Disease Manangement 疾病管理
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.
Output
* Line 1: M, the maximum number of cows which can be milked.
Sample Input 1
6 3 2
0———第一头牛患0种病
1 1——第二头牛患一种病,为第一种病.
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]
只是稍微接触过一点状压dp，没有一点思路。
首先预处理出 num[i] 表示状态i有几种病，用a[i]存每头牛的状态。
然后dp[i] 表示状态为i的最多牛数
那么转移方程为 dp[s] = max(dp[s] , dp[S] + 1); 其中s=a[i] | S;
#include<bits/stdc++.h>
using namespace std;
const int maxn=; const int M=<<;
int a[maxn],num[M],dp[M];
int main() {
for(int i=;i<M;i++) num[i]=num[i>>]+(i&);
int n,m,k;
while(~scanf("%d%d%d",&n,&m,&k)) {
memset(a,,sizeof(a));
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++) {
int x;
scanf("%d",&x);
while(x--) {
int val;
scanf("%d",&val);
a[i]|=<<(val-);//将第val位变为1
}
}
for(int i=;i<=n;i++) {
for(int S=(<<m)-;S>=;S--) {//枚举状态
int s=a[i]|S;//由S可以到达的状态s
if(num[s]>k) continue;
dp[s]=max(dp[s],dp[S]+);//那么s壮态数量 就可能更新
}
}
int ans=-;
for(int i=;i<M;i++) {
if(num[i]>k) continue;
ans=max(ans,dp[i]);
}
printf("%d\n",ans);
}
}
#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);//将第val位变为1 } } for(int i=1;i<=n;i++) { for(int S=(1<<m)-1;S>=0;S–) {//枚举状态 int s=a[i]|S;//由S可以到达的状态s if(num[s]>k) continue; dp[s]=max(dp[s],dp[S]+1);//那么s壮态数量 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); }}
