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답안 #205989

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
205989 2020-03-01T18:48:22 Z stefdasca Skyscraper (JOI16_skyscraper) C++14
100 / 100
 109 ms 23928 KB 
#include<bits/stdc++.h>
#define god dimasi5eks
#pragma GCC optimize("O3")
#define fi first
#define se second
#define pb push_back
#define pf push_front
#define mod 1000000007
#define dancila 3.14159265359
#define eps 1e-9

// #define fisier 1

using namespace std;

typedef long long ll;


int add(int a, int b)
{
    int x = a+b;
    if(x >= mod)
        x -= mod;
    if(x < 0)
        x += mod;
    return x;
}
ll mul(ll a, ll b)
{
    return (a*b) % mod;
}

ll pw(ll a, ll b)
{
    ll ans = 1;
    while(b)
    {
        if(b & 1)
            ans = (ans * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand_seed()
{
    long long a = rng();
    return a;
}

ll dp[101][101][1001][3];
/*
    dp[i][j][k][l] :
    i - number of numbers placed
    j - number of connected components
    k - total sum currently (filling empty spaces with a_{i} (0-indexed)
    l - number of endpoints that are filled
*/
ll v[101];
int main()
{

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n, l;
    cin >> n >> l;
    for(int i = 0; i < n; i++)
        cin >> v[i];
    sort(v, v + n);
    if(n == 1)
    {
        cout << 1;
        return 0;
    }
    v[n] = (1<<20);
    if(v[1] - v[0] <= l)
        dp[1][1][v[1] - v[0]][1] = 2; //fill a[0] at one of the endpoints, there are 2 endpoints to fill.
    if(2 * (v[1] - v[0]) <= l)
        dp[1][1][2 * (v[1] - v[0])][0] = 1; //fill a[0] in the middle, positions doesn't matter.
    for(int i = 1; i < n; i++)
        for(int j = 1; j <= i; j++)
            for(int k = 0; k <= l; k++)
                for(int z = 0; z < 3; z++)
                {
                    if(!dp[i][j][k][z])
                        continue;
                    int diff = v[i + 1] - v[i];
                    //First, we try to fill one of the ends
                    if(z < 2 && k + diff * (2 * j - z - 1) <= l) //there are 2*j - z - 1 positions that we're supposed to "upgrade" (-1 because one of the positions is merged with the endpoints after this move)
                    {
                        if(i == n - 1)
                            dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1] = add(dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1], mul(dp[i][j][k][z], (2-z) * j)); //we have j con. comp. to choose to merge with
                        else
                            if(z == 0 || j > 1) //otherwise this coincides with i == n - 1
                                dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1] = add(dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1], mul(dp[i][j][k][z], (2-z)*(j-z))); //can only merge with the con comp. that are not connected to ends.
                        if(k + diff * (2 * j - z + 1) <= l) //now we create a new cc.
                            dp[i + 1][j + 1][k + diff * (2 * j - z + 1)][z + 1] = add(dp[i + 1][j + 1][k + diff * (2 * j - z + 1)][z + 1], mul(dp[i][j][k][z], (2-z))); //we can choose one of the ends to create
                    }
                    //Next, we dont fill the ends.
                    //Part 1 : Create new cc
                    if(k + diff*(2*j - z + 2) <= l) //2 new positions to "upgrade"
                    {
                        dp[i + 1][j + 1][k + diff*(2*j - z + 2)][z] = add(dp[i + 1][j + 1][k + diff*(2*j - z + 2)][z], dp[i][j][k][z]); //nothing new happens
                    }
                    //Part 2 : Stick to one cc
                    if(k + diff*(2*j - z) <= l) //no new positions to "upgrade"
                    {
                        dp[i + 1][j][k + diff*(2*j - z)][z] = add(dp[i + 1][j][k + diff*(2*j - z)][z], mul(dp[i][j][k][z], (2*j - z))); //we can merge in 2*j - z possible positions
                    }
                    //Part 3 : Merge two ccs together
                    if((k + diff*(2*j - z - 2) <= l) && (j >= 2) && (i == n - 1 || j > 2 || z < 2))
                    {
                        if(z == 0)
                        {
                            dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = add(dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z], mul(dp[i][j][k][z], j*(j-1))); //there are jP2 possible merges
                        }
                        if(z == 1)
                        {
                            dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = add(dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z], mul(dp[i][j][k][z], (j-1)*(j-1))); //there are (j-1)P2+(j-1) merges
                        }
                        if(z == 2)
                        {
                            if(i == n - 1)
                            {
                                dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = add(dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z], dp[i][j][k][z]); //there's only 1 place it can go.
                            }
                            else
                            {
                                dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = add(dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z], mul(dp[i][j][k][z], (j-2)*(j-1))); //there're (j-2)P2 + 2(j-2) possiblilities
                            }
                        }
                    }
                }
    ll answer = 0;
    for(int i = 0; i <= l; i++)
        answer = add(answer, dp[n][1][i][2]);
    cout << answer << '\n';
    return 0;
}

