UVaLive 7143 Room Assignment (组合数+DP)

Posted 2020-08-20 dwtfukgv

题意：有 n 个客人，m个房间，每个房间可住ci个人，这 n 个人中有 t 对双胞胎，sum{ci}  = n 问你有多少种住房方法。

析：计数DP，dp[i][j] 表示前 i 个房间，还剩下 j 对双胞胎未住，第 i+1 个房间，就从剩下的 j 对双胞胎中选 k 对，然后再从不是双胞胎的人选剩下的，每对先选一个，然后再从剩下的选全部的，

求组合数过程可能要用到逆元，可以提前打表。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int t;
LL fact[maxn], inv[maxn];
int c[15];
LL dp[15][105];

inline LL f(int x){  return 1 == x ? 1LL : (mod - mod/x) * f(mod % x) % mod;  }

void init(){
    fact[0] = 1;
    for(int i = 1; i < maxn; ++i)  fact[i] = fact[i-1] * i % mod;
    for(int i = 0; i < maxn; ++i)  inv[i] = f(fact[i]);
}

inline LL C(int a, int b){  return (a < b || a < 0 || b < 0) ? 0 : fact[a] * inv[b] % mod * inv[a - b] % mod;  }

LL solve(){
    memset(dp, 0, sizeof dp);
    dp[0][t] = 1;
    for(int i = 0, sum = n; i < m; ++i, sum -= c[i])
        for(int j = 0; j <= t; ++j)  if(dp[i][j])
            for(int k = 0; k <= j && c[i+1] >= k; ++k)
                dp[i+1][j-k] = (dp[i+1][j-k] + dp[i][j] * C(j, k) % mod * C(sum-2*j+k, c[i+1]-k)) % mod;
    return dp[m][0];
}

int main(){
    init();
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d %d", &n, &m, &t);
        for(int i = 1; i <= m; ++i)  scanf("%d", c+i);
        printf("Case #%d: %lld\n", kase, solve());
    }
    return 0;
}