zoukankan      html  css  js  c++  java
  • [UVALive 7143]Room Assignment(Dp)

    Description

    There are N guests checking in at the front desk of the hotel. 2K (0 ≤ 2K ≤ N) of them are twins.
    There are M rooms available. Each room has capacity ci which means how many guests it can hold.
    It happens that the total room capacity is N, i.e. c1 + c2 + . . . + cM = N.
    The hotel receptionist wonders how many different room assignments to accommodate all guests.
    Since the, receptionist cannot tell the two twins in any pair of twins apart, two room assignments are
    considered the same if one can be generated from the other by swapping the two twins in each of some
    number of pairs. For rooms with capacity greater than 1, it only matters which people are in the room;
    they are not considered to be in any particular order within the room.

    Solution

    题意:m个房间,每个房间有容量ci(总容量为n),n位客人,其中有k对双胞胎,双胞胎被看做同样的人,求方案数

    用f[i][j]表示分配到i个房子,还剩j对完整的双胞胎没有分配:

    f[i][j-k]+=f[i-1][j]*C(j,k)*C(sum-(j-k)*2-k,c[i]-k) sum表示剩下的还需分配的人

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #define Mod 1000000007
    #define N 100005
    typedef long long LL;
    using namespace std;
    int T,kase=0,n,m,K,c[20],p[20];
    LL f[20][110],fac[N],inv[N];
    void init()
    {
        fac[0]=1,inv[1]=1;
        for(int i=1;i<N;i++)
        fac[i]=(fac[i-1]*i)%Mod;
        for(int i=2;i<N;i++)
        inv[i]=(inv[Mod%i]*(Mod-Mod/i))%Mod;
        inv[0]=1;
        for(int i=1;i<N;i++)
        inv[i]=(inv[i-1]*inv[i])%Mod;
    }
    LL C(int m,int n)
    {
        if(m<n||m<0||n<0)return 0;
        return ((fac[m]*inv[n])%Mod*inv[m-n])%Mod;
    }
    int main()
    {
        init();
        scanf("%d",&T);
        while(T--)
        {
            ++kase;
            memset(f,0,sizeof(f));
            scanf("%d%d%d",&n,&m,&K);
            for(int i=1;i<=m;i++)
            {scanf("%d",&c[i]);p[i]=p[i-1]+c[i];}
            f[0][K]=1;
            for(int i=1;i<=m;i++)
            {
                int sum=p[m]-p[i-1];
                for(int j=0;j<=K;j++)
                {
                    for(int k=0;k<=j;k++)
                    {
                        f[i][j-k]+=(f[i-1][j]*C(j,k)%Mod)*C(sum-(j-k)*2-k,c[i]-k)%Mod;
                        f[i][j-k]%=Mod;
                    } 
                }
            }
            printf("Case #%d: %d
    ",kase,f[m][0]);
        }
        return 0;
    }
  • 相关阅读:
    如何完成看似不可能完成的任务
    SQL Server 2008 数据挖掘算法
    混在北京
    09年的一个方案,很遗憾没有采纳,回头看看,我还认为我是对的
    Metro Home
    InputScope
    Mozart Update 3 (7392…)
    搏斗Mango beta…
    Mango 7712 is coming
    HD2 update NODO
  • 原文地址:https://www.cnblogs.com/Zars19/p/6915665.html
Copyright © 2011-2022 走看看