LightOj 1332 矩阵快速幂

LightOj 1332
题目链接：
http://www.bnuoj.com/v3/problem_show.php?pid=13251
题意：
给一个10*n(n<=1e9)的矩阵，现在每列需要放两个国王。
已知国王会在他所在的九宫格范围内实施攻击，问有多少种放置方法使得国王之间不相互攻击。
思路：
刚开始拿到手是一点思路没有。
强行打表找规律，失败。
因为n很大，不是公式就是有递推关系。公式的话肯定由前几项可以算的，那么体现在打的表上就会呈现一定规律。然而失败了，说明一定是递推的关系，因为n很大所以猜用快速幂优化。
然后强行往快速幂那边靠，发现每次取的点进行编号，下一次可以取得点只能由上一次取得一些点得到，所以可以构成一个关系矩阵。
然而我并不相信这是题解，随便写一发没过样例，看了别人代码后悔莫及啊……
源码：
赶时间版本

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;
#define LL unsigned long long
typedef pair<int,int> pii;
const LL mod = ((LL)1) << 32;
int vis[15][15];
LL g[40][40];
LL a[40];
map<pii, int>M;
int cnt;
int cal_pos(int x, int y){return (19 - x) * (x - 1) / 2 ;}
bool valid(int row)
{
    for(int i = 1 ; i <= 10 ; i++){
        if(vis[row][i] && vis[row][i - 1]) return false;
        if(vis[row][i] && (vis[row - 1][i - 1] || vis[row - 1][i] || vis[row - 1][i + 1])) return false;
    }
    return true;
}
void dfs1(int row)
{
// if(row == 2){
// int x1, y1;
// x1 = -1;
// for(int i = 1 ; i <= 10 ; i++){
// if(vis[1][i]){
// if(x1 == -1) x1 = i;
// else y1 = i;
// }
// }
//// pii temp = make_pair(x1, y1);
//// if(M[temp] == 0)
// }
    if(row == 3){
        int x1, y1;
        x1 = -1;
        for(int i = 1 ; i <= 10 ; i++){
            if(vis[1][i]){
                if(x1 == -1) x1 = i;
                else y1 = i;
            }
        }
        int x2, y2;
        x2 = -1;
        for(int i = 1 ; i <= 10 ; i++){
            if(vis[2][i]){
                if(x2 == -1) x2 = i;
                else y2 = i;
            }
        }
// printf("x1 = %d, y1 = %d, x2 = %d, y2 = %d\n,cal1 = %d, cal2 = %d", x1, y1, x2, y2, cal_pos(x1, y1), cal_pos(x2, y2));
// system("pause");
        g[M[make_pair(x1, y1)]][M[make_pair(x2, y2)]] = 1;
        return;
    }
    for(int i = 1 ; i <= 10 ; i++){
        vis[row][i] = 1;
        for(int j = i + 1 ; j <= 10 ; j++){
            vis[row][j] = 1;
            if(valid(row)) dfs1(row + 1);
            vis[row][j] = 0;
        }
        vis[row][i] = 0;
    }
}
void init()
{
    M.clear();
    cnt = 0;
    memset(vis, 0, sizeof(vis));
    memset(g, 0, sizeof(g));
    for(int i = 1 ; i <= 10 ; i++){
        for(int j = i + 2 ; j <= 10 ; j++)
            M[make_pair(i, j)] = ++cnt;
    }
// printf("%d\n", cnt);
// memset(a, 0, sizeof(a));
    dfs1(1);
}
struct Matrix
{
    LL d[40][40];
    void init(){for(int i = 1 ; i <= 36 ; i++) for(int j = 1 ; j <= 36 ; j++) if(i == j) d[i][j] = 1; else d[i][j] = 0;}
    Matrix operator * (const Matrix& b)const{
        Matrix ans;
        for(int i = 1 ; i <= 36 ; i++){
            for(int j = 1 ; j <= 36 ; j++){
                ans.d[i][j] = 0;
                for(int k = 1 ; k <= 36 ; k++)
                    ans.d[i][j] = (ans.d[i][j] + d[i][k] * b.d[k][j] % mod) % mod;
            }
        }
        return ans;
    }
}org;
Matrix ppow(Matrix a, int x)
{
    Matrix ans;
    ans.init();
    while(x){
        if(x & 1) ans = (ans * a);
        a = (a * a);
        x = x >> 1;
    }
    return ans;
}
int main()
{
    init();
// for(int j = 1 ; j <= 36 ; j++) printf("%d ", g[i][j]);
// printf("\n");
// }
// for(int i = 1 ; i <= 36 ; i++) printf("%d ", a[i]);
    int T;
    scanf("%d", &T);
    for(int cas = 1 ; cas <= T ; cas++){
        int n;
        scanf("%d", &n);
        for(int i = 1 ; i <= 36 ; i++) for(int j = 1 ; j <= 36 ; j++) org.d[i][j] = g[i][j];
        Matrix ans = ppow(org, n - 1);
// for(int i = 1 ; i <= 36 ; i++){
// for(int j = 1 ; j <= 36 ; j++) printf("%d ", g[i][j]);
// printf("\n");
// }
        LL res = 0;
// printf("%I64d\n", mod);
        for(int i = 1 ; i <= 36 ; i++) for(int j = 1 ; j <= 36 ; j++){
            LL tres = res;
            res = (res + ans.d[i][j]) % mod;
// if(res != tres){
// printf("res = %llu, ans.d[i][j] = %llu\n", res, ans.d[i][j]);
// system("pause");
// }
        }
        printf("Case %d: %llu\n", cas, res);
    }
    return 0;
}

