HDU 4332 Constructing Chimney 矩阵快速幂

算法：

1.状态压缩

2.搜索

3.矩阵快速幂

#include<iostream>
using namespace std;

typedef long long LL;
const int mod = 1000000007;
const int mask = 255;

struct Matrix{
    LL mat[100][100];
    void unit(){
        memset(mat,0,sizeof(mat));
        for(int i = 0; i < 100; i++) mat[i][i]=1;    
    }    
    void zero(){
        memset(mat,0,sizeof(mat));    
    }
}T,A,U;
Matrix operator * (const Matrix &a,const Matrix &b){
    Matrix tmp; tmp.zero();
    for(int i = 0; i <= 70; i++){
        for(int k = 0; k <= 70; k++){
            if(!a.mat[i][k]) continue;
            for(int j = 0; j <= 70; j++){
                tmp.mat[i][j] += (a.mat[i][k]*b.mat[k][j])%mod;
                tmp.mat[i][j] %= mod;    
            }    
        }    
    }    
    return tmp;
}
Matrix operator ^(Matrix x,int k){
    Matrix tmp; tmp.unit();
    while(k){
        if(k&1) tmp = tmp*x;
        x = x*x; k = (k>>1);    
    }    
    return tmp;
}
Matrix Pow(Matrix x,int k){
    if(k == 0) return U;
    if(k&1) return (Pow(x,k/2)*Pow(x,k/2)*x);
    else    return (Pow(x,k/2)*(Pow(x,k/2)));
    
}
int vis[mask+10],id[mask+10],legal[mask+10],idx;
/*
判断横放是否合法
0 1 2
7   3
6 5 4
只要判断01234567，65432170是否有两个连续1
*/
    
bool check(int x){
    int y=0,t = x>>1;
    for(int i = 0; i < 7; i++){
        y = (y<<1) | (t&1);
        t >>= 1;    
    }    
    y = (y<<1) | (x&1);
    bool flag = true;
    for(int i = 0; i < 8; i++){
        if( !( x&(1<<i) ) ) continue;
        if( i == 7 || !( x&(1<<(i+1))) ){
            flag = false; break;    
        }
        i++;    
    }
    if(flag) return true;
    flag = true;
    for(int i = 0; i < 8; i++){
        if( !( y&(1<<i) ) ) continue;
        if( i == 7 || !( y&(1<<(i+1)) ) ){
            flag = false; break;    
        }    
        i++;
    }
    return flag;
}
/*
搜索出合法状态，并且再把状态压缩。。
*/ 
void DFS(int x){
    if( vis[x] ) return;  vis[x] = 1;
    
    int c = ~x & mask;
    if( id[x] == -1 ) id[x] = idx++;
    for(int i = 0; i <= mask; i++){
        if( (i&c)==c && check( i&x ) ){    
            if( id[i] == -1 ) id[i] = idx++;    
            T.mat[ id[i] ][ id[x] ] = 1;
            if( !vis[i] ) DFS(i);
        }    
    }    
}

void init(){
    T.zero(); U.unit();
    memset(vis,0,sizeof(vis));
    memset(id,0xff,sizeof(id));
    idx = 0; 
    DFS(mask); T.mat[0][0]+=1; 
    memset(legal,0,sizeof(legal));
    for(int i = 0; i <= mask; i++){
        if( ~id[i] && T.mat[id[i]][0] ) 
            legal[ id[i] ] = 1;    
    } 
    legal[0] += 1;
}
void solve(int k){
    A = T^(k-1);
    LL res = 0;
    for(int i = 0; i < idx; i++){
        res = (res+legal[i]*A.mat[i][0]%mod)%mod;
    }    
    printf("%I64d\n", res);
}
int main(){
    //freopen("1002.in","r",stdin);
    //freopen("1.out","w",stdout);
    init();    
    int t, k;
    scanf("%d", &t);
    for(int ncase = 1; ncase <= t; ncase++){
        scanf("%d",&k);
        printf("Case %d: ", ncase);
        solve(k);    
    }
}