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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> HDU 3364 Lanterns 高斯消元(水 異或方程

HDU 3364 Lanterns 高斯消元(水 異或方程

編輯:關於C++

 

題意:

給定n盞燈,m個開關

下面m行給出每個開關可以控制哪些燈(即按下此開關,這些燈的狀態會改變)

下面q個詢問:一行一個詢問,一個詢問n個數字表示燈的最終狀態

問從全暗到這個狀態的方案數(一個開關只能按一次)

n條方程,等式右邊就是輸入的燈的狀態。

m個未知數,表示每個開關是否按下,系數就是這個開關能否影響到那盞燈

解完方程後首先判斷系數矩陣的秩是否和增廣矩陣的秩相同,若相同則必有解,否則則無解。


因為求的是方案數,所以求出解完方程後的自由元數量即可。

方案數=2^(自由元數量)

 

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Queue;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;

public class Main {
	int[][] A = new int[N][N], init = new int[N][N];
	int n, m;
	long Gauss(int[][] mat, int row, int col){
		int r, c, i, j;
		for(r = c = 0; r < row && c < col; r++, c++){
			for(i = r; i < row; i++)
				if(mat[i][c]>0)break;
			if(i == row){r--; continue;}
			if(i!=r)
				for(j = c; j <= col; j++){
					int tmp = mat[r][j]; mat[r][j] = mat[i][j]; mat[i][j] = tmp;
				}
			for(i = r+1; i < row; i++)
				if(mat[i][c]>0)
					for(j = c; j <= col; j++)
						mat[i][j] ^= mat[r][j];
		}
		for(i = r; i < row; i++)
			if(mat[i][col]!=0)return 0L;
		
		return 1L<<(col-r);
	}
	void work() throws Exception {
		int T = Int(), Cas = 1;
		while (T-->0){
			n = Int(); m = Int();
			for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)init[i][j] = 0;
			for(int i = 0, j, num; i < m; i++)
			{
				num = Int(); 
				while(num-->0)init[Int()-1][i] = 1;
			}
			int q = Int();
			out.println(Case +(Cas++)+:);
			while(q-->0){
				for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)A[i][j] = init[i][j];
				for(int i = 0; i < n; i++)A[i][m] = Int();
				out.println(Gauss(A, n, m));
			}
		}
	}
	public static void main(String[] args) throws Exception {
		Main wo = new Main();
        in = new BufferedReader(new InputStreamReader(System.in));  
        out = new PrintWriter(System.out);  
		// in = new BufferedReader(new InputStreamReader(new FileInputStream(new File(input.txt))));
		// out = new PrintWriter(new File(output.txt));
		wo.work();
		out.close();
	}

	static int N = 56;
	static int M = N * 2;
	DecimalFormat df = new DecimalFormat(0.0000);
	static int inf = (int) 1e9;
	static long inf64 = (long) 1e18;
	static double eps = 1e-8;
	static double Pi = Math.PI;
	static int mod = (int) 1e9 + 7;

	private String Next() throws Exception {
		while (str == null || !str.hasMoreElements())
			str = new StringTokenizer(in.readLine());
		return str.nextToken();
	}

	private int Int() throws Exception {
		return Integer.parseInt(Next());
	}

	private long Long() throws Exception {
		return Long.parseLong(Next());
	}

	private double Double() throws Exception {
		return Double.parseDouble(Next());
	}

	StringTokenizer str;
	static Scanner cin = new Scanner(System.in);
	static BufferedReader in;
	static PrintWriter out;

	
	class Edge{
		int from, to, dis, nex;
		Edge(){} 
		Edge(int from, int to, int	 dis, int nex)
		{
		this.from = from; this.to = to; this.dis = dis; this.nex =	nex; 
		}
	} 
	Edge[] edge = new Edge[M<<1]; 
	int[] head = new int[N]; int edgenum; 
	void init_edge(){ for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;} 
	void add(int u, int v, int dis){ 
		edge[edgenum] = new Edge(u, v, dis, head[u]); 
		head[u] = edgenum++; 
	}
	 /*
	 */
	int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;
		int pos = r;
		r--;
		while (l <= r) {
			int mid = (l + r) >> 1;
			if (A[mid] <= val) {
				l = mid + 1;
			} else {
				pos = mid;
				r = mid - 1;
			}
		}
		return pos;
	}

	int lower_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;
		int pos = r;
		r--;
		while (l <= r) {
			int mid = (l + r) >> 1;
			if (A[mid] < val) {
				l = mid + 1;
			} else {
				pos = mid;
				r = mid - 1;
			}
		}
		return pos;
	}

	int Pow(int x, int y) {
		int ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}

	double Pow(double x, int y) {
		double ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}

	int Pow_Mod(int x, int y, int mod) {
		int ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			ans %= mod;
			y >>= 1;
			x = x * x;
			x %= mod;
		}
		return ans;
	}

	long Pow(long x, long y) {
		long ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}

	long Pow_Mod(long x, long y, long mod) {
		long ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			ans %= mod;
			y >>= 1;
			x = x * x;
			x %= mod;
		}
		return ans;
	}

	int gcd(int x, int y) {
		if (x > y) {
			int tmp = x;
			x = y;
			y = tmp;
		}
		while (x > 0) {
			y %= x;
			int tmp = x;
			x = y;
			y = tmp;
		}
		return y;
	}

	int max(int x, int y) {
		return x > y ? x : y;
	}

	int min(int x, int y) {
		return x < y ? x : y;
	}

	double max(double x, double y) {
		return x > y ? x : y;
	}

	double min(double x, double y) {
		return x < y ? x : y;
	}

	long max(long x, long y) {
		return x > y ? x : y;
	}

	long min(long x, long y) {
		return x < y ? x : y;
	}

	int abs(int x) {
		return x > 0 ? x : -x;
	}

	double abs(double x) {
		return x > 0 ? x : -x;
	}

	long abs(long x) {
		return x > 0 ? x : -x;
	}

	boolean zero(double x) {
		return abs(x) < eps;
	}

	double sin(double x) {
		return Math.sin(x);
	}

	double cos(double x) {
		return Math.cos(x);
	}

	double tan(double x) {
		return Math.tan(x);
	}

	double sqrt(double x) {
		return Math.sqrt(x);
	}
	double fabs(double x){return x>0?x:-x;}
}


 

因為求的是方案數,所以求出解完方程後的自由元數量即可。
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