Zoj 1100-mondriaan ' s Dream

Title: On the floor of the m*n the same 1*2 tile, ask how many kinds of paving method.

Analysis: DP, combination, count. Classic DP problem, state compression.

Status: Set F (i,j) for the front i-1 line, the first row of the state of the position expressed as J when the number of paving bricks;

Transfer: Because only horizontal or vertical paving, then a brick shop before the state only two kinds;

And if the current vertical position will affect the next line, establish two adjacent lines of state correspondence;

Here, Dfs is used to find all the pre-state f (i-1,k) plus and i,j of all the previous lines of f (a).

F (i,j) = SUM (f (i-1,k)) {where F (i-1,k) can produce an F (i,j) state};

(Rhubarb's three-dimensional DP implementation is simple and inefficient.) ）

Combinatorial formula: π (4cos (pi+i/(h+1)) ^2+4cos (pi+j/(w+1)) ^2) {1<=I<=H/2,1<=J<=W/2}.

Description: Tangled n long last found%i64d has been WA. The%LLD is over. (2011-09-27 19:15).

#include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct Node {int s,l;}
Seg

SEG s[10];

Long long f[[1<<11];
int v[1<<11 [99];

int count[1<<11]; Use DFS to find the status that can be reached in void Dfs (int A, int B, int C) {if (!
        A) {v[c] [+ count[c]] = B;
    Return 
        }else {int V = a&-a;//Gets the last 1 position of Dfs (A&~v, b&~v, C);
    if (a& (v<<1)) Dfs (a&~ (3*v), B, C);
    }} int main () {int n,m; while (scanf ("%d%d", &n,&m)! = EOF && m) {if (n%2&&m%2) {printf ("0\n"); contin
        UE;}
        
        if (m>n) {int t = M;m = N;n = t;}
        int M = (1<<m)-1;
            for (int i = 0; I <= M; + + i) {count[i] = 0;
        DFS (i, M, i);  } for (int i = 0; I <= N; + + i) for (int j = 0; J <= M; + + j) f[I [j]
      ] = 0LL;  f[0] [M] = 1LL; for (int i = 1; I <= n; + + i) for (int j = M; J >= 0;--j) for (int k = count[j]; K &G T;= 1;
        
        --K) f[I [j] + = f[i-1 [v[j] [K]];
    printf ("%lld\n", f[N [M]);
} return 0;
 }