hdu 6006 Engineer Assignment(状压DP)

Engineer Assignment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 456    Accepted Submission(s): 149

Problem Description

In Google, there are many experts of different areas. For example, MapReduce experts, Bigtable experts, SQL experts, etc. Directors need to properly assign experts to various projects in order to make the projects going smoothly.
There are N projects owned by a director. For the ith project, it needs Ci different areas of experts, ai,0,ai,1,⋅⋅⋅,ai,Ci−1 respective. There are M engineers reporting to the director. For the ith engineer, he is an expert of Di different areas, bi,0,bi,1,...,bi,Di−1.
Each engineer can only be assigned to one project and the director can assign several engineers to a project. A project can only be finished successfully if the engineers expert areas covers the project areas, which means, for each necessary area of the project, there is at least one engineer
masters it.
The director wants to know how many projects can be successfully finished.

 

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line consisting of 2 integers, N the number of projects and M the number of engineers. Then N lines follow. The ith line containing the information of the ith project starts
with an integer Ci then Ci integers follow, ai,0,ai,1,...,ai,Ci−1 representing the expert areas needed for the ith project. Then another M lines follow. The ithline containing the information of the ith engineer starts with an integer Di then Di integers follow, bi,0,bi,1,...,bi,Di−1 representing the expert areas mastered by ith engineer.

 

Output

For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the maximum number of projects can be successfully finished.

limits

∙1≤T≤100.
∙1≤N,M≤10.
∙1≤Ci≤3.
∙1≤Di≤2.
∙1≤ai,j,bi,j≤100.

 

Sample Input

13 43 40 77 643 10 40 203 40 20 772 40 772 77 642 40 102 20 77

 

Sample Output

Case #1: 2
Hint
For the first test case, there are 3 projects and 4 engineers. One of the optimal solution is to assign the first(40 77) and second engineer(77 64) to project 1, which could cover the necessary areas 40, 77, 64. Assign the third(40 10) and forth(20 77) engineer to project 2, which could cover the necessary areas 10, 40, 20. There are other solutions, but none of them can finish all 3 projects.So the answer is 2.

题意：一共有n个任务，完成某个任务需要会一些领域的人，一共有m个工程师，每个工程师会一些领域，问这些工程师最多完成多少任务
题解：先预处理出能第i个任务可以由哪些人一起完成，因为人数不多，所以可以用状态压缩，然后dp[i][j]来表示表前i个任务，状态为j时，最多完成的任务数

一开始用一维表状态一直wa   改成二维 第一维是前i个工程的最大能完成数目的最优状态  当考虑第i+1个时 可能选或者不选

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>#include <bits/stdc++.h>using namespace std;const int N = 120;typedef long long LL;int vis[N];int  dp[20][1<<15], a[20][30], b[20][30];void init(){    memset(dp,0,sizeof(dp));    return ;}vector<int>p[N];int main(){    //cout<<(1<<30)<<endl;    int t, ncase=1;    scanf("%d", &t);    while(t--)    {        int n,m;        scanf("%d %d", &n, &m);        for(int i=0; i<=n; i++) p[i].clear();        init();        int  x, y;        for(int i=1; i<=n; i++)        {            scanf("%d", &x);            a[i][0]=x;            for(int j=1; j<=x; j++)  scanf("%d", &a[i][j]);        }        for(int i=0; i<m; i++)        {            scanf("%d", &x);            b[i][0]=x;            for(int j=1; j<=x; j++)  scanf("%d", &b[i][j]);        }        for(int j=1; j<=n; j++)        {            for(int i=0; i<(1<<m); i++)            {                memset(vis,0,sizeof(vis));                int cnt=0;                for(int j=0; j<m; j++)                {                    if(i&(1<<j))                    {                        cnt++;                        for(int k=1; k<=b[j][0]; k++)                        {                            vis[b[j][k]]=1;                        }                    }                }                int flag=0;                for(int k=1; k<=a[j][0]; k++)                {                    if(!vis[a[j][k]])                    {                        flag=1;                        break;                    }                }                if(flag==0) p[j].push_back(i);            }        }        int ans=0;        for(int i=1; i<=n; i++)        {            for(int j=0; j<(1<<m); j++)            {                for(int k=0; k<p[i].size(); k++)                {                    if((j|p[i][k])==j)dp[i][j]=max(dp[i][j],dp[i-1][j-p[i][k]]+1);                }                dp[i][j]=max(dp[i][j],dp[i-1][j]);                ans=max(ans,dp[i][j]);            }        }        printf("Case #%d: %d\n",ncase++, ans);    }    return 0;}