Codeforces 567F Mausoleum dp

Posted 2022-02-13 cjlhy

Mausoleum

考虑将数字从大到小一种一种填进去， 新加入的数字只会在之前组成序列的两侧， 所以形状一直是连续的一段。

dp[ k ][ i ][ j ] 表示填入了前 k 大的数字， 它们位于i - j 的合法方案数， 转移的时候check一下能否转移。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 700 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) a += b; if(a >= mod) a -= mod;
template<class T, class S> inline void sub(T& a, S b) a -= b; if(a < 0) a += mod;
template<class T, class S> inline bool chkmax(T& a, S b) return a < b ? a = b, true : false;
template<class T, class S> inline bool chkmin(T& a, S b) return a > b ? a = b, true : false;

int n, k;

LL dp[36][71][71];

int lh[101], rh[101], op[101];
char s[10];

bool check(int l, int r, int p1, int p2) 
    for(int i = 1; i <= k; i++) 
        if(lh[i] == rh[i]) continue;
        if(lh[i] == p1 && rh[i] == p2) 
            if(op[i] == 1 || op[i] == 2) return false;
         else if(lh[i] == p1 || lh[i] == p2) 
            if(l <= rh[i] && rh[i] <= r) 
                if(op[i] != 1 && op[i] != 3) return false;
             else 
                if(op[i] != 2 && op[i] != 4) return false;
            
         else if(rh[i] == p2 || rh[i] == p1) 
            if(l <= lh[i] && lh[i] <= r) 
                if(op[i] != 2 && op[i] != 4) return false;
             else 
                if(op[i] != 1 && op[i] != 3) return false;
            
        
        else 
            if(rh[i] < l || lh[i] > r) continue;
            if(l <= lh[i] && rh[i] <= r) continue;
            if(lh[i] < l && rh[i] > r) continue;
            if(lh[i] < l) 
                if(op[i] != 1 && op[i] != 3) return false;
             else 
                if(op[i] != 2 && op[i] != 4) return false;
            
        
    
    return true;


int main() 
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= k; i++) 
        scanf("%d%s%d", &lh[i], s, &rh[i]);
        if(strlen(s) == 1) 
            if(s[0] == ‘=‘) op[i] = 0;
            else if(s[0] == ‘<‘) op[i] = 1;
            else if(s[0] == ‘>‘) op[i] = 2;
         else 
            if(s[0] == ‘<‘) op[i] = 3;
            else if(s[0] == ‘>‘) op[i] = 4;
        
        if(lh[i] > rh[i]) 
            if(op[i] == 1) op[i] = 2;
            else if(op[i] == 2) op[i] = 1;
            else if(op[i] == 3) op[i] = 4;
            else if(op[i] == 4) op[i] = 3;
            swap(lh[i], rh[i]);
        
    
    for(int i = 1; i < 2 * n; i++) 
        dp[n][i][i + 1] = 1;
        for(int j = 1; j <= k; j++) 
            if(lh[j] == i && rh[j] == i + 1 && (op[j] == 1 || op[j] == 2)) 
                dp[n][i][i + 1] = 0;
                break;
            
        
    
    for(int k = n; k > 1; k--) 
        int len = (n - k + 1) * 2;
        for(int i = 1; i + len - 1 <= 2 * n; i++) 
            int j = i + len - 1;
            if(!dp[k][i][j]) continue;

            if(i >= 3 && check(i, j, i - 2, i - 1)) 
                dp[k - 1][i - 2][j] += dp[k][i][j];
            

            if(j <= 2 * n - 2 && check(i, j, j + 1, j + 2)) 
                dp[k - 1][i][j + 2] += dp[k][i][j];
            

            if(i >= 2 && j <= 2 * n - 1 && check(i, j, i - 1, j + 1)) 
                dp[k - 1][i - 1][j + 1] += dp[k][i][j];
            
        
    
    printf("%lld\n", dp[1][1][2 * n]);
    return 0;


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