1087: [SCOI2005]互不侵犯King
Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 4772 Solved: 2769
[Submit][Status][Discuss]
Description
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。
Input
只有一行,包含两个数N,K ( 1 <=N <=9, 0 <= K <= N * N)
Output
方案数。
Sample Input
3 2
Sample Output
16
HINT
Source
套路,预处理合法状态并标号,state[i]表示标号为i的合法状态,g[i][j]表示标号为i的合法状态与标号为j的合法状态是否可以作为相邻行
dp[i][j][l]表示第i行,前i行有j个国王,第i行状态标号为l
刷表转移即可
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <cmath> 7 #include <queue> 8 #include <vector> 9 #define min(a, b) ((a) < (b) ? (a) : (b)) 10 #define max(a, b) ((a) > (b) ? (a) : (b)) 11 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a)) 12 inline void swap(long long &a, long long &b) 13 { 14 long long tmp = a;a = b;b = tmp; 15 } 16 inline void read(long long &x) 17 { 18 x = 0;char ch = getchar(), c = ch; 19 while(ch < ‘0‘ || ch > ‘9‘) c = ch, ch = getchar(); 20 while(ch <= ‘9‘ && ch >= ‘0‘) x = x * 10 + ch - ‘0‘, ch = getchar(); 21 if(c == ‘-‘)x = -x; 22 } 23 24 const long long INF = 0x3f3f3f3f; 25 26 long long n,k,dp[10][82][1 << 9]; 27 //dp[i][j][l]表示第i行放j个国王状态为state[l]的方案数 28 29 long long state[1 << 9], tot, g[1 << 9][1 << 9], num[1 << 9]; 30 31 int main() 32 { 33 read(n), read(k); 34 long long ma = 1 << n; 35 for(register long long i = 0;i < ma;++ i) 36 if(!(i & (i << 1)) && !(i & (i >> 1))) 37 { 38 state[++ tot] = i; 39 long long tmp = i; 40 for(;tmp;tmp >>= 1) num[tot] += (tmp & 1); 41 } 42 for(register long long i = 1;i <= tot;++ i) 43 for(register long long j = i;j <= tot;++ j) 44 if(!(state[i] & (state[j] << 1)) && !(state[i] & (state[j] >> 1)) && !(state[i] & state[j])) 45 g[i][j] = g[j][i] = 1; 46 for(register long long i = 1;i <= tot;++ i) if(num[i] <= k) dp[1][num[i]][i] = 1; 47 for(register long long i = 1;i < n;++ i) 48 for(register long long j = 0;j <= k;++ j) 49 for(register long long l = 1;l <= tot;++ l) 50 if(num[l] <= j) 51 for(register long long r = 1;r <= tot;++ r) 52 if(g[l][r] && j + num[r] <= k) 53 dp[i + 1][j + num[r]][r] += dp[i][j][l]; 54 long long ans = 0; 55 for(register long long i = 1;i <= tot;++ i) 56 ans += dp[n][k][i]; 57 printf("%lld", ans); 58 return 0; 59 }