Question

假设我有一个数组1,2,2,10。

长度为3的增加的子序列是1,2,4和1,3,4（基于指数）。

所以答案是2. Problem LINK

我想要一个更好的解决方案，使用BIT树，这可以改善我的解决方案。我尝试过使用BIT树，但是给了我超出时限的错误。

这是BIT implementation Code。

我也试过直接的方法

for (i = 1; i<n;i++) 
  dp[i, 1] = 1

for (i = 1; i<n;i++) 
  for (j = 0; j<i-1;j++) 
    if array[i] > array[j]
     for (p = 2; p<k;p++) 
        dp[i, p] += dp[j, p - 1]

请帮帮我

Answer 1

希望这会有所帮助..

int dp[51][100001];

void update(int bit[], int idx, int val){
for(int x = idx;x <= 100000;x += x & -x){
    bit[x] += val;
    if(bit[x] >= MOD) bit[x] -= MOD;
}
}

int query(int bit[], int idx){
int ret = 0;

    for(int x = idx;x > 0;x -= x & -x){
        ret += bit[x];
        if(ret >= MOD) ret -= MOD;
    }

return ret;
}

int main(){
    int N,K;

    scanf("%d %d",&N,&K);

int ans = 0;

    for(int i = 0,x;i < N;++i){
        scanf("%d",&x);

        for(int k = K;k > 1;--k)
            update(dp[k],x + 1,query(dp[k - 1],x));

        update(dp[1],x + 1,1);
    }

    printf("%d\n",query(dp[K],100000));

    return 0;
}

Explanation:

input: 1
For input 1:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0   
0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 // update for X=2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

input: 1 2
For input 2:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 1 1 2 0 0 0 2 0 0 0 0 0 0 0  // update for X=3, length 1; got 2 increasing subsequence  with length 1
0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0  // update for X=3, length 2;  got 1 increasing subsequence  with length 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

input: 1 2 2
For input 2:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 1 2 3 0 0 0 3 0 0 0 0 0 0 0  // update for X=3, length 1; got 3 increasing subsequence  with length 1
0 0 0 2 2 0 0 0 2 0 0 0 0 0 0 0  // update for X=3, length 2; got 2 increasing subsequence  with length 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0  // But you have no increasing subsequence with length 3

input 1 2 2 10
For input 10:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0  
0 0 1 2 3 0 0 0 3 0 0 1 1 0 0 0  // update for X=11, length 1
0 0 0 2 2 0 0 0 2 0 0 3 3 0 0 0  // update for X=11, length 2
0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0  // update for X=11, length 3;  got 2 increasing subsequence  with length 3; tihs is calculate with help of length 2

每次，你取一个值..计算你找到多少个增加的子序列并逐渐更新长度为3,2,1

求出长度为k的递增子序列的总数

1 个答案: