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zc
2019-12-21 19:47:00
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首先二分出最大一段的最小值,然后dp找出方案数

f[i][j]表示前j个分成i组的方案数,求出的最小值为x

f[i][j] = \sum(f[i-1][k])(S_j-S_k \le x,k \le j-1)

\Theta(n^3m)当然会TLE

我们发现只要找到最小的k之后[k,j-1]都是合法的

这时我们就可以前缀和优化dp

sum[i][j] = \sum_{k=0}^j f[i][k]

f[i][j]=sum[i-1][j-1] - sum[i-1][k-1]

可是\Theta(n^2m)还是TLE

我们可以先预处理出第一个S_j-S_k \le xk

因为S_i具有单调性,所以k不需要每次都重新枚举

还没结束

因为f[i][j]每次更新的时候只需要用到f[i-1][...]这一维,我们可以用滚动数组滚掉数组的一维

但是\Theta(nm+n \log \sum L_i)的复杂度还是觉得很悬

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;++i)
    #define Fdr(i,x,y) for(int i=x;i>=y;--i)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 50011
#define mod 10007
int n,m,a[N];
il int mo(int x){
    if(x>=mod)x-=mod;
    return x;
}
il bool chk(int w){
    int s=0,t=0;
    Fur(i,1,n){
        s+=a[i];
        if(s>w){
            s=a[i],++t;
            if(t>m+1)return 0;
        }
    }
    if(s)++t;
    if(t>m+1)return 0;
    return 1;
}
int f[N],s[N],S[N],pre[N];
int main(){
    fin("in");
    in>>n>>m;
    int l=0,r=0,ans=0;
    Fur(i,1,n)
        in>>a[i],
        l=MAX(l,a[i]),
        S[i]=S[i-1]+a[i];

    r=S[n];
    while(l<=r){
        int mid=(l+r)>>1;
        if(chk(mid))ans=mid,r=mid-1;
        else l=mid+1;
    }
    out<<ans<<" ";

    int k=0,res=(S[n]<=ans);
    Fur(i,1,n){
        while(S[i]-S[k]>ans)++k;
        pre[i]=k;
    }
    Fur(j,1,n)
        f[j]=(S[j]<=ans),
        s[j]=mo(s[j-1]+f[j]);

    Fur(i,2,m+1){
        Fur(j,1,n){
            f[j]=s[j-1];
            if(pre[j]-1>=0)f[j]=mo(f[j]-s[pre[j]-1]+mod);
        }
        Fur(j,1,n)
            s[j]=mo(s[j-1]+f[j]);
        res=mo(res+f[n]);
    }

    out<<res<<ln;
}
LG 2511 [HAOI2008]木棍分割
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