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POJ - 3977 Subset(二分+折半枚举)

题意:有一个N(N <= 35)个数的集合,每个数的绝对值小于等于1015,找一个非空子集,使该子集中所有元素的和的绝对值最小,若有多个,则输出个数最小的那个。

分析:

1、将集合中的元素分成两半,分别二进制枚举子集并记录子集所对应的和以及元素个数。

2、枚举其中一半,二分查找另一半,不断取最小值。

#pragma comment(linker, "/STACK:102400000, 102400000")#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define Min(a, b) ((a < b) ? a : b)#define Max(a, b) ((a < b) ? b : a)const double eps = 1e-10;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 35 + 10;const int MAXT = 100000 + 10;using namespace std;LL a[MAXN];map<LL, LL> mp1;map<LL, LL> mp2;vector<LL> v1;vector<LL> v2;LL ans, num;LL Abs(LL x){    return x >= 0 ? x : -x;}void init(){    mp1.clear();    mp2.clear();    v1.clear();    v2.clear();}void solve(int l, int r, map<LL, LL> &mp, vector<LL> &v){    int n = r - l + 1;    for(int i = 1; i < (1 << n); ++i){        LL sum = 0;        LL cnt = 0;        for(int j = 0; j < n; ++j){            if(i & (1 << j)){                sum += a[l + j];                ++cnt;            }        }        if(mp.count(sum))            mp[sum] = Min(mp[sum], cnt);        else            mp[sum] = cnt;    }    for(map<LL, LL>::iterator it = mp.begin(); it != mp.end(); ++it){        v.push_back((*it).first);        if(Abs((*it).first) < ans){            ans = Abs((*it).first);            num = (*it).second;        }        else if(ans == Abs((*it).first)){            num = Min(num, (*it).second);        }    }}void judge(LL x){    int l = 0, r = v2.size() - 1;    while(l <= r){        int mid = l + (r - l) / 2;        if(Abs(x + v2[mid]) < ans){            ans = Abs(x + v2[mid]);            num = mp1[x] + mp2[v2[mid]];        }        else if(Abs(x + v2[mid]) == ans){            num = Min(num, mp1[x] + mp2[v2[mid]]);        }        if(x + v2[mid] < 0) l = mid + 1;        else r = mid - 1;    }}int main(){    int N;    while(scanf("%d", &N) == 1){        if(!N) return 0;        init();        for(int i = 0; i < N; ++i){            scanf("%lld", &a[i]);        }        if(N == 1){            printf("%lld 1\n", Abs(a[0]));            continue;        }        ans = LL_INF, num = LL_INF;        solve(0, N / 2 - 1, mp1, v1);        solve(N / 2, N - 1, mp2, v2);        int len = v1.size();        sort(v2.begin(), v2.end());        for(int i = 0; i < len; ++i){            judge(v1[i]);        }        printf("%lld %lld\n", ans, num);    }    return 0;}

  

POJ - 3977 Subset(二分+折半枚举)