QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#668064#5576. Advertising ICPCenze114514TL 1ms3604kbC++203.8kb2024-10-23 11:01:532024-10-23 11:02:04

Judging History

你现在查看的是最新测评结果

  • [2024-10-23 11:02:04]
  • 评测
  • 测评结果:TL
  • 用时:1ms
  • 内存:3604kb
  • [2024-10-23 11:01:53]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int MOD = 998244353;

ll powmod_func(ll a, ll b) {
    ll res = 1;
    a %= MOD;
    while(b > 0){
        if(b & 1){
            res = res * a % MOD;
        }
        a = a * a % MOD;
        b >>= 1;
    }
    return res;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n, m;
    cin >> n >> m;
    vector<string> grid(n);
    for(int i=0; i<n; i++) cin >> grid[i];
    
    int k = 0;
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            if(grid[i][j] == '?') k++;
        }
    }
    
    bool has_pre_filled_pattern = false;
    for(int i=0; i<n-1 && !has_pre_filled_pattern; i++) {
        for(int j=0; j<m-1 && !has_pre_filled_pattern; j++) {
            if(grid[i][j] != '?' && grid[i][j+1] != '?' && grid[i+1][j] != '?' && grid[i+1][j+1] != '?'){
                if(grid[i][j] == 'I' && grid[i][j+1] == 'C' && grid[i+1][j] == 'P' && grid[i+1][j+1] == 'C'){
                    has_pre_filled_pattern = true;
                }
            }
        }
    }
    
    ll T = powmod_func(3, k);
    
    if(has_pre_filled_pattern){
        cout << T % MOD;
        return 0;
    }
    
    int pow3_m = 1;
    for(int i=0; i<m; i++) pow3_m *= 3;
    
    vector<vector<int>> allowed_states(n, vector<int>());
    for(int r=0; r<n; r++){
        for(int s=0; s<pow3_m; s++){
            bool valid = true;
            int temp = s;
            for(int j=0; j<m; j++){
                int val = temp % 3;
                temp /= 3;
                if(grid[r][j] != '?'){
                    if((grid[r][j] == 'C' && val != 0) || 
                       (grid[r][j] == 'I' && val != 1) || 
                       (grid[r][j] == 'P' && val != 2)){
                        valid = false;
                        break;
                    }
                }
            }
            if(valid){
                allowed_states[r].push_back(s);
            }
        }
    }
    
    vector< vector< vector<int> > > row_digits(n, vector< vector<int> >(pow3_m, vector<int>(m, 0)));
    for(int r=0; r<n; r++){
        for(auto &s: allowed_states[r]){
            int temp = s;
            for(int j=0; j<m; j++){
                row_digits[r][s][j] = temp % 3;
                temp /= 3;
            }
        }
    }
    
    vector< vector< vector<int> > > transitions(n-1, vector< vector<int> >(pow3_m, vector<int>()));
    for(int r=1; r<n; r++){
        for(auto &s_prev: allowed_states[r-1]){
            for(auto &s_curr: allowed_states[r]){
                bool forbidden = false;
                for(int j=0; j<m-1; j++){
                    if(row_digits[r-1][s_prev][j] == 1 &&         
                       row_digits[r-1][s_prev][j+1] == 0 &&   
                       row_digits[r][s_curr][j] == 2 &&          
                       row_digits[r][s_curr][j+1] == 0){       
                        forbidden = true;
                        break;
                    }
                }
                if(!forbidden){
                    transitions[r-1][s_prev].push_back(s_curr);
                }
            }
        }
    }
    
    vector<ll> dp_prev(pow3_m, 0);
    for(auto &s: allowed_states[0]){
        dp_prev[s] = 1;
    }
    
    for(int r=1; r<n; r++){
        vector<ll> dp_curr(pow3_m, 0);
        for(auto &s_prev: allowed_states[r-1]){
            if(dp_prev[s_prev] == 0) continue;
            for(auto &s_curr: transitions[r-1][s_prev]){
                dp_curr[s_curr] = (dp_curr[s_curr] + dp_prev[s_prev]) % MOD;
            }
        }
        dp_prev = dp_curr;
    }
    
    ll A = 0;
    for(auto &s: allowed_states[n-1]){
        A = (A + dp_prev[s]) % MOD;
    }
    
    ll answer = (T - A + MOD) % MOD;
    cout << answer;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3536kb

input:

3 3
???
?I?
???

output:

243

result:

ok single line: '243'

Test #2:

score: 0
Accepted
time: 0ms
memory: 3604kb

input:

2 2
IC
PC

output:

1

result:

ok single line: '1'

Test #3:

score: -100
Time Limit Exceeded

input:

8 8
????????
????????
????????
????????
????????
????????
????????
????????

output:


result: