='1' && S[i+n][j-d]=='1'){
					ct_[i][d]++;
					ct_[i][d]%=mod;
				}
			}
		}
	}
#ifdef LOCAL
    FOR(i,1,n){
    	FOR(j,0,N-1){
    		cerr<<ct[i][j]<<" ";
    	}
    	cerr<<"\n";
    }
    FOR(i,1,n){
    	FOR(j,0,N-1){
    		cerr<<ct_[i][j]<<" ";
    	}
    	cerr<<"\n";
    }
#endif
	FOR(i,1,N/2){
		FOR(d,-(N-1),N-1){
			FOR(j,0,N/2*N/2){
				if(j+d<0) continue;
				if(d<=0){
					dp[i][j]+=dp[i-1][j+d]*ct_[i][abs(d)];
					dp[i][j]%=mod;
				}else{
					dp[i][j]+=dp[i-1][j+d]*ct[i][d];
					dp[i][j]%=mod;
				}
				
			}
		}
	}
#ifdef LOCAL
    FOR(i,1,n){
    	FOR(j,0,N/2*N/2){
    		cerr<<dp[i][j]<<" ";
    	}
    	cerr<<"\n";
    }
#endif
	ans-=dp[n][0];
	ans%=mod;
	if(ans<0) ans+=mod;
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>lT;
	while(lT--)
		solve();
	return 0;
}
/*
AC
https://www.codechef.com/viewsolution/1172894021
*/

心路历程（WA，TLE，MLE……）