【JZOJ 4345】【WC2016模拟】Fountain

Description

Solution

考虑暴力n!枚举顺序，如何快速求答案，
设：sum=∑n−1i=1max(ai,ai+1)
答案显然是Cnm−sum+n

写出这个式子考虑转成DP：
设：f[i][j][k][l]，表示从大到小做到第i个，当前有j个空隙必须放点，sum为k，l为0/1/2，表示确认了几个端点（最左和最右确认没有），

转移就比较好推了，

复杂度：O(n4)

Code

#include <cstdio>#include <algorithm>#define fo(i,a,b) for(int i=a;i<=b;++i)#define fod(i,a,b) for(int i=a;i>=b;--i)#define min(q,w) ((q)>(w)?(w):(q))#define max(q,w) ((q)<(w)?(w):(q))using namespace std;typedef long long LL;const int N=43,mo=1e9+7;int read(int &n){    char ch=' ';int q=0,w=1;    for(;(ch!='-')&&((ch<'0')||(ch>'9'));ch=getchar());    if(ch=='-')w=-1,ch=getchar();    for(;ch>='0' && ch<='9';ch=getchar())q=q*10+ch-48;n=q*w;return n;}int m,n;LL ans;int a[N];int f[N][N][N*N*2][3];LL jc[100500],jcn[100500];LL ksm(LL q,int w){    LL ans=1;q=q%mo;    for(;w;w>>=1,q=q*q%mo)if(w&1)ans=ans*q%mo;    return ans;}LL C(int m,int n){return jc[m]*jcn[n]%mo*jcn[m-n]%mo;}int main(){    freopen("!.in","r",stdin);//  freopen(".out","w",stdout);    int q,Alls=0;    read(n),read(m);    jc[0]=1;fo(i,1,m+1)jc[i]=jc[i-1]*(LL)i%mo;    jcn[m+1]=ksm(jc[m+1],mo-2);fod(i,m,0)jcn[i]=jcn[i+1]*(i+1LL)%mo;    fo(i,1,n)Alls+=(a[i]=read(q)-1);    sort(a+1,a+1+n);    fo(i,1,n/2)swap(a[i],a[n-i+1]);    if(n==1){printf("%d\n",m);return 0;}    if(n==2){m-=a[1];printf("%d\n",m*(m-1)%mo);return 0;}    f[1][0][a[1]*2][0]=1;    f[1][0][a[1]][1]=2;    fo(i,1,n-1)    {        fo(j,0,n+2)        {            fo(k,0,Alls*2)            {                fo(l,0,2)if(f[i][j][k][l])                {                    f[i+1][j+1][k+a[i+1]*2][l]=((LL)f[i+1][j+1][k+a[i+1]*2][l]+(LL)f[i][j][k][l]*(LL)(j+2LL-l))%mo;                    f[i+1][j][k+a[i+1]][l]=((LL)f[i+1][j][k+a[i+1]][l]+(LL)f[i][j][k][l]*(LL)(j*2+2LL-l))%mo;                    if(j)f[i+1][j-1][k][l]=((LL)f[i+1][j-1][k][l]+(LL)f[i][j][k][l]*(LL)(j))%mo;                    if(l<2)                    {                        f[i+1][j+1][k+a[i+1]][l+1]=((LL)f[i+1][j+1][k+a[i+1]][l+1]+(LL)f[i][j][k][l]*(2LL-l))%mo;                        f[i+1][j][k][l+1]=((LL)f[i+1][j][k][l+1]+(LL)f[i][j][k][l]*(2LL-l))%mo;                    }                }            }        }    }    fo(i,0,min(m,Alls*2))ans=(ans+(LL)f[n][0][i][2]*C(m-i,n))%mo;    printf("%lld\n",ans);    return 0;}