【codeforces 821E】Okabe and El Psy Kongroo

【题目链接】:http://codeforces.com/problemset/problem/821/E

【题意】

一开始位于(0,0)的位置;
然后你每次可以往右上,右，右下3走一步;
(x+1,y+1),(x+1,y),(x+1,y-1)
然后有n条横线,限制你在横线与x轴之间;
问你从(0,0)到(k,0)有多少种行走方案;
k可以很大;
保证横线是连续的,即一条横线完,马上接另外一条横线;

【题解】

如果不考虑那么大的横坐标的话;
用最简单的DP
设f[i][j]表示到(i,j)这个点的方案数;
则有f[i][j] = f[i-1][j+1] + f[i-1][j] + f[i-1][j-1];
在这个DP的基础上,写一个矩阵优化;
构造这么一个矩阵

    (1 1 0    ....0)
    (1 1 1 0   ...0)
    (0 1 1 1 0 ...0)
    (0 0 1 1 1 ...0)
    ....
    (0 0 0 0 0...1 1)
    c[i]+1行，c[i]+1列

然后用一个一列的矩阵

f[1]
f[2]
f[3]
...
f[c[i]+1]

去左乘它;
这里f[i]表示当前这一段横线;
到达纵坐标为i的方案有多少种;
把那个矩阵的min(k,b[i])-a[i]次幂求出来；
然后用那个f矩阵去左乘它得到新的f矩阵就好;
每段都这样做;
最后输出f[1];

【Number Of WA】

0

【反思】

没仔细去想矩阵的构造方法,就跳过去了;
还是不敢去做题啊。被吓到了。

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;

const int G = 16;       //矩阵大小
const LL MOD = 1e9 + 7;    //模数
struct MX
{
    int v[G+5][G+5];
    void O() { ms(v, 0); }
    void E() { ms(v, 0); for (int i = 1; i <= G; ++i)v[i][i] = 1; }
    void P()
    {
        for (int i = 1; i <= G; ++i)
        {
            for (int j = 1; j <= G; ++j)printf("%d ", v[i][j]); puts("");
        }
    }
    MX operator * (const MX &b) const
    {
        MX c; c.O();
        for (int k = 1; k <= G; ++k)
        {
            for (int i = 1; i <= G; ++i) if (v[i][k])
            {
                for (int j = 1; j <= G; ++j)
                {
                    c.v[i][j] = (c.v[i][j] + (LL)v[i][k] * b.v[k][j]) % MOD;
                }
            }
        }
        return c;
    }
    MX operator + (const MX &b) const
    {
        MX c; c.O();
        for (int i = 1; i <= G; ++i)
        {
            for (int j = 1; j <= G; ++j)
            {
                c.v[i][j] = (v[i][j] + b.v[i][j]) % MOD;
            }
        }
        return c;
    }
    MX operator ^ (LL p) const
    {
        MX y; y.E();
        MX x; memcpy(x.v, v, sizeof(v));
        while (p)
        {
            if (p&1) y = y*x;
            x = x*x;
            p>>=1;
        }
        return y;
    }
}xishu;

struct abc{
    LL a,b;
    int c;
};

int n;
LL k;
LL f[20],nf[20];
abc a[N];

int main(){
    //Open();
    Close();
    cin >> n >> k;
    rep1(i,1,n){
        cin >> a[i].a >> a[i].b >> a[i].c;
    }
    f[1] = 1;
    rep1(i,1,n){
        xishu.O();
        rep1(j,1,a[i].c+1){
            int now =j-1;
            for (int jj = 1;jj <= 3;jj++){
                xishu.v[j][now++] = 1;
            }
        }
        xishu = xishu^(min(k,a[i].b)-a[i].a);
        rep1(j,1,16) nf[j] = 0;
        rep1(j,1,a[i].c+1){
            rep1(k,1,a[i].c+1){
                nf[j] = (nf[j]+f[k]*xishu.v[j][k])%MOD;
            }
        }
        rep1(j,1,16) f[j] = nf[j];
    }
    cout << f[1] << endl;
    return 0;
}