oj.uz

답안 #140136

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
140136 2019-08-02T07:35:55 Z thoughtFlow Skyscraper (JOI16_skyscraper) C++14
100 / 100
 94 ms 23996 KB 
#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef long double ld;
 
#define fi first
#define se second
#define pb push_back
#define mp make_pair
 
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 a[101];
const ll MOD = 1e9 + 7;
 
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n, l;
    cin>>n>>l;
    for(int i = 0; i < n; i++)
    {
        cin>>a[i];
    }
    sort(a, a + n);
    if(n == 1) //special case
    {
        cout << 1;
        return 0;
    }
    a[n] = 10000; //inf for simplicity
    if(a[1] - a[0] <= l) dp[1][1][a[1] - a[0]][1] = 2; //fill a[0] at one of the endpoints, there are 2 endpoints to fill.
    if(2*(a[1] - a[0]) <= l) dp[1][1][2*(a[1] - a[0])][0] = 1; //fill a[0] in the middle, positions doesn't matter.
 
    for(int i = 1; i < n; i++)
    {
        int diff = a[i + 1] - a[i]; //this thing is "INF" if i = n - 1.
        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; //this value does not exist
                    //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] = (dp[i + 1][j][k + diff*(2*j - z - 1)][z + 1] + dp[i][j][k][z]*(2-z)*j)%MOD; //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] = (dp[i + 1][j][k + diff*(2*j - z - 1)][z + 1] + dp[i][j][k][z]*(2-z)*(j-z))%MOD; //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] = (dp[i + 1][j + 1][k + diff*(2*j - z + 1)][z + 1] + dp[i][j][k][z]*(2-z))%MOD; //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] = (dp[i + 1][j + 1][k + diff*(2*j - z + 2)][z] + dp[i][j][k][z])%MOD; //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] = (dp[i + 1][j][k + diff*(2*j - z)][z] + dp[i][j][k][z]*(2*j - z))%MOD; //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] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z]*j*(j-1))%MOD; //there are jP2 possible merges
                        }
                        if(z == 1)
                        {
                            dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z]*(j-1)*(j-1))%MOD; //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] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z])%MOD; //there's only 1 place it can go.
                            }
                            else
                            {
                                dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z]*(j-2)*(j-1))%MOD; //there're (j-2)P2 + 2(j-2) possiblilities
                            }
                        }
                    }
                }
            }
        }
    }
 
    ll answer = 0;
    for(int i = 0; i <= l; i++)
    {
        answer = (answer + dp[n][1][i][2])%MOD; //sum the dp values for all possible sums
    }
    cout << answer << '\n';
    return 0;
}

Subtask #1
5.0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 376 KB Output is correct
3 Correct 2 ms 376 KB Output is correct
4 Correct 2 ms 376 KB Output is correct
5 Correct 2 ms 504 KB Output is correct
6 Correct 2 ms 504 KB Output is correct
7 Correct 2 ms 504 KB Output is correct
8 Correct 3 ms 504 KB Output is correct
9 Correct 2 ms 632 KB Output is correct
10 Correct 2 ms 504 KB Output is correct

Subtask #2
15.0 / 15.0

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

Subtask #3
80.0 / 80.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 376 KB Output is correct
3 Correct 2 ms 376 KB Output is correct
4 Correct 2 ms 376 KB Output is correct
5 Correct 2 ms 504 KB Output is correct
6 Correct 2 ms 504 KB Output is correct
7 Correct 2 ms 504 KB Output is correct
8 Correct 3 ms 504 KB Output is correct
9 Correct 2 ms 632 KB Output is correct
10 Correct 2 ms 504 KB Output is correct
11 Correct 2 ms 632 KB Output is correct
12 Correct 3 ms 632 KB Output is correct
13 Correct 3 ms 632 KB Output is correct
14 Correct 3 ms 632 KB Output is correct
15 Correct 2 ms 632 KB Output is correct
16 Correct 3 ms 632 KB Output is correct
17 Correct 3 ms 504 KB Output is correct
18 Correct 3 ms 632 KB Output is correct
19 Correct 3 ms 760 KB Output is correct
20 Correct 3 ms 632 KB Output is correct
21 Correct 4 ms 1272 KB Output is correct
22 Correct 94 ms 23996 KB Output is correct
23 Correct 72 ms 8076 KB Output is correct
24 Correct 71 ms 12252 KB Output is correct
25 Correct 73 ms 9232 KB Output is correct
26 Correct 69 ms 8696 KB Output is correct
27 Correct 36 ms 9720 KB Output is correct
28 Correct 46 ms 12024 KB Output is correct
29 Correct 72 ms 16504 KB Output is correct
30 Correct 73 ms 9464 KB Output is correct