标签：bool ++ stream i+1 linker stdout stack eof char
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;
}
UVaLive 7143 Room Assignment (组合数+DP)
标签：bool ++ stream i+1 linker stdout stack eof char
原文地址：http://www.cnblogs.com/dwtfukgv/p/6090699.html
