sgu223
223. Little Kings
time limit per test: 1 sec.
memory limit per test: 65536 KB
memory limit per test: 65536 KB
input: standard
output: standard
output: standard
After solving nice problems about bishops and rooks, Petya decided that he would like to learn to play chess. He started to learn the rules and found out that the most important piece in the game is the king.
The king can move to any adjacent cell (there are up to eight such cells). Thus, two kings are in the attacking position, if they are located on the adjacent cells.
Of course, the first thing Petya wants to know is the number of ways one can position k kings on a chessboard of size n × n so that no two of them are in the attacking position. Help him!
The king can move to any adjacent cell (there are up to eight such cells). Thus, two kings are in the attacking position, if they are located on the adjacent cells.
Of course, the first thing Petya wants to know is the number of ways one can position k kings on a chessboard of size n × n so that no two of them are in the attacking position. Help him!
Input
The input file contains two integers n (1 ≤ n ≤ 10) and k (0 ≤ k ≤ n2).
Output
Print a line containing the total number of ways one can put the given number of kings on a chessboard of the given size so that no two of them are in attacking positions.
Sample test(s)
Input
Test #1
3 2
Test #2
4 4
3 2
Test #2
4 4
Output
16
Test #2
79
题目大意：给你一个N*N的棋盘，要你在其中放入K个国王，每个国王会攻击到以它为中心的九宫格的相邻8个位置，求方案总数.
这道题可以用状态压缩的思想来做.
我们在进行DP的时候,我们可以通过已知前 i-1 行的最优解来得出前i行的最优解,那么这就涉及到一个问题,我们如何来表示每一行的状态呢?
这也就是状态压缩的动机.如果我们不能够很方便的来表示状态,我们可以把很多维的状态,每一维的范围相同,那么可以把这些维合并成一个X进制数.
用X进制数来表示这个状态.
要了解详情的，可以去参考 郑州101中学/天津大学 周伟的论文《状态压缩》
这道题目我们也可以用SCR来做。
具体思路：
1、求出每一行可能的所有状态数（也就是用二进制来表示这一行某一列是否放入国王）
2、进行动态规划：
用F [ i ] [ j ] [ k ]来表示前i行放入k个国王，并且第 i 行的状态为第 j 个状态。
那么F[ i ] [ j ] [ k ] += F [ i-1][ p ][ k - c[i]],也就是对于前i-1行，我们枚举第i-1行的所有可能状态，
并且保证状态 j和状态p没有同一位放置国王（可以用位运算来完成），
由于这个国王可以攻击8个周围的点，所以我们有必要将i-1行的状态进行左移和右移的操作，来进行判断
并且k-c[i]>=c[p].
这样就可以进行转移了，枚举每一种状态，得到的所有求个和便得到答案了。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
const int MAXN = 200,MAXS = 300,MAXPN = 200;
int s[MAXS],c[MAXS];
LL f[2][MAXS][MAXPN],ans;
int n,pn;
void dfs(int p,int last,int now,int cnt){
if (p == n){s[++s[0]]=now;c[s[0]]=cnt;return ;}
dfs(p+1,0,now*2,cnt);
if (!last) dfs(p+1,1,now*2+1,cnt+1);
}
int main(){
cin >> n >> pn;
if (pn>pow((n+1)/2,2)){
cout << 0 << endl;
return 0;
}
dfs(0,0,0,0);
f[0][1][0]=1;
for (int i=1;i<=n;++i){
for (int j=1;j<=s[0];++j)
for (int k=c[j];k<=pn;++k)
for (int p=1;p<=s[0];++p)
if (!(s[p] & s[j]) && ! (s[p] & (s[j] << 1)) && !(s[p] &(s[j] >> 1)) && (k-c[j]>=c[p]))
f[i & 1][j][k] += f[1-i&1][p][k-c[j]];
memset(f[1-i&1],0,sizeof(f[1-i&1]));
}
for (int i=1;i<=s[0];++i) ans += f[n&1][i][pn];
cout << ans << endl;
return 0;
}