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Codeforces 327E Axis Walking 状压dp(水

时间:2015-03-20 23:55:19      阅读:519      评论:0      收藏:0      [点我收藏+]

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题目链接:点击打开链接

题意:

给定n个数,随意排列。

给定k个违禁数b[]。

问:有多少个排列使得这个排列的 n项前缀和中不出现违禁数。

(formally,if it‘s a legal permutation, sum[i] != b[j] (1<=i<=n, 1<=j<=k))

sum[0] = 0; sum[i] = sum[i-1]+a[permutaion[i]];

==java党表示被tle,心疼自己T^T

#include <stdio.h>
const int N = 24;
int dp[1<<N], a[N], b[N];
int n, k;
const int mod = 1e9+7;
int main(){
	scanf("%d", &n);
    for(int i = 0; i < n; i++)scanf("%d", &a[i]);
    scanf("%d", &k);
    for(int i = 0; i < k; i++)scanf("%d", &b[i]);
    
    dp[0] = 1;int all = 1<<n;
    for(int i = 1, sum; i < all; i++){
        sum = 0;
        for(int j = 0; j < n; j++)
            if((i&(1<<j))>0)
            {
                dp[i] += dp[i^(1<<j)];
                if(dp[i]>=mod)dp[i] -= mod;
                sum += a[j];
        }
        for(int j = 0; j < k; j++)if(sum == b[j])dp[i] = 0;
	}       
	printf("%d\n",dp[(1<<n)-1]);
	return 0;
}

任性附上java tle代码

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[] dp = new int[1<<N], a = new int[N], b = new int[N];
	int n, k;
	void work() throws Exception {
		n = Int();
		for(int i = 0; i < n; i++)a[i] = Int();
		k = Int();
		for(int i = 0; i < k; i++)b[i] = Int();
		
		dp[0] = 1;int all = 1<<n;
		for(int i = 1, sum; i < all; i++){
			sum = 0;
			for(int j = 0; j < n; j++)
				if((i&(1<<j))>0)
				{
					dp[i] += dp[i^(1<<j)];
					if(dp[i]>=mod)dp[i] -= mod;
					sum += a[j];
				}
			for(int j = 0; j < k; j++)if(sum == b[j])dp[i] = 0;
		}		
		out.println(dp[(1<<n)-1]);
	}

	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 = 24;
	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);
	}
}


Codeforces 327E Axis Walking 状压dp(水

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原文地址:http://blog.csdn.net/qq574857122/article/details/44498103

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