【bzoj3886】[Usaco2015 Jan]Moovie Mooving 状态压缩dp+二分

题目描述

Bessie is out at the movies. Being mischievous as always, she has decided to hide from Farmer John for L (1 <= L <= 100,000,000) minutes, during which time she wants to watch movies continuously. She has N (1 <= N <= 20) movies to choose from, each of which has a certain duration and a set of showtimes during the day. Bessie may enter and exit a movie at any time during one if its showtimes, but she does not want to ever visit the same movie twice, and she cannot switch to another showtime of the same movie that overlaps the current showtime. Help Bessie by determining if it is possible for her to achieve her goal of watching movies continuously from time 0 through time L. If it is, determine the minimum number of movies she needs to see to achieve this goal (Bessie gets confused with plot lines if she watches too many movies).
PoPoQQQ要在电影院里呆L分钟，这段时间他要看小型电影度过。电影一共N部，每部都播放于若干段可能重叠的区间，PoPoQQQ决不会看同一部电影两次。现在问他要看最少几部电影才能度过这段时间？ 注：必须看电影才能在电影院里呆着，同时一场电影可以在其播放区间内任意时间入场出场。

输入

The first line of input contains N and L. The next N lines each describe a movie. They begin with its integer duration, D (1 <= D <= L) and the number of showtimes, C (1 <= C <= 1000). The remaining C integers on the same line are each in the range 0..L, and give the starting time of one of the showings of the movie. Showtimes are distinct, in the range 0..L, and given in increasing order. 

输出

A single integer indicating the minimum number of movies that Bessie

needs to see to achieve her goal.  If this is impossible output -1

instead.

样例输入

4 100
50 3 15 30 55
40 2 0 65
30 2 20 90
20 1 0

样例输出

3

---

题解

状态压缩dp

设f[i]为状态i时最多能持续的时间。

那么就有f[i] = max(f[i] , a[j][k] + t[j]) (i&(1<<(j-1))==1,k为第一个使得a[j][k]>=f[i ^ (1 << (j - 1))]的数)。

于是二分查找即可，时间复杂度O(2^n*n*logd)，看似很大，而亲测可过。

#include <cstdio>
#include <algorithm>
using namespace std;
int f[1050000] , t[21] , d[21] , a[21][1001] , num[1050000];
int search(int p , int x)
{
	int l = 1 , r = d[p] , mid , ans = -1;
	while(l <= r)
	{
		mid = (l + r) >> 1;
		if(a[p][mid] <= x)
			ans = mid , l = mid + 1;
		else r = mid - 1;
	}
	return ans;
}
int main()
{
	int n , m , i , j , k , ans = 0x7fffffff;
	scanf("%d%d" , &n , &m);
	for(i = 1 ; i <= n ; i ++ )
	{
		scanf("%d%d" , &t[i] , &d[i]);
		for(j = 1 ; j <= d[i] ; j ++ )
			scanf("%d" , &a[i][j]);
	}
	for(i = 1 ; i < 1 << n ; i ++ )
	{
		for(j = 1 ; j <= n ; j ++ )
		{
			if(i & (1 << (j - 1)))
			{
				k = search(j , f[i ^ (1 << (j - 1))]);
				if(k != -1)
					f[i] = max(f[i] , a[j][k] + t[j]);
			}
		}
	}
	for(i = 1 ; i < 1 << n ; i ++ )
	{
		num[i] = num[i - (i & (-i))] + 1;
		if(f[i] >= m)
			ans = min(ans , num[i]);
	}
	printf("%d\n" , ans == 0x7fffffff ? -1 : ans);
	return 0;
}