oj.uz

답안 #61241

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
61241 2018-07-25T12:26:33 Z SpaimaCarpatilor parentrises (BOI18_parentrises) C++17
100 / 100
 96 ms 3528 KB 
#include<bits/stdc++.h>

using namespace std;

int N;
char sir[1000009], ans[1000009];
const int mod = 1e9 + 7;
const int maxN = 300;
int cnt[maxN + 5], dp[maxN + 5][maxN + 5][2][4];

void adto (int &x, int y) {x += y; if (x >= mod) x -= mod;}

void addGreens (int l, int r)
{
    while (1)
    {
        while (sir[l] == ')' && l <= N) l ++;
        while (sir[r] == '(' && r >= 1) r --;
        if (l == N + 1 || r == 0) break;
        if (l < r) ans[l] = ans[r] = 'G', l ++, r --;
        else break;
    }
}

void fillOutRB (bool &ok)
{
    int j = 0, k = 0;
    for (int i=1; i<=N; i++)
    {
        assert (j <= k);
        if (ans[i] == 'G')
        {
            if (sir[i] == '(') j ++, k ++;
            else j --, k --;
            if (j < 0 || k < 0) ok = 0;
            continue;
        }
        if (sir[i] == '(')
        {
            if (j >= k) k ++, ans[i] = 'B';
            else j ++, ans[i] = 'R';
            continue;
        }
        if (j < k) k --, ans[i] = 'B';
        else j --, ans[i] = 'R';
        if (j < 0) ok = 0;
    }
    if (!ok) assert (j + k == 0);
}

namespace generator {
    const int N = 3;
    bool canDo = 0;
    void back (int pos, int i, int j)
    {
        if (canDo) return ;
        if (i < 0 || j < 0) return ;
        if (pos == N + 1)
        {
            if (i + j == 0)
                canDo = 1;
            return ;
        }
        back (pos + 1, i + sir[pos], j + sir[pos]);
        back (pos + 1, i + sir[pos], j);
        back (pos + 1, i, j + sir[pos]);
    }

    void genTests ()
    {
        vector < string > input;
        for (int msk = 0; msk < (1 << N); msk ++)
        {
            for (int i=0; i<N; i++)
                sir[i + 1] = (((msk & (1 << i)) > 0) ? +1 : -1);
            canDo = 0, back (1, 0, 0);
            if (canDo)
            {
                string curr;
                for (int i=1; i<=N; i++)
                    if (sir[i] == +1) curr.push_back ('(');
                    else curr.push_back (')');
                input.push_back (curr);
            }
        }
        printf ("%d\n", input.size ());
        for (auto s : input)
            printf ("%s\n", s.c_str ());
        exit (0);
    }
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

//generator::genTests ();

int type;
scanf ("%d", &type);
if (type == 1)
{
    int tests;
    scanf ("%d\n", &tests);
    while (tests --)
    {
        scanf ("%s\n", sir + 1), N = strlen (sir + 1);
        int balance = 0;
        for (int i=1; i<=N; i++)
            if (sir[i] == '(') balance ++;
            else balance --;
        for (int i=1; i<=N + 1; i++)
            ans[i] = 0;
        bool ok = 1;
        if (balance > 0)
        {
            ///I'll use first X + balance open brackets and last X closed ones where X is maximal so that all the closed brackets come after the open ones
            int l = 1, r = N;
            while (balance > 0)
            {
                while (sir[r] == '(' && r >= 1) r --;
                if (r == 0)
                {
                    ok = 0;
                    break;
                }
                ans[r] = 'G', r --;
                balance --;
            }
            if (ok)
                addGreens (l, r);
        }
        else
        {
            balance = -balance;
            ///same as above
            int l = 1, r = N;
            while (balance > 0)
            {
                while (sir[l] == ')' && l <= N) l ++;
                if (l == N + 1)
                {
                    ok = 0;
                    break;
                }
                ans[l] = 'G', l ++;
                balance --;
            }
            if (ok)
                addGreens (l, r);
        }
        if (ok)
            fillOutRB (ok);
        if (!ok) printf ("impossible\n");
        else
        {
            for (int i=1; i<=N; i++)
                printf ("%c", ans[i]);
            printf ("\n");
        }
    }
    return 0;
}
///a state is defined by:
///j, kj (which means (j, k) = (j, j + kj))
///the current phase: 0/1/2/3
///in phase 0 we need to take any ( and put it as green

///in phases 1 and 2 we can't put any green and we can't have the subsequence () -
///phase 1 is as long as we have )
///phase 2 is as long as we have ( after phase 1 (and after that both phases must stop since we can't have () )

