CodeForces 417D Cunning Gena 状压dp

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题目链接：点击打开链接

题意：主角请他的n个朋友帮他做m道题，常数b

下面每两行代表一个朋友：

请这个朋友的花费 请这个朋友的前提是要拥有k台电脑 这个朋友能解的题目数量

下面是这个朋友能解的具体题目号。

买一台电脑花费是b，开始主角手里没有电脑。

思路：

先按k值小到大排序

问题数<=20, 所以状压问题然后把一个个人依次更新到状态里即可。

import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
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.TreeMap;
import java.util.TreeSet;
import java.util.Queue;

public class Main {	
	class Node implements Comparable<Node>{
		long x, k; int m, t;
		Node(){}
		Node(long x, long k, int m){
			this.x = x; this.k = k; this.m = m;
			this.t = 0;
		}
		public int compareTo(Node o) {
			return Long.compare(k, o.k);
		}		
	}
	int n, m;long b;
	long[] dp = new long[1<<N];
	Node[] o = new Node[105];
	void input(){
		n = cin.nextInt(); m = cin.nextInt(); b = cin.nextInt();
		o[0] = new Node(0,0,0);
		for(int i = 1; i <= n; i++){
			o[i] = new Node(cin.nextInt(), cin.nextInt(), cin.nextInt());
			for(int j = 0; j < o[i].m; j ++)
				o[i].t |= (1<<(cin.nextInt()-1));
		}
	}
	
	void work()	{
		input();
		Arrays.sort(o, 1, n+1);
		for(int i = 0; i < (1<<m); i++)dp[i] = inf64;
		dp[0] = 0; 
		long ans = inf64;
		for(int i = 1; i <= n; i++){
			for(int j = 0; j < (1<<m); j++){
				if(dp[j] == inf)continue;
				int now = j|o[i].t;
				dp[now] = min(dp[now], dp[j] + o[i].x);
			}
			ans = min(ans, dp[(1<<m)-1]+o[i].k*b);
		}
		if(ans>=inf64)
			out.println("-1");
		else
			out.println(ans);
	}
	
	Main() {
		cin = new Scanner(System.in);
		out = new PrintWriter(System.out);
	}

	public static void main(String[] args) {
		Main e = new Main();
		e.work();
		out.close(); 
	}

	public Scanner cin;
	public static PrintWriter out;
	static int N = 20;
	static int M = 2005;
	DecimalFormat df=new DecimalFormat("0.0000");
	static int inf = (int) 1e9 + 7;
	static long inf64 = (long) 1e18*2;
	static double eps = 1e-8;
	static double Pi = Math.PI;
	static int mod = 1000000007 ;
/*
	class Edge{
		int from, to, nex;
		Edge(){}
		Edge(int from, int to, int nex){
			this.from = from;
			this.to = to;
			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){
		edge[edgenum] = new Edge(u, v, 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 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 417D Cunning Gena 状压dp

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原文地址：http://blog.csdn.net/qq574857122/article/details/43610277