BZOJ4321 queue2(动态规划)

  考虑套路地将1~n依次加入排列。设f[i][j]为已将1~i加入排列,有j对不合法的方案数。加入i+1时可能减少一对不合法的,可能不变,可能增加一对,对于i+1与i的关系再增设0/1/2状态表示i与左边/右边的数是否构成不合法对即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 1010
#define P 7777777
int n,f[N][N][3];
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4321.in","r",stdin);
    freopen("bzoj4321.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    f[1][0][0]=1;
    for (int i=2;i<=n;i++)
        for (int j=0;j<i;j++)
        {
            inc(f[i][j][0],(1ll*f[i-1][j][0]*(i-2-j)+1ll*(f[i-1][j][1]+f[i-1][j][2])*(i-1-j))%P);
            inc(f[i][j][0],(1ll*f[i-1][j+1][0]*(j+1)+1ll*(f[i-1][j+1][1]+f[i-1][j+1][2])*j)%P);
            if (j) inc(f[i][j][1],(f[i-1][j-1][0]+f[i-1][j-1][1])%P);inc(f[i][j][1],f[i-1][j][2]);
            if (j) inc(f[i][j][2],(f[i-1][j-1][0]+f[i-1][j-1][2])%P);inc(f[i][j][2],f[i-1][j][1]);
        }
    cout<<f[n][0][0];
    return 0;
}

 

posted @ 2018-10-27 02:01  Gloid  阅读(165)  评论(0编辑  收藏  举报