Luogu P5469 [NOI2018]机器人 (DP、多项式)
作者：互联网
不用FFT的多项式(大雾)
题目链接: https://www.luogu.org/problemnew/show/P5469
(这题在洛谷都成绿题了海星)
题解: 首先我们考虑，一个序列位置最右边的最大值可以走遍整个序列，并且其余任何点都不能跨过这个位置。
所以我们可以区间dp, \(dp[l][r][x]\)表示区间\([l,r]\)最大值不超过\(x\)的方案数，枚举最大值点\(mid\)及其值\(k\), \(dp[l][r][x]=\sum_{mid}\sum_{k}dp[l][mid-1][k]\times dp[mid+1][r][k-1]\), 也可以设\(dp[l][r][x]\)表示区间\([l,r]\)的最大值恰好为\(x\)的方案数，枚举最大值点\(mid\)则有\(dp[l][r][x]=\sum_{mid}\sum_{k\le x}dp[l][mid-1][k]\sum_{k<x}dp[mid+1][r][k]\).
可获得\(35\)分，当然如果你有梦想数组开大点卡卡常就有\(50\)分了。（然而我在考场上没梦想\(35\)分滚粗了）
然后正解的话，恰好为\(x\)那种状态比较好。
首先离散化，那么我们发现当\(k\)在每一段区间内时，转移是类似的。
进一步，归纳易证当\(k\)在某一段区间内时\(dp[l][r][k]\)是关于\(k\)的不超过\((r-l)\)次多项式。
于是记忆化搜索一波，使用多项式前缀和进行转移，这样枚举\(mid\)之后复杂度为多项式次数的平方。
多项式前缀和需要预处理\(s_k(x)=\sum^{x}_{i=0}x^k\), 这是一个\((k+1)\)次多项式，所以Lagrange插值求出来系数。
裸做\(80\)分起步(我裸做了一波得了\(85\))
剪枝优化可获得\(100\)分。
好难写啊，我好菜啊……
代码
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<vector>
#include<algorithm>
#define llong long long
using namespace std;
const int P = 1e9+7;
const int N = 301;
llong quickpow(llong x,llong y)
{
llong cur = x,ret = 1ll;
for(int i=0; y; i++)
{
if(y&(1ll<<i)) {y-=(1ll<<i); ret = ret*cur%P;}
cur = cur*cur%P;
}
return ret;
}
llong mulinv(llong x) {return quickpow(x,P-2);}
llong aux[N+4],aux2[N+4];
struct Polynomial
{
vector<llong> a; int n;
Polynomial() {}
Polynomial(int _n) {n = _n; for(int i=0; i<=n; i++) a.push_back(0ll);}
void clear() {n = 0; a.clear(); a.push_back(0ll);}
void output() {printf("deg%d, ",n); for(int i=0; i<=n; i++) printf("%lld ",a[i]); puts("");}
void simplify()
{
while(n>=1 && a[n]==0)
{
a.pop_back();
n--;
}
}
Polynomial operator +(Polynomial &arg) const
{
Polynomial ret(max(n,arg.n));
for(int i=0; i<=min(n,arg.n); i++)
{
ret.a[i] = (a[i]+arg.a[i])%P;
}
for(int i=min(n,arg.n)+1; i<=n; i++) ret.a[i] = a[i];
for(int i=min(n,arg.n)+1; i<=arg.n; i++) ret.a[i] = arg.a[i];
return ret;
}
Polynomial operator -(Polynomial &arg) const
{
Polynomial ret(max(n,arg.n));
for(int i=0; i<=min(n,arg.n); i++)
{
ret.a[i] = (a[i]-arg.a[i]+P)%P;
}
for(int i=min(n,arg.n)+1; i<=n; i++) ret.a[i] = a[i];
for(int i=min(n,arg.n)+1; i<=arg.n; i++) ret.a[i] = P-arg.a[i];
return ret;
}
Polynomial operator *(Polynomial &arg) const
{
Polynomial ret(n+arg.n);
for(int i=0; i<=n; i++)
{
// if(a[i]==0) continue;
for(int j=0; j<=arg.n; j++)
{
ret.a[i+j] = (ret.a[i+j]+a[i]*arg.a[j])%P;
}
}
return ret;
}
llong calc(llong x)
{
llong ret = 0ll;
for(int i=n; i>=0; i--)
{
ret = (ret*x+a[i])%P;
}
return ret;
}
void interpoly(int _n,llong ax[],llong ay[])
{
n = _n; for(int i=0; i<=n; i++) a.push_back(0ll);
for(int i=0; i<=n+1; i++) aux[i] = 0ll;
aux[0] = 1ll;
for(int i=0; i<=n; i++)
{
for(int j=i+1; j>0; j--)
{
aux[j] = (aux[j-1]-aux[j]*ax[i]%P+P)%P;
}
aux[0] = P-aux[0]*ax[i]%P;
}
for(int i=0; i<=n; i++)
{
llong tmp = 1ll;
for(int j=0; j<=n; j++)
{
if(i==j) continue;
tmp = tmp*(ax[i]-ax[j]+P)%P;
}
llong coe = mulinv(tmp);
for(int j=n+1; j>=0; j--) {aux2[j] = aux[j];}
for(int j=n; j>=0; j--)
{
a[j] = (a[j]+aux2[j+1]*coe%P*ay[i])%P;
aux2[j] = (aux2[j]+ax[i]*aux2[j+1])%P;
}
}
}
};
Polynomial tmp1,tmp2,tmp3;
Polynomial dp[2661][(N<<1)+3],sdp[2661][(N<<1)+3];
llong lval[2661][(N<<1)+3],rval[2661][(N<<1)+3];
