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幹頭來,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 1

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 1

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).

Source

[BZOJ1688][Usaco2005 Open]

只是稍微接觸過一點狀壓dp，沒有一點思路。

首先預處理出 num[i] 表示狀態i有幾種病，用a[i]存每頭牛的狀態。

然後dp[i] 表示狀態為i的最多牛數

那麽轉移方程為 dp[s] = max(dp[s] , dp[S] + 1); 其中s=a[i] | S;

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1010;
 4 
 5 const int M=1<<15;
 6 int a[maxn],num[M],dp[M];
 7 int main() {
 8     for(int i=0;i<M;i++) num[i]=num[i>>1]+(i&1);
 9     int n,m,k;
10     while(~scanf("%d%d%d",&n,&m,&k)) {
11         memset(a,0,sizeof(a));
12         memset(dp,0,sizeof(dp));
13         for(int i=1;i<=n;i++) {
14             int x;
15             scanf("%d",&x);
16             while(x--) {
17                 int val;
18                 scanf("%d",&val);
19                 a[i]|=1<<(val-1);//將第val位變為1
20             }
21         }
22         for(int i=1;i<=n;i++) {
23             for(int S=(1<<m)-1;S>=0;S--) {//枚舉狀態
24                 int s=a[i]|S;//由S可以到達的狀態s
25                 if(num[s]>k) continue;
26                 dp[s]=max(dp[s],dp[S]+1);//那麽s壯態數量 就可能更新
27             }
28         }
29         int ans=-1;
30         for(int i=0;i<M;i++) {
31             if(num[i]>k) continue;
32             ans=max(ans,dp[i]);
33         }
34         printf("%d\n",ans);
35     }
36 }

#include<bits/stdc++.h>using namespace std;const int maxn=1010;
const int M=1<<15;int a[maxn],num[M],dp[M];int main() { for(int i=0;i<M;i++) num[i]=num[i>>1]+(i&1); int n,m,k; while(~scanf("%d%d%d",&n,&m,&k)) { memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { int x; scanf("%d",&x); while(x--) { int val; scanf("%d",&val); a[i]|=1<<(val-1);//將第val位變為1 } } for(int i=1;i<=n;i++) { for(int S=(1<<m)-1;S>=0;S--) {//枚舉狀態 int s=a[i]|S;//由S可以到達的狀態s if(num[s]>k) continue; dp[s]=max(dp[s],dp[S]+1);//那麽s壯態數量 就可能更新 } } int ans=-1; for(int i=0;i<M;i++) { if(num[i]>k) continue; ans=max(ans,dp[i]); } printf("%d\n",ans); }}

BZOJ1688 Disease Manangement 疾病管理