题目链接

题面:

题解:
d p [ i ] [ j ] [ k ] dp[i][j][k] dp[i][j][k]为考虑了前 i i i 个位置,其中有 j j j 个位置放置了传送器,到 i i i 为止且包括 i i i 一共连续放置了 k k k 的传送器的方案数。

代码:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<set>
#include<list>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
#define lc (cnt<<1)
#define rc (cnt<<1|1)
//#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
using namespace std;

const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const int mod=1e9+7;
const double eps=1e-5;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=1010;
const int maxp=1100;
const int maxm=4000100;
const int up=1000;
const int N=26;

int dp[maxn][maxn][11];

int main(void)
{
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=m;j++)
                memset(dp[i][j],0,sizeof(dp[i][j]));
        }
        dp[1][0][0]=1;
        for(int i=2;i<=n;i++)
        {
            for(int j=0;j<i&&j<=m&&j<=n-2;j++)
            {
                for(int k=0;k<=10;k++)
                {
                    dp[i][j][0]=(dp[i][j][0]+dp[i-1][j][k])%mod;
                    if(j>0&&k>0)
                        dp[i][j][k]=(dp[i][j][k]+1ll*dp[i-1][j-1][k-1]*(i-1))%mod;
                }

            }
        }
        printf("%d\n",dp[n][m][0]==0?-1:dp[n][m][0]);
    }
    return 0;
}