CF183D T-shirt

CF183D T-shirt 

考虑每次的选择，一定是选择一件衣服使得能送出的概率最大

f[i][j]表示，至少有j个人能穿第i个的概率

选择最大的n个f[i][j]就是答案。

直接dpO(m*n^2)会TLE

我们只要前n大，而f[i]显然单调

加入f[i][1]，选择最大的。再计算f[i][2]

O(n^2+nlogn)

#include<bits/stdc++.h>
#define reg register int
#define il inline
#define fi first
#define se second
#define mk(a,b) make_pair(a,b)
#define numb (ch^'0')
#define pb push_back
#define solid const auto &
#define enter cout<<endl
#define pii pair<int,int>
using namespace std;
typedef long long ll;
template<class T>il void rd(T &x){
    char ch;x=0;bool fl=false;while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);
    for(x=numb;isdigit(ch=getchar());x=x*10+numb);(fl==true)&&(x=-x);}
template<class T>il void output(T x){if(x/10)output(x/10);putchar(x%10+'0');}
template<class T>il void ot(T x){if(x<0) putchar('-'),x=-x;output(x);putchar(' ');}
template<class T>il void prt(T a[],int st,int nd){for(reg i=st;i<=nd;++i) ot(a[i]);putchar('\n');}
namespace Modulo{
const int mod=998244353;
il int ad(int x,int y){return x+y>=mod?x+y-mod:x+y;}
il int sub(int x,int y){return ad(x,mod-y);}
il int mul(int x,int y){return (ll)x*y%mod;}
il void inc(int &x,int y){x=ad(x,y);}
il void inc2(int &x,int y){x=mul(x,y);}
il int qm(int x,int y=mod-2){int ret=1;while(y){if(y&1) ret=mul(x,ret);x=mul(x,x);y>>=1;}return ret;}
template<class ...Args>il int ad(const int a,const int b,const Args &...args) {return ad(ad(a,b),args...);}
template<class ...Args>il int mul(const int a,const int b,const Args &...args) {return mul(mul(a,b),args...);}
}
// using namespace Modulo;
#define ld long double
namespace Miracle{
const int N=3003;
const int M=3003;
int n,m;
ld f[M][N];
ld p[M][N];
int get[M];
struct po{
    ld val;
    int id;
    po(){}
    po(ld v,int d){
        val=v;id=d;
    }
    bool friend operator <(po a,po b){
        return a.val<b.val;
    }
};
priority_queue<po>q;

int main(){
    rd(n);rd(m);
    for(reg i=1;i<=n;++i){
        for(reg j=1;j<=m;++j){
            int x;rd(x);p[j][i]=(ld)x/1000;
        }
    }
    for(reg j=1;j<=m;++j){
        f[j][0]=1;
        for(reg i=1;i<=n;++i){
            f[j][i]=f[j][i-1]*(1-p[j][i]);
        }
    }
    for(reg j=1;j<=m;++j){
        ld lp=1.00;
        for(reg i=1;i<=n;++i) lp*=(1.00-p[j][i]);
        lp=1-lp;
//        cout<<" j "<<j<<" "<<lp<<endl;
        q.push(po(lp,j));
    }
    int nd=n;
    ld ans=0.0;
    while(nd--){
        po now=q.top();q.pop();
//        cout<<" now "<<now.val<<" "<<now.id<<endl;
        ans+=now.val;
        int id=now.id;
        ld las=f[id][0];//get[id]?0.0:1.0;
        f[id][0]=0;
        for(reg i=1;i<=n;++i){
            ld tp=f[id][i];
//            cout<<" tp "<<tp<<endl;
            f[id][i]=f[id][i-1]*(1.00-p[id][i])+las*p[id][i];
//            cout<<" i "<<f[id][i]<<endl;
            las=tp;
        }
        f[id][0]=0;
        ++get[id];
        q.push(po(now.val-f[id][n],id));
    }
//    printf("%.10Lf",ans);
    cout<<fixed<<setprecision(10)<<ans<<endl;
    return 0;
}

}
signed main(){
    Miracle::main();
    return 0;
}

/*
   Author: *Miracle*
*/