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  • BZOJ4000 [TJOI2015]棋盘 【状压dp + 矩阵优化】

    题目链接

    BZOJ4000

    题解

    注意题目中的编号均从(0)开始= =

    (m)特别小,考虑状压
    (f[i][s])为第(i)行为(s)的方案数
    每个棋子能攻击的只有本行,上一行,下一行,
    我们能迅速找出哪些状态是合法的,以及每个状态所对应的上一行攻击位置的并和下一行攻击位置的并
    如果两个状态上下相互攻击不到,就是合法的转移

    我们弄一个(2^m * 2^m)的转移矩阵,就可以矩阵优化了

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #define uint unsigned int
    #define LL long long int
    #define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
    #define REP(i,n) for (int i = 1; i <= (n); i++)
    #define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
    using namespace std;
    const int maxn = 65,maxm = 100005,INF = 1000000000;
    inline int read(){
    	int out = 0,flag = 1; char c = getchar();
    	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    	return out * flag;
    }
    int s1,s2,s3,n,m,p,k;
    struct Matrix{
    	uint s[maxn][maxn]; int n,m;
    	Matrix(){memset(s,0,sizeof(s)); n = m = 0;}
    }A,F,Fn;
    inline Matrix operator *(const Matrix& a,const Matrix& b){
    	Matrix c;
    	if (a.m != b.n) return c;
    	c.n = a.n; c.m = b.m;
    	for (int i = 0; i < c.n; i++)
    		for (int j = 0; j < c.m; j++)
    			for (int k = 0; k < a.m; k++)
    				c.s[i][j] += a.s[i][k] * b.s[k][j];
    	return c;
    }
    inline Matrix qpow(Matrix a,int b){
    	Matrix ans; ans.n = ans.m = a.n;
    	for (int i = 0; i < ans.n; i++) ans.s[i][i] = 1;
    	for (; b; b >>= 1,a = a * a)
    		if (b & 1) ans = ans * a;
    	return ans;
    }
    int f[maxn];
    bool check(int s){
    	for (int i = 0; i < m; i++){
    		if (s & (1 << i)){
    			if (i + 1 >= p - k){
    				if (s & (s2 << (i + 1 - (p - k)))) return false;
    			}
    			else if (s & (s2 >> ((p - k) - i - 1))) return false;
    		}
    	}
    	return true;
    }
    int getu(int s){
    	int re = 0;
    	for (int i = 0; i < m; i++){
    		if (s & (1 << i)){
    			if (i + 1 >= p - k) re |= s1 << (i + 1 - (p - k));
    			else re |= s1 >> ((p - k) - i - 1);
    		}
    	}
    	return re;
    }
    int getd(int s){
    	int re = 0;
    	for (int i = 0; i < m; i++){
    		if (s & (1 << i)){
    			if (i + 1 >= p - k) re |= s3 << (i + 1 - (p - k));
    			else re |= s3 >> ((p - k) - i - 1);
    		}
    	}
    	return re;
    }
    void print(int x){
    	for (int i = 4; i >= 0; i--)
    		printf("%d",(x & (1 << i)) != 0);
    }
    int main(){
    	n = read(); m = read();
    	p = read(); k = read();
    	REP(i,p) s1 = (s1 << 1) + read();
    	REP(i,p){
    		if (i == k + 1) s2 <<= 1,read();
    		else s2 = (s2 << 1) + read();
    	}
    	REP(i,p) s3 = (s3 << 1) + read();
    	int N = 1 << m;
    	F.n = N; F.m = 1;
    	for (int s = 0; s < N; s++)
    		if (check(s)) F.s[s][0] = 1;
    	A.n = A.m = N;
    	for (int s = 0; s < N; s++){
    		if (!check(s)) continue;
    		for (int e = 0; e < N; e++){
    			if (!check(e)) continue;
    			int u = getu(s),d = getd(e);
    			if (!(s & d) && !(e & u))
    				A.s[s][e] = 1;
    		}
    	}
    	Fn = qpow(A,n - 1) * F;
    	uint ans = 0;
    	for (int i = 0; i < N; i++) ans += Fn.s[i][0];
    	cout << ans << endl;
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Mychael/p/9007679.html
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