POJ 2441 Arrange the Bulls (状压DP)
The DP poj 状压
2023-09-11 14:17:18 时间
题意:n头牛,m个位置,每头牛有各自喜欢的位置,问安排这n头牛使得每头牛都在各自喜欢的位置有几种安排方法。
析:dp[i][s] 表示前 i 头牛,已经占的位置是 s,有多少种安排方法,其他的就很简单了,注意用滚动数组 。
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> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #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 unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e6 + 10; const int mod = 100000000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int dp[2][1<<20]; int a[30][30]; int main(){ while(scanf("%d %d", &n, &m) == 2){ for(int i = 1; i <= n; ++i){ scanf("%d", &a[i][0]); for(int j = 1; j <= a[i][0]; ++j){ scanf("%d", &a[i][j]); --a[i][j]; } } memset(dp[0], 0, sizeof dp[0]); dp[0][0] = 1; int cnt = 1; int all = 1 << m; for(int i = 1; i <= n; ++i, cnt ^= 1){ memset(dp[cnt], 0, sizeof dp[cnt]); for(int j = 0; j < all; ++j){ if(!dp[cnt^1][j]) continue; for(int k = 1; k <= a[i][0]; ++k){ if(j&(1<<a[i][k])) continue; dp[cnt][j|(1<<a[i][k])] += dp[cnt^1][j]; } } } int ans = 0; for(int i = 0; i < all; ++i) ans += dp[cnt^1][i]; printf("%d\n", ans); } return 0; }
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