状压dp

//Poj    3254

 

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20258		Accepted: 10633

Description

Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant.

Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant.

Input

Line 1: Two space-separated integers: M and N 
Lines 2..M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)

Output

Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.

Sample Input

2 3
1 1 1
0 1 0

Sample Output

9

Hint

Number the squares as follows:

1 2 3
  4  

There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.

Source

USACO 2006 November Gold

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <utility>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#define max(x,y) x>=y?x:y
#define lowbit(x) x&(-x)
#define ll long long 
const int mod=1e8;
int  m,n,cnt;
int r[20],dp[20][400],a[20];
int state[400];
bool  check1(int x)
{
    return  !(x&(x<<1));//x的二进制有没有相邻的1
}
bool check2(int x,int k)
{
return      !((x&(a[k])));//x和第k+1行的逆有没有重合
/*
如  k+1行为 0 1 0 1 
            1 0 1 0
        x :    1 0 0 1 虽然x满足没有相邻的1，但是他的第一个1的位置为0(不能放)  
*/
}
void init()
{
     cnt=0;
    for(int i=0;i<(1<<n);i++){
        if(check1(i))  state[cnt++]=i;//找出=所有的可行状态
    }
}
int  main()
{
    scanf("%d%d",&m,&n);
    init();
    int x;
    for(int i=0;i<m;i++){
        for(int  j=0;j<n;j++){
            scanf("%d",&x);
            if(!x){
                a[i]+=(1<<(n-j-1));//逆
            }
        }
        
    }
    for(int i=0;i<cnt;i++){
        if(check2(state[i],0))  dp[0][i]=1;//初始化
    }
    for(int  i=1;i<m;i++){//要从1开始
        for(int  j=0;j<cnt;j++){
            if(!check2(state[j],i))  continue;
            for(int k=0;k<cnt;k++){
                if(!check2(state[k],i-1)) continue;
                if(state[j]&state[k]) continue;//上下不能有相邻的1
                dp[i][j]=(dp[i][j]+dp[i-1][k])%mod;
            }
        }
    }
    int  ans=0;
    for(int  i=0;i<cnt;i++){
        ans=(ans+dp[m-1][i])%mod;
    }
    printf("%d\n",ans);
    return  0;
}

 

 

ACM-ICPC 2018 南京赛区网络预赛    E. AC Challenge

•  128536K

 

Dlsj is competing in a contest with n (0 < n \le 20)n(0<n≤20) problems. And he knows the answer of all of these problems.

However, he can submit ii-th problem if and only if he has submitted (and passed, of course) s_isi​ problems, the p_{i, 1}pi,1​-th, p_{i, 2}pi,2​-th, ......, p_{i, s_i}pi,si​​-th problem before.(0 < p_{i, j} \le n,0 < j \le s_i,0 < i \le n)(0<pi,j​≤n,0<j≤si​,0<i≤n) After the submit of a problem, he has to wait for one minute, or cooling down time to submit another problem. As soon as the cooling down phase ended, he will submit his solution (and get "Accepted" of course) for the next problem he selected to solve or he will say that the contest is too easy and leave the arena.

"I wonder if I can leave the contest arena when the problems are too easy for me."
"No problem."
—— CCF NOI Problem set

If he submits and passes the ii-th problem on tt-th minute(or the tt-th problem he solve is problem ii), he can get t \times a_i + b_it×ai​+bi​ points. (|a_i|, |b_i| \le 10^9)(∣ai​∣,∣bi​∣≤109).

Your task is to calculate the maximum number of points he can get in the contest.

Input

The first line of input contains an integer, nn, which is the number of problems.

Then follows nn lines, the ii-th line contains s_i + 3si​+3 integers, a_i,b_i,s_i,p_1,p_2,...,p_{s_i}ai​,bi​,si​,p1​,p2​,...,psi​​as described in the description above.

Output

Output one line with one integer, the maximum number of points he can get in the contest.

Hint

In the first sample.

On the first minute, Dlsj submitted the first problem, and get 1 \times 5 + 6 = 111×5+6=11 points.

On the second minute, Dlsj submitted the second problem, and get 2 \times 4 + 5 = 132×4+5=13 points.

On the third minute, Dlsj submitted the third problem, and get 3 \times 3 + 4 = 133×3+4=13 points.

On the forth minute, Dlsj submitted the forth problem, and get 4 \times 2 + 3 = 114×2+3=11 points.

On the fifth minute, Dlsj submitted the fifth problem, and get 5 \times 1 + 2 = 75×1+2=7 points.

So he can get 11+13+13+11+7=5511+13+13+11+7=55 points in total.

In the second sample, you should note that he doesn't have to solve all the problems.

样例输入1复制

5
5 6 0
4 5 1 1
3 4 1 2
2 3 1 3
1 2 1 4

样例输出1复制

55

样例输入2复制

1
-100 0 0

样例输出2复制

0

题目来源

ACM-ICPC 2018 南京赛区网络预赛

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <utility>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#define max(x,y) x>=y?x:y
#define lowbit(x) x&(-x)
#define ll long long 
const  ll inf=9e18;
#define   N  50
#define   M 1<<21
int n,k,pre[N];
ll a[N],b[N],dp[M];//dp的大小
int  main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld%lld%d",&a[i],&b[i],&k);
        int x;
        for(int j=0;j<k;j++)
        {
            scanf("%d",&x);
            pre[i]+=(1<<(x-1));
            //在提交第i个之前要完成的
            //如 提交第1个之前要完成的 2 3
            //那么 pre[1]=110
        }
    }
    int cnt=(1<<n);
    for(int i=1;i<cnt;i++ ) dp[i]=-inf;
    dp[0]=0ll;
    ll ans=0;//题意最小0
    //dp[i] :按照i的方法进行可以获得的最大价值
    for(int i=1;i<cnt;i++)//枚举所有情况
        //0 1 1 就包含了1 2|2 1
    {
        for(int j=0;j<n;j++)
            
            {
                if(i>>j&1){
                
                    int u=i-(1<<j);//提交第j+1个之前已经做的
                    if((u&pre[j+1])==pre[j+1]){//把提交第j+1个之前要做的都做了。要加(),==高于&
                        
                        ll val=dp[u]+__builtin_popcount(i)*a[j+1]+b[j+1];
                        dp[i]=max(dp[i],val);
                        ans=max(ans,dp[i]);
                    }
                }
            }
    }
    printf("%lld\n",ans);
    return 0;
}