Atcoder Grand Contest 017 F Zigzag
2018-02-25 18:22
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Zigzag
Problem Statement
有一些点,共n行,第i行有i个点,记第i行第j个点为(i,j),(i,j)每次可以往左下走到(i+1,j),或往右下走到(i+1,j+1),现在有m条由(1,1)走到第n行的路径,设Xp,i表示第p条路径经过了点(i,Xp,i),要求对于1<=p<m,1< i <=n,必须要有Xp,i<=Xp+1,i,接下来给出若干条件形如(a,b,c)表示强制Xa,b+1=Xa,b+c,求这m条路径在满足条件下的方案数模10^9+7。Data Constraint
n,m≤20Solution
我们显然可以把一条路径表示成一个n-1位的二进制数。假设已经求出第i-1条的状态S的方案数fi−1,S,再确认第i条路径的状态S′,设它们的前j-1位都相同,接着第j位开始就不同了,S的第j位为0,S′的为1,那么显然我们可以把S在第j位后的第一个1变成0,把第j位的0变为1,设得到的新状态为T,那么就能同层转移到fi−1,T,显然这样变动S是正确的。
若在第j位后没有1了,那接下来可以乱走了都没问题了,根据这个就可dp了。
时间复杂度O(nm2n−1),详细实现可以参考代码。
Code
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define fo(i,j,l) for(int i=j;i<=l;++i) #define fd(i,j,l) for(int i=j;i>=l;--i) using namespace std; typedef long long LL; const LL N=21,M=(1<<19)+3,mo=1e9+7; int dir ,m2[N<<1],nex[N<<1]; int n,m,k,a,b,c; LL f[2] [M],g [M]; int main() { cin>>n>>m>>k; --n; fo(i,1,k){ scanf("%d%d%d",&a,&b,&c); dir[a][b]=c+1; } int u=0,v; f[0][0][0]=1; m2[1]=1; fo(i,2,n+1)m2[i]=m2[i-1]<<1; int zd=(m2 <<1)-1; fo(p,1,m){ v=u^1; fo(l,0,n)fo(i,0,zd)f[v][l][i]=0; fo(l,0,n)fo(i,0,zd)g[l][i]=0; fd(i,zd,0){ LL ok=f[u][0][i]%mo; nex =n+1; fd(l,n,2)if(i&m2[l])nex[l-1]=l;else nex[l-1]=nex[l]; if(!i)g[0][0]=ok; fo(l,1,n){ if((i&m2[l])==0&&dir[p][l]!=1&&nex[l]!=n+1) f[u][l][(i^m2[nex[l]])^m2[l]]=f[u][l][(i^m2[nex[l]])^m2[l]]+ok; ok=(f[u][l][i]+ok)%mo; if(dir[p][l]&&(((i&m2[l])>0)-(dir[p][l]==2)))break; if(nex[l]==n+1&&(i&m2[l])>0)g[l][i]=ok; } } fo(i,0,n-1) fo(l,0,zd)if(g[i][l]){ g[i][l]=g[i][l]%mo; if(dir[p][i+1]!=2)g[i+1][l]=g[i+1][l]+g[i][l]; if(dir[p][i+1]!=1)g[i+1][l^m2[i+1]]=g[i+1][l^m2[i+1]]+g[i][l]; } fo(i,0,zd)f[v][0][i]=g [i]; u=v; } LL ans=0; fo(i,0,zd)ans=(ans+f[v][0][i])%mo; cout<<ans; }
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