///in phase 3 we need to take any ) and put it as green
///complexity: N^2
///dp[i][j][kj][phase]
dp[0][0][0][0] = 1;
for (int i=0; i<maxN; i++)
{
    for (int j=0; j<=i; j++)
        for (int kj=0; kj<2; kj++)
            for (int phase=0; phase<4; phase++)
                if (dp[i][j][kj][phase] > 0)
                    for (int br = 0; br < 2; br ++)
                        for (int nPhase = phase; nPhase < 4; nPhase ++)
                        {
                            int ni = i + 1, nj = j, nk = j + kj;
                            if (phase == 1 && br == 0 && nPhase == 1) continue;
                            if (phase == 2 && br == 1 && nPhase == 2) continue;
                            if (phase == 1 && nPhase == 2 && br != 0) continue;
                            if (phase == 0 && br == 0 && nPhase == 1) continue;
                            if (phase == 0 && 1 <= nPhase && nPhase <= 2 && br == 1) continue;
                            if (nPhase == 3 && 0 <= phase && phase <= 2 && br == 0) continue;
                            if ((nPhase == 3 && br == 1) || (nPhase == 0 && br == 0))
                            {
                                if (br == 0) nj ++, nk ++;
                                else nj --, nk --;
                            }
                            else
                            {
                                if (br == 0)
                                {
                                    if (nj >= nk) nk ++;
                                    else nj ++;
                                }
                                else
                                {
                                    if (nj < nk) nk --;
                                    else nj --;
                                }
                            }
                            if (nj >= 0 && nk >= 0)
                                adto (dp[ni][nj][nk - nj][nPhase], dp[i][j][kj][phase]);
                        }
    for (int phase = 0; phase < 4; phase ++)
        adto (cnt[i + 1], dp[i + 1][0][0][phase]);
}
int tests;
scanf ("%d", &tests);
while (tests --)
{
    scanf ("%d", &N);
    printf ("%d\n", cnt[N]);
}
return 0;
}

Compilation message

parentrises.cpp: In function 'void generator::genTests()':
parentrises.cpp:86:38: warning: format '%d' expects argument of type 'int', but argument 2 has type 'std::vector<std::__cxx11::basic_string<char> >::size_type {aka long unsigned int}' [-Wformat=]
         printf ("%d\n", input.size ());
                         ~~~~~~~~~~~~~^
parentrises.cpp: In function 'int main()':
parentrises.cpp:101:7: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
 scanf ("%d", &type);
 ~~~~~~^~~~~~~~~~~~~
parentrises.cpp:105:11: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     scanf ("%d\n", &tests);
     ~~~~~~^~~~~~~~~~~~~~~~
parentrises.cpp:108:32: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
         scanf ("%s\n", sir + 1), N = strlen (sir + 1);
         ~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~
parentrises.cpp:219:7: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
 scanf ("%d", &tests);
 ~~~~~~^~~~~~~~~~~~~~
parentrises.cpp:222:11: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     scanf ("%d", &N);
     ~~~~~~^~~~~~~~~~

Subtask #1
5.0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Correct 4 ms 436 KB Output is correct
2 Correct 3 ms 436 KB Output is correct
3 Correct 3 ms 436 KB Output is correct
4 Correct 2 ms 556 KB Output is correct
5 Correct 3 ms 556 KB Output is correct

Subtask #2
11.0 / 11.0

# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 556 KB Output is correct
2 Correct 3 ms 556 KB Output is correct
3 Correct 2 ms 556 KB Output is correct
4 Correct 3 ms 556 KB Output is correct
5 Correct 2 ms 556 KB Output is correct

Subtask #3
6.0 / 6.0

# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 556 KB Output is correct
2 Correct 3 ms 556 KB Output is correct
3 Correct 2 ms 556 KB Output is correct
4 Correct 3 ms 556 KB Output is correct
5 Correct 2 ms 556 KB Output is correct
6 Correct 2 ms 620 KB Output is correct
7 Correct 4 ms 620 KB Output is correct
8 Correct 4 ms 620 KB Output is correct
9 Correct 3 ms 620 KB Output is correct
10 Correct 3 ms 620 KB Output is correct

Subtask #4
28.0 / 28.0

# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 556 KB Output is correct
2 Correct 3 ms 556 KB Output is correct
3 Correct 2 ms 556 KB Output is correct
4 Correct 3 ms 556 KB Output is correct
5 Correct 2 ms 556 KB Output is correct
6 Correct 2 ms 620 KB Output is correct
7 Correct 4 ms 620 KB Output is correct
8 Correct 4 ms 620 KB Output is correct
9 Correct 3 ms 620 KB Output is correct
10 Correct 3 ms 620 KB Output is correct
11 Correct 4 ms 620 KB Output is correct
12 Correct 5 ms 620 KB Output is correct
13 Correct 3 ms 636 KB Output is correct
14 Correct 4 ms 640 KB Output is correct
15 Correct 3 ms 640 KB Output is correct
16 Correct 8 ms 764 KB Output is correct
17 Correct 12 ms 1048 KB Output is correct
18 Correct 11 ms 1048 KB Output is correct
19 Correct 10 ms 1048 KB Output is correct
20 Correct 15 ms 1048 KB Output is correct
21 Correct 52 ms 1332 KB Output is correct
22 Correct 96 ms 3500 KB Output is correct
23 Correct 81 ms 3500 KB Output is correct
24 Correct 72 ms 3500 KB Output is correct
25 Correct 88 ms 3528 KB Output is correct

Subtask #5
6.0 / 6.0

# 결과 실행 시간 메모리 Grader output
1 Correct 18 ms 3528 KB Output is correct

Subtask #6
16.0 / 16.0

# 결과 실행 시간 메모리 Grader output
1 Correct 18 ms 3528 KB Output is correct
2 Correct 16 ms 3528 KB Output is correct

Subtask #7
28.0 / 28.0

# 결과 실행 시간 메모리 Grader output
1 Correct 18 ms 3528 KB Output is correct
2 Correct 16 ms 3528 KB Output is correct
3 Correct 17 ms 3528 KB Output is correct