1688: [Usaco2005 Open]Disease Manangement 疾病管理

Time Limit:?5 Sec??Memory Limit:?64 MB
Submit:?727??Solved:?468
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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頭牛,它們可能患有D種病,現在從這些牛中選出若干頭來,但選出來的牛患病的集合中不過超過K種病.

Output

* Line 1: M, the maximum number of cows which can be milked.

Sample Input

6 3 2
0---------第一頭牛患0種病
1 1------第二頭牛患一種病,為第一種病.
1 2
1 3
2 2 1
2 2 1

Sample Output

5

OUTPUT DETAILS:

If FJ milks cows 1, 2, 3, 5, and 6, then the milk will have only two
diseases (#1 and #2), which is no greater than K (2).

HINT

Source

Silver

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Analysis

不明不白的

和狀態壓縮DP打了場遭遇戰

= =

你好，狀壓DP！

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Code

1 /************************************************************** 2 Problem: 1688 3 User: child 4 Language: C++ 5 Result: Accepted 6 Time:140 ms 7 Memory:1456 kb 8 ****************************************************************/ 9 10 #include<cstdio> 11 #include<iostream> 12 #define maxn 10100 13 using namespace std; 14 15 int n,d,k,num,tot,sick[maxn],DP[1<<15],ret; 16 17 bool check(int code){ 18 int cnt = 0; 19 for(int i = 0;i < d;i++){ 20 if((code>>i)&1) cnt++; 21 }return (cnt<=k)?1:0; 22 } 23 24 int main(){ 25 scanf("%d%d%d",&n,&d,&k); 26 27 tot = (1<<d)-1; 28 29 for(int i = 1;i <= n;i++){ 30 scanf("%d",&num); 31 int cnt; 32 for(int j = 1;j <= num;j++){ 33 scanf("%d",&cnt); 34 sick[i] |= (1<<(cnt-1)); 35 } 36 } 37 38 for(int i = 1;i <= n;i++) 39 for(int j = tot;j >= 0;j--) 40 DP[sick[i]|j] = max(DP[sick[i]|j],DP[j]+1); 41 42 for(int i = 0;i <= tot;i++) if(check(i)) ret = max(ret,DP[i]); 43 44 printf("%d",ret); 45 46 return 0; 47 } = =

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轉載于:https://www.cnblogs.com/Chorolop/p/7569993.html

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