int dpid[N+4][N+4];
Polynomial spw[N+4];
struct Interval
{
llong lb,rb; //[1,2n]
} a[N+3];
vector<llong> disc;
llong spwx[N+3],spwy[N+3];
int mx[N+3][N+3];
int n,nsta;
llong ans;
int getid(llong x) {return lower_bound(disc.begin(),disc.end(),x)-disc.begin();} //no +1
Polynomial prefixsum(Polynomial poly)
{
Polynomial ret(poly.n+1);
for(int i=0; i<=poly.n; i++)
{
for(int j=0; j<=i+1; j++)
{
ret.a[j] = (ret.a[j]+poly.a[i]*spw[i].a[j])%P;
}
}
return ret;
}
void dfs(int l,int r,int x)
{
Polynomial tmp1,tmp2,tmp3;
if(dpid[l][r] && dp[dpid[l][r]][x].a.size()>0) {return;}
if(!dpid[l][r]) {nsta++; dpid[l][r] = nsta;}
if(l==r)
{
if(!dpid[l][r]) {nsta++; dpid[l][r] = nsta;}
dp[dpid[l][r]][x] = Polynomial(0); dp[dpid[l][r]][x].a[0] = (x>=a[l].lb&&x<=a[l].rb) ? 1ll : 0ll;
sdp[dpid[l][r]][x] = prefixsum(dp[dpid[l][r]][x]);
lval[dpid[l][r]][x] = sdp[dpid[l][r]][x].calc(disc[x-1]);
rval[dpid[l][r]][x] = sdp[dpid[l][r]][x].calc(disc[x]);
return;
}
dp[dpid[l][r]][x].clear(); sdp[dpid[l][r]][x].clear();
if(mx[l][r]>x) return;
for(int lenl=(r-l+1)>>1; lenl<=(r-l+1)+1-((r-l+1)>>1); lenl++)
{
int mid = l+lenl-1;
if(!(x>=a[mid].lb && x<=a[mid].rb)) {continue;} //´Ë´¦ÒªÌØÅÐ
tmp1.clear(); tmp2.clear();
if(mid>l)
{
for(int k=1; k<=x; k++)
{
dfs(l,mid-1,k);
if(k<x)
{
tmp1.a[0] = (tmp1.a[0]+rval[dpid[l][mid-1]][k]-lval[dpid[l][mid-1]][k]+P)%P;
}
else
{
tmp1 = tmp1+sdp[dpid[l][mid-1]][k];
tmp1.a[0] = (tmp1.a[0]-lval[dpid[l][mid-1]][k]+P)%P;
}
}
}
else
{
tmp1 = Polynomial(0); tmp1.a[0] = 1ll;
}
if(mid<r)
{
for(int k=0; k<=x; k++)
{
dfs(mid+1,r,k);
if(k<x)
{
tmp2.a[0] = (tmp2.a[0]+rval[dpid[mid+1][r]][k]-lval[dpid[mid+1][r]][k]+P)%P;
}
else
{
tmp2 = tmp2+sdp[dpid[mid+1][r]][k];
tmp2 = tmp2-dp[dpid[mid+1][r]][k];
tmp2.a[0] = (tmp2.a[0]-lval[dpid[mid+1][r]][k]+P)%P;
}
}
}
else
{
tmp2 = Polynomial(0); tmp2.a[0] = 1ll;
}
tmp3 = tmp1*tmp2;
dp[dpid[l][r]][x] = dp[dpid[l][r]][x]+tmp3;
}
sdp[dpid[l][r]][x] = prefixsum(dp[dpid[l][r]][x]);
lval[dpid[l][r]][x] = sdp[dpid[l][r]][x].calc(disc[x-1]);
rval[dpid[l][r]][x] = sdp[dpid[l][r]][x].calc(disc[x]);
}
int main()
{
scanf("%d",&n);
for(int i=1; i<=n; i++) {scanf("%lld%lld",&a[i].lb,&a[i].rb); disc.push_back(a[i].lb-1); disc.push_back(a[i].rb);}
for(int i=0; i<=n; i++)
{
spwx[0] = 0ll; spwy[0] = 0ll;
for(int j=1; j<=i+1; j++)
{
spwx[j] = j;
spwy[j] = (spwy[j-1]+quickpow(j,i))%P;
}
spw[i].interpoly(i+1,spwx,spwy);
}
sort(disc.begin(),disc.end()); disc.erase(unique(disc.begin(),disc.end()),disc.end());
for(int i=1; i<=n; i++) {a[i].lb = getid(a[i].lb); a[i].rb = getid(a[i].rb);}
nsta = 1; for(int i=0; i<disc.size(); i++)
{
dp[1][i] = Polynomial(0); dp[1][i].a[0] = 1ll;
sdp[1][i] = prefixsum(dp[1][i]);
lval[1][i] = sdp[1][i].calc(disc[i-1]);
rval[1][i] = sdp[1][i].calc(disc[i]);
}
for(int i=1; i<=n; i++)
{
mx[i][i] = a[i].lb;
for(int j=i+1; j<=n; j++)
{
mx[i][j] = max(mx[i][j-1],(int)a[j].lb);
}
}
ans = 0ll;
for(int i=1; i<disc.size(); i++)
{
dfs(1,n,i);
ans = (ans+sdp[dpid[1][n]][i].calc(disc[i])-sdp[dpid[1][n]][i].calc(disc[i-1])+P)%P;
}
printf("%lld\n",ans);
return 0;
}标签：llong,int,Luogu,mid,P5469,Polynomial,aux,NOI2018,dp 来源： https://www.cnblogs.com/suncongbo/p/11217236.html