核发电站 (dp前缀优化)

Posted 2022-03-07 uid001

大意: $n$个城市, $m$种核电站, 第$i$种假设要建在第$x$个城市, 必须满足$[x-i,x+i]$范围内无其他核电站, 求建核电站的方案数.

 

简单$dp$题, 设$dp[i][j]$为位置$i$建第$j$种核电站的方案数.

枚举上一个核电站的位置来转移, 有:

$dp[i][1]=1+dp[i-2][1]+\sum\limits_k=1^2 dp[i-3][k]+\sum\limits_k=1^3dp[i-4][k]+...$

$dp[i][j]=dp[i][j-1]-\sum\limits_k=1^j-1dp[i-j][k],\space j>1$.

前缀优化一下即可$O(nm)$.

#include <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;
const int P = 1e9+7;
int dp[10010][110], s[10010];

int main() 
	int t;
	scanf("%d", &t);
	REP(cas,1,t) 
		int n, m;
		scanf("%d%d", &n, &m);
		REP(i,1,n) 
			int now = 1;
			dp[i][1] = 1;
			PER(j,1,i-2)  
				if (now==m) 
					(dp[i][1] += s[j]) %= P;
					break;
				
				(dp[i][1] += dp[j][now++]) %= P;
			
			REP(j,2,m) dp[i][j] = (dp[i][j-1]-(i-j>0?dp[i-j][j-1]:0))%P;
			REP(j,2,m) (dp[i][j] += dp[i][j-1]) %= P;
			s[i] = (s[i-1]+dp[i][m])%P;
		
		int ans = (s[n]+1)%P;
		if (ans<0) ans += P;
		printf("Case %d: %d\n", cas, ans);