HDU 4152 ZZYs Dilemma（枚举 or dfs）

ZZY’s Dilemma

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 648    Accepted Submission(s): 276

Problem Description

ZZY has many habits like seeing movie, listening music, playing PC game and football and so on. Besides he has many goals, as not every habit is good for achieving his goals, he falls into the dilemma of making a choice between his goals and habits.
Now ,we define the effect that one habit has on each goal is represented as a vector,and the elements of the vector are integers,ex.vector(100,90,-10,80)means it has 100 point effect on goal 1,90 point effect on goal 2,-10 point effect on goal 3 and 80 point effect on goal 4(the positive point means good effect and the negative point means bad effect),and the given requirement of each goal is represented as integer.please help ZZY to achieve his goals as well as keeps his habits as many as possible.

 

Input

There are multi cases , read the data until EOF.( No more than 10 cases )
Line 1: The number of ZZY’s goals N(0<N<=20)
Line 2: The requirement of each goals (0 < w <= 1000)
Line 3: The number of ZZY’s habits M(0 < M <= 16)
Line 4-M+4: Each line contains N integers, the i^th integer represents the effect on i^th goal (-1000 <= data <= 1000).

 

Output

For each case: The output is a single line that contains:
* the maximum number of habits ZZY can keep, followed by:
* a SORTED list (from smallest to largest) of the habit ZZY can keep. If more than one set of habits could meet the requirement, choose the set with the smallest habit numbers.
Just please output 0 if there is no way to achieve ZZY’s goals.

 

Sample Input

4100 200 300 4003100 100 400 500100 -10 50  300100 100 -50  -50

 

Sample Output

2 1 3

 

Author

ZZY@USC

题解：总共有M个习惯，如果用0 1两种状态表明选还是不选，那么共有2^m种情况（2^16=65536）,时间复杂度为O（(2^m )*m*N）

AC代码：

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include<algorithm>typedef long long LL;using namespace std;int goal[30]; //表示N个目标要达到的分数值 int f[25][25];int g[30];int main(){int n,m,ans1,ans2;while(~scanf("%d",&n)){ans1=-1;for(int i=0;i<n;i++){cin>>goal[i];}scanf("%d",&m);for(int i=0;i<m;i++)  for(int j=0;j<n;j++)    scanf("%d",&f[i][j]);        for(int i=1;i<(1<<m);i++){    //利用二进制枚举每个习惯要还是不要     memset(g,0,sizeof(g));    int num=0;        for(int j=0;j<m;j++)       if(i&(1<<j)){          //如果要j这个习惯        num++;     for(int k=0;k<n;k++)        g[k]+=f[j][k];        }              int k1;  for(k1=0;k1<n;k1++)       //判断是否每个目标都达到了给定的分数值     if(g[k1]<goal[k1]) break;    if(k1>=n){             //更新答案     if(num>ans1) ans1=num,ans2=i;    else if(num==ans1&&ans2>i)    ans1=num,ans2=i;    }    }    if(ans1==-1)puts("0");else{printf("%d",ans1);for(int i=0;i<n;i++)  if(ans2&(1<<i))    printf(" %d",i+1);printf("\n"); }  }     return 0;}

 DFS：

#include <cstdio>#include <cstring>#define M 25int a[M],b[M],mp[M][M],n,m,vis[M],ans;bool judge(){for(int i=0;i<n;i++){int sum=0;for(int j=0;j<m;j++){if(vis[j])sum+=mp[j][i];}if(sum<a[i])return false;}return true;}void dfs(int pos){if(pos==m){if(judge()){int tot=0;for(int i=0;i<m;i++)if(vis[i])tot++;if(ans<tot){ans=tot;for(int i=0;i<m;i++)b[i]=vis[i];}}return;}vis[pos]=1;dfs(pos+1);vis[pos]=0;dfs(pos+1);}int main(void){while(~scanf("%d",&n)){for(int i=0;i<n;i++)scanf("%d",&a[i]);scanf("%d",&m);for(int i=0;i<m;i++)for(int j=0;j<n;j++)scanf("%d",&mp[i][j]);memset(vis,0,sizeof(vis));memset(b,0,sizeof(b));ans=0;dfs(0);printf("%d",ans);for(int i=0;i<m;i++)if(b[i])printf(" %d",i+1);puts("");}return 0;}