在N×N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案。国王能攻击到它上下左右,以及左上
左下右上右下八个方向上附近的各一个格子,共8个格子。
标签:submit hint www. mon algo sample stdout for 左右
在N×N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案。国王能攻击到它上下左右,以及左上
左下右上右下八个方向上附近的各一个格子,共8个格子。
只有一行,包含两个数N,K ( 1 <=N <=9, 0 <= K <= N * N)
方案数。
析:dp[i][j][s] 表示前 i 行放 j 个king,状态为 s 时有多少情况,然后很简单的转移。
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> #include <list> #include <assert.h> #include <bitset> #include <numeric> #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 fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() //#define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #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 = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 2e5 + 10; const int maxm = 3e5 + 10; const ULL mod = 3; 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; } LL dp[13][100][600]; vector<P> state[600]; bool judge(int s1, int s2){ for(int k = 0; k < n; ++k) if(s2&1<<k){ if(s1&1<<k) return false; if(k && s2&1<<k-1) return false; if(k && s1&1<<k-1) return false; if(k + 1 < n && s1&1<<k+1) return false; } return true; } int main(){ scanf("%d %d", &n, &m); int all = 1<<n; FOR(i, 0, all) FOR(j, 0, all)if(judge(i, j)){ int cnt = 0; for(int k = 0; k < n; ++k) if(j&1<<k) ++cnt; state[i].pb(P(j, cnt)); } dp[0][0][0] = 1; for(int i = 1; i <= n; ++i) FOR(j, 0, all){ for(int k = 0; k < state[j].sz; ++k){ P &p = state[j][k]; for(int l = 0; l + p.se <= m; ++l) dp[i][l+p.se][p.fi] += dp[i-1][l][j]; } } LL ans = 0; for(int i = 0; i < all; ++i) ans += dp[n][m][i]; printf("%lld\n", ans); return 0; }
BZOJ 1087 [SCOI2005]互不侵犯King (状压DP)
标签:submit hint www. mon algo sample stdout for 左右
原文地址:http://www.cnblogs.com/dwtfukgv/p/7819350.html