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POJ 3977 折半搜索

2024-11-03 14:29:02   205人阅读

Subset
Time Limit: 30000MS   Memory Limit: 65536K
Total Submissions: 4128   Accepted: 796

Description

Given a list of N integers with absolute values no larger than 1015, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.

Input

The input contains multiple data sets, the first line of each data set contains N <= 35, the number of elements, the next line contains N numbers no larger than 1015 in absolute value and separated by a single space. The input is terminated with N = 0

Output

For each data set in the input print two integers, the minimum absolute sum and the number of elements in the optimal subset.

Sample Input

1
10
3
20 100 -100
0

Sample Output

10 1
0 2

Source

Seventh ACM Egyptian National Programming Contest

题意： 给n（n<=35）个数，求一个它的子集，使其中元素和的绝对值最小。

思路：分成两半来枚举。
代码：
#include "cstdio"
#include "stdlib.h"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "set"
#define db double
#define inf 0x3f3f3f
#define mj
typedef long long ll;
using  namespace std;
const int N=4000*4000+5;
const int maxN=40;
typedef pair<ll,int> pii;
ll ab(ll x){
    return x>0?x:-x;
}
ll a[maxN];
int main()
{
    int n;
    while(cin>>n&&n){
        for(int i=0;i<n;++i) cin>>a[i];
        map<ll,int> m;
        pii res(ab(a[0]),1);
        for(int i=0;i<1<<(n/2);++i){
            ll sum=0;
            int num=0;
            for(int j=0;j<n/2;++j)
                if((i>>j)&1){
                    sum+=a[j];
                    ++num;
                }
            if(!num) continue;
            res=min(res,make_pair(ab(sum),num));//取最小值,优先级从前到后
            map<ll,int>::iterator iter=m.find(sum);
            if(iter!=m.end())
                iter->second=min(iter->second,num);
            else
                m[sum]=num;
        }
        for(int i=0;i<1<<(n-n/2);++i){
            ll sum=0;
            int num=0;
            for(int j=0;j<(n-n/2);++j)
                if((i>>j)&1){
                    sum+=a[n/2+j];
                    ++num;
                }
            if(!num) continue;
            res=min(res,make_pair(ab(sum),num));
            map<ll,int>::iterator iter=m.lower_bound(-sum);
            if(iter!=m.end())
                res=min(res,make_pair(ab(sum+iter->first),num+iter->second));
            if(iter!=m.begin()){
                --iter;
                res=min(res,make_pair(ab(sum+iter->first),num+iter->second));
            }
        }
        printf("%lld %d\n",res.first,res.second);
    }
    return 0;
}

 

POJ 3977 折半搜索


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