Subtask #1
5.0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Correct 5 ms 376 KB Output is correct
2 Correct 5 ms 504 KB Output is correct
3 Correct 5 ms 380 KB Output is correct
4 Correct 5 ms 376 KB Output is correct
5 Correct 5 ms 504 KB Output is correct
6 Correct 5 ms 504 KB Output is correct
7 Correct 5 ms 504 KB Output is correct
8 Correct 5 ms 504 KB Output is correct
9 Correct 5 ms 636 KB Output is correct
10 Correct 5 ms 504 KB Output is correct

Subtask #2
15.0 / 15.0

# 결과 실행 시간 메모리 Grader output
1 Correct 5 ms 632 KB Output is correct
2 Correct 5 ms 632 KB Output is correct
3 Correct 5 ms 632 KB Output is correct
4 Correct 5 ms 632 KB Output is correct
5 Correct 5 ms 504 KB Output is correct
6 Correct 5 ms 632 KB Output is correct
7 Correct 5 ms 504 KB Output is correct
8 Correct 5 ms 632 KB Output is correct
9 Correct 5 ms 760 KB Output is correct
10 Correct 5 ms 632 KB Output is correct

Subtask #3
80.0 / 80.0

# 결과 실행 시간 메모리 Grader output
1 Correct 5 ms 376 KB Output is correct
2 Correct 5 ms 504 KB Output is correct
3 Correct 5 ms 380 KB Output is correct
4 Correct 5 ms 376 KB Output is correct
5 Correct 5 ms 504 KB Output is correct
6 Correct 5 ms 504 KB Output is correct
7 Correct 5 ms 504 KB Output is correct
8 Correct 5 ms 504 KB Output is correct
9 Correct 5 ms 636 KB Output is correct
10 Correct 5 ms 504 KB Output is correct
11 Correct 5 ms 632 KB Output is correct
12 Correct 5 ms 632 KB Output is correct
13 Correct 5 ms 632 KB Output is correct
14 Correct 5 ms 632 KB Output is correct
15 Correct 5 ms 504 KB Output is correct
16 Correct 5 ms 632 KB Output is correct
17 Correct 5 ms 504 KB Output is correct
18 Correct 5 ms 632 KB Output is correct
19 Correct 5 ms 760 KB Output is correct
20 Correct 5 ms 632 KB Output is correct
21 Correct 6 ms 1272 KB Output is correct
22 Correct 109 ms 23928 KB Output is correct
23 Correct 53 ms 8056 KB Output is correct
24 Correct 60 ms 12152 KB Output is correct
25 Correct 58 ms 9336 KB Output is correct
26 Correct 54 ms 8568 KB Output is correct
27 Correct 35 ms 9720 KB Output is correct
28 Correct 44 ms 11896 KB Output is correct
29 Correct 72 ms 16504 KB Output is correct
30 Correct 56 ms 9464 KB Output is correct