1688: [Usaco2005 Open]Disease Manangement 疾病管理( 枚举 )

最新推荐文章于 2017-02-13 18:35:19 发布

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最新推荐文章于 2017-02-13 18:35:19 发布

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原文链接：http://www.cnblogs.com/JSZX11556/p/4652537.html

本文介绍了一道USACO竞赛题目“疾病管理”的解题思路与代码实现。目标是在不超过K种疾病的限制下，尽可能多地选择可以挤奶的牛。通过状态压缩和枚举策略，最终求解出最大可挤奶的牛的数量。

我一开始写了个状压dp..然后没有滚动就MLE了...

其实这道题直接暴力就行了... 2^15枚举每个状态, 然后检查每头牛是否能被选中, 这样是O( 2^15*1000 ), 也是和dp一样的时间复杂度....有点贪心的feel

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#include<cstdio>

#include<algorithm>

#include<iostream>

#include<cstring>

#define rep(i, n) for(int i = 0; i < n; ++i)

#define clr(x, c) memset(x, c, sizeof(x))

#define b(i) (1 << (i))

using namespace std;

const int maxn = 1005, maxd = 16;

int cow[maxn], lim, n, d;

bool jud(int s) {

int tot = 0;

rep(i, d) if(b(i) & s) tot++;

return tot <= lim;

}

int main() {

freopen("test.in", "r", stdin);

cin >> n >> d >> lim;

rep(i, n) {

cow[i] = 0;

int t, v;

scanf("%d", &t);

while(t--) {

scanf("%d", &v);

cow[i] |= b(v - 1);

}

int tot, ans = 0;

rep(s, b(d)) if(jud(s)) {

tot = 0;

rep(i, n) if((s | cow[i]) == s) tot++;

ans = max(tot, ans);

}

cout << ans << "\n";

return 0;

}

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1688: [Usaco2005 Open]Disease Manangement 疾病管理

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 423 Solved: 281
[ Submit][ Status][ Discuss]

Description

Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) different diseases among them, then the milk will be too contaminated and will have to be discarded in its entirety. Please help determine the largest number of cows FJ can milk without having to discard the milk.

Input

* Line 1: Three space-separated integers: N, D, and K * Lines 2..N+1: Line i+1 describes the diseases of cow i with a list of 1 or more space-separated integers. The first integer, d_i, is the count of cow i's diseases; the next d_i integers enumerate the actual diseases. Of course, the list is empty if d_i is 0. 有N头牛,它们可能患有D种病,现在从这些牛中选出若干头来,但选出来的牛患病的集合中不过超过K种病.