uva11270 - Tiling Dominoes 插头DP

A. Tiling Dominoes

Problem

Given a rectangular grid, with dimensions m x n, compute the number of ways of completely tiling it with dominoes. Note that if the rotation of one tiling matches another, they still count as different ones. A domino is a shape formed by the union of two unit squares meeting edge-to-edge. Equivalently, it is a matching in the grid graph formed by placing a vertex at the center of each square of the region and connecting two vertices when they correspond to adjacent squares. An example of a tiling is shown below.

The Input

The input will consist of a set of lines with m n, given the restriction n*m<101.

The Output

For each line of input, output the number of tilings in a separate line.

Sample Input

2 10
4 10
8 8

Sample Output

89
18061
12988816

  N*M的棋盘用1*2的骨牌完全覆盖有多少种方法。

  第一次做插头DP。插头DP适合在比较窄的棋盘上进行操作。假设行N比较大，列M比较小，这里用一个M位的状态表示当前状态，设第i行j列的格子编号是i*M+j，那么如果上一次的状态是s[i-M],s[i-M+1]...s[i-1]，那么这次的状态就是s[i-M+1],s[i-M+2]...s[i]，当前状态S的第j位就是s[i-j]，s[i]为0代表编号为i的格子还没有被覆盖，1代表覆盖了。对于每个格子，可以横着放，竖着放，或者暂时不放。要注意判断一些不合法的情况的判断。因为每一步的状态只和上一步的状态有关，所以用滚动数组就行，最后答案是dp[cur][2^M-1]。复杂度O（2^M*N*M）。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int MAXN=11;
const int MAXM=20010;

int N,M,cur;
LL dp[2][1<<MAXN];

//a是旧状态，b是新状态
void update(int a,int b){
    if(b&(1<<M)) dp[cur][b^(1<<M)]+=dp[!cur][a];
}

int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&N,&M)!=EOF){
        if(N<M) swap(N,M);
        cur=0;
        memset(dp,0,sizeof(dp));
        dp[0][(1<<M)-1]=1;
        for(int i=0;i<N;i++)
            for(int j=0;j<M;j++){
                cur=!cur;
                memset(dp[cur],0,sizeof(dp[cur]));
                for(int k=0;k<(1<<M);k++){
                    update(k,k<<1);     //不放
                    if(i&&!(k&(1<<(M-1)))) update(k,(k<<1)^(1<<M)^1);   //竖着放
                    if(j&&!(k&1)) update(k,(k<<1)^3);   //横着放
                }
            }
        printf("%lld\n",dp[cur][(1<<M)-1]);
    }
    return 0;
}