sgu220Little Bishops(dp)
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题意:
在n*n的棋盘上放K个象,使得两两之间不互相攻击。有多少种放法?
tip:
棋盘按照(i+j)的奇偶黑白染色,则在黑色格子中放的象不可能攻击到白色格子,分开考虑并将棋盘翻转45度。(旋转后就是一行一列只能放一个)
f[i][j]=f[i-1][j]+f[i-1][j-1]*(a[i]-(j-1))
f[i][j]表示前i行放j个车的方案数,a[i]表示第i行的可放位置数。最后枚举在黑色格子中放置的象的个数,剩下的象放在白色格子中。
因为格子先增后减:
一个是:cnt = (i+1)/2 * 2-1;
另一个是:cnt = i/2 * 2;且少一行(2~n)
#include <cstdio>#include <iostream>#include <cmath>#include <cstring>using namespace std;typedef long long LL;const int maxn = 12;const int maxm = maxn*maxn;LL n,m,dp1[maxn][maxm],dp2[maxn][maxm];void init(){ memset(dp1,0,sizeof(dp1)); memset(dp2,0,sizeof(dp2)); dp1[0][0] = dp2[1][0] = 1;}void sov(){ for(LL i = 1; i <= n ; i++){ dp1[i][0] = dp1[i-1][0]; LL cnt = (i+1)/2 * 2-1; for(LL j = 1; j <= (min(cnt,m)) ; j++ ) dp1[i][j] = dp1[i-1][j]+dp1[i-1][j-1] * (cnt-j+1); } for(LL i = 2; i <= n ; i++){ dp2[i][0] = dp2[i-1][0]; LL cnt = i/2 * 2; for(LL j = 1; j <= min(cnt,m) ; j++) dp2[i][j] = dp2[i-1][j] + dp2[i-1][j-1] * (cnt-j+1); }}void print(){ LL ans = 0; for(int i = 0 ; i <= m ; i++){ ans += dp1[n][i] * dp2[n][m-i]; } printf("%lld\n",ans);}int main(){ while(~scanf("%lld%lld",&n,&m)){ if(n == 0 && m == 0) break; init(); sov(); print(); }}
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