标签：date set long people oms mod not while art
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\n",kase,f[m][0]); } return 0; }
[UVALive 7143]Room Assignment（Dp）
标签：date set long people oms mod not while art
原文地址：http://www.cnblogs.com/Zars19/p/6915665.html
