The last question of winter vacation is finished. ^∀^

Test instructions

1 to n each number has two, ranked first does not drop after the sequence of non-ascending, such as 112332, and satisfies the K shape such as 3 <= 6 for the third number to ≤ the sixth number of such a constraint requirements, how many kinds of row.

Analysis

Interval Dp,dp[i][j] Indicates the number of scenarios where only the interval [i,j] has not been filled.

B[I][J] represents the number constraint relationship for position I and J.

Then start the platoon from both ends and place the two number T on either side or the same side at a time.

dp[i+1][j-1]+=dp[i][j];//on both sides.

dp[i+2][j]+=dp[i][j];//on the left.

dp[i][j-2]+=dp[i][j];//on the right.

and to satisfy the constraint conditions. If I position is less than J position, then I position must be placed first.

That is, the number of,< (≤) placed in the position at the current position on the Count of > (≥) has been placed.

and two positions each time if there is a constraint relationship, can only contain = (=, ≥, ≤).

When J==i+1, there is only one way to put the law, this time you can accumulate the answer.

When we put the number T, the length of the interval [i,j] is 2*n-2* (t-1), so j=i+2*n-(t-1) -1=2*n-2*t+i+1.

The number of scenarios is larger, so a long long is used.

code

#include <cstdio> #include <cstring> #define LL long long#define N 75int n,k;ll dp[n][n],ans;int b[n][n];//b[i ][J] -2-1 3 1 2//i s J > >= = <= <int ch (int fl,int fr,int a,int c)//The number of positions in the row now is smaller than the free part { for (int i=fl; i<=fr; i++) if (b[a][i]==3| | b[c][i]==3| | b[a][i]<0| | B[C][I]<0) return 0; return 1;} int cc (int bl,int br,int a,int c) {//Now the number of positions in the row is greater than the portion of the large for (int i=1;i<=bl;i++) if (b[a][i]>0| | B[C][I]>0) return 0; for (int i=br;i<=2*n;i++) if (b[a][i]>0| | B[C][I]>0) return 0; return 1;} int check (int i,int j)//cannot be placed on two digits not allowed in the same position {return b[i][j]!=-2&&b[i][j]!=2;} int main () {scanf ("%d%d", &n,&k); for (int i=1; i<=k; i++) {int l,r,f=3; Char s[5]; scanf ("%d%s%d", &l,s,&r); if (s[0]== ' < ') {f=2; if (s[1]) f--; } else if (s[0]== ' > ') {f=-2; if (s[1]) f++; } b[l][R]=f; B[r][l]=f==3?f:-f; } dp[1][2*n]=1; for (int t=1, t<=n; t++) for (int i=1; i<=2*t-1; i++) {int j=2*n-2*t+i+1; if (Dp[i][j]) {if (j==i+1) {if (check (I,J)) ANS+=DP[I][J]; } else {if (ch (i+1,j-1,i,j) &&cc (i-1,j+1,i,j) &&check (i,j)) DP[I+1][J-1]+=DP[I][J]; if (CH (i+2,j,i,i+1) &&cc (i-1,j+1,i,i+1) &&check (i,i+1)) dp[i+2][j]+=dp[i][j]; if (CH (i,j-2,j-1,j) &&cc (i-1,j+1,j-1,j) &&check (j-1,j)) dp[i][j-2]+=dp[i][ J]; }}} printf ("%lld", ans); return 0;}

"Codeforces 567F" mausoleum