好看的版本

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;
#define LL unsigned long long
typedef pair<int,int> pii;
const LL mod = ((LL)1) << 32;
int vis[15][15];
LL g[40][40];
LL a[40];
map<pii, int>M;
int cnt;
int cal_pos(int x, int y){return (19 - x) * (x - 1) / 2 ;}
bool valid(int row)
{
    for(int i = 1 ; i <= 10 ; i++){
        if(vis[row][i] && vis[row][i - 1]) return false;
        if(vis[row][i] && (vis[row - 1][i - 1] || vis[row - 1][i] || vis[row - 1][i + 1])) return false;
    }
    return true;
}
void dfs1(int row)
{
    if(row == 3){
        int x1, y1;
        x1 = -1;
        for(int i = 1 ; i <= 10 ; i++){
            if(vis[1][i]){
                if(x1 == -1) x1 = i;
                else y1 = i;
            }
        }
        int x2, y2;
        x2 = -1;
        for(int i = 1 ; i <= 10 ; i++){
            if(vis[2][i]){
                if(x2 == -1) x2 = i;
                else y2 = i;
            }
        }
        g[M[make_pair(x1, y1)]][M[make_pair(x2, y2)]] = 1;
        return;
    }
    for(int i = 1 ; i <= 10 ; i++){
        vis[row][i] = 1;
        for(int j = i + 1 ; j <= 10 ; j++){
            vis[row][j] = 1;
            if(valid(row)) dfs1(row + 1);
            vis[row][j] = 0;
        }
        vis[row][i] = 0;
    }
}
void init()
{
    M.clear();
    cnt = 0;
    memset(vis, 0, sizeof(vis));
    memset(g, 0, sizeof(g));
    for(int i = 1 ; i <= 10 ; i++){
        for(int j = i + 2 ; j <= 10 ; j++)
            M[make_pair(i, j)] = ++cnt;
    }
    dfs1(1);
}
struct Matrix
{
    LL d[40][40];
    void init(){for(int i = 1 ; i <= 36 ; i++) for(int j = 1 ; j <= 36 ; j++) if(i == j) d[i][j] = 1; else d[i][j] = 0;}
    Matrix operator * (const Matrix& b)const{
        Matrix ans;
        for(int i = 1 ; i <= 36 ; i++){
            for(int j = 1 ; j <= 36 ; j++){
                ans.d[i][j] = 0;
                for(int k = 1 ; k <= 36 ; k++)
                    ans.d[i][j] = (ans.d[i][j] + d[i][k] * b.d[k][j] % mod) % mod;
            }
        }
        return ans;
    }
}org;
Matrix ppow(Matrix a, int x)
{
    Matrix ans;
    ans.init();
    while(x){
        if(x & 1) ans = (ans * a);
        a = (a * a);
        x = x >> 1;
    }
    return ans;
}
int main()
{
    init();
    int T;
    scanf("%d", &T);
    for(int cas = 1 ; cas <= T ; cas++){
        int n;
        scanf("%d", &n);
        for(int i = 1 ; i <= 36 ; i++) for(int j = 1 ; j <= 36 ; j++) org.d[i][j] = g[i][j];
        Matrix ans = ppow(org, n - 1);
        LL res = 0;
        for(int i = 1 ; i <= 36 ; i++) for(int j = 1 ; j <= 36 ; j++){
            LL tres = res;
            res = (res + ans.d[i][j]) % mod;
        }
        printf("Case %d: %llu\n", cas, res);
    }
    return 0;
}