Poj 3977 Subset 折半枚举 超大背包

http://poj.org/problem?id=3977

题意：从一个集合（有正有负）里挑选若干个数字（数字个数不能为0），求挑选出来的数字的和的绝对值的最小值，数字个数<=35,数据范围<=10^15

注意事项：注意这道题的数据范围和对空集的处理，且运用了离散化的思想

AC代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const long long  inf = 1e17;
struct Node{
ll v, l;
};
Node node1[3000000], node2[3000000],nod[3000000];
ll a[40];
ll absx(ll a)
{
return a > 0 ? a : -a;
}
bool cmp(Node a, Node b)
{
if(a.v!= b.v)
        return a.v<b.v;
    else return a.l<b.l;
}
ll binary(ll m, ll n)
{
ll l = 0, r = n;
while (r - l>1)
{
ll mid = (l + r) >> 1;
if (nod[mid].v >= m)
r = mid;
else l = mid;
}
return r;
}
int main()
{
ll n;
while (~scanf("%lld", &n)&&n)
{
for (ll i = 1; i <= n; i++)
scanf("%lld", &a[i]);
ll n1 = n / 2,n2=n-n1,minn = inf, mlen = 40;;
ll cnt1 = (1<<n1)-1, cnt2 = (1<<n2)-1;
for (ll i = 1; i <=cnt1; i++)
{
node1[i].v = node1[i].l = 0;
for (ll j = 0; j <=n1-1; j++)
if (i&(1<<j))
{
node1[i].v += a[j+1];
node1[i].l++;
}
            ll temp=absx(node1[i].v);
            if(minn>temp)
            {
                minn=temp;
                mlen=node1[i].l;
            }
            else if(minn==temp&&mlen>node1[i].l)
                mlen=node1[i].l;
}
for (ll i = 1; i <=cnt2; i++)
{
node2[i].v = node2[i].l = 0;
for (ll j = 0; j <=n2-1; j++)
if (i&(1<<j))
{
node2[i].v += a[n1+j+1];
node2[i].l++;
}
            ll temp=absx(node2[i].v);
            if(minn>temp)
            {
                minn=temp;
                mlen=node2[i].l;
            }
            else if(minn==temp&&mlen>node2[i].l)
                mlen=node2[i].l;
}
sort(node2 + 1,node2 +cnt2+1, cmp);
nod[1].v=node2[1].v;
nod[1].l=node2[1].l;
ll p=1;
for(ll i=2;i<=cnt2;i++)
            if(node2[i].v!=nod[p].v)
            {
                ++p;
                nod[p].v=node2[i].v;
                nod[p].l=node2[i].l;
            }
for (ll i = 1; i <=cnt1; i++)
{
ll m = node1[i].v;
ll j = binary(-m,p);
if (absx(m + nod[j].v)<minn)
{
minn = absx(m + nod[j].v);
mlen = node1[i].l + nod[j].l;
}
else if (minn ==absx(m + nod[j].v))
if (mlen>node1[i].l + nod[j].l)
mlen = node1[i].l + nod[j].l;
j--;
if(j==0) continue;
if (absx(m + nod[j].v)<minn)
{
minn = absx(m + nod[j].v);
mlen = node1[i].l + nod[j].l;
}
else if (minn == absx(m + nod[j].v))
if (mlen>node1[i].l + nod[j].l)
mlen = node1[i].l + nod[j].l;
}
printf("%lld %lld\n", minn, mlen);
}
return 0;
}

第一次wa代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
struct Node{
ll v, l;
};
Node node1[3000000], node2[3000000];
int a[40];
ll absx(ll a)
{
return a > 0 ? a : -a;
}
bool cmp(Node a, Node b)
{
return a.v < b.v;
}
ll binary(int m, int n)
{
ll l = 0, r = n;
while (r - l>1)
{
ll mid = (l + r) >> 1;
if (node2[mid].v >= m)
r = mid;
else l = mid;
}
return r;
}
int main()
{
ll n;
while (~scanf("%lld", &n)&&n)
{
for (ll i = 1; i <= n; i++)
scanf("%lld", &a[i]);
ll n1 = n / 2,n2=n-n1;
ll cnt1 = (1<<n1), cnt2 = (1<<n2);
for (ll i = 0; i <cnt1; i++)
{
node1[i].v = node1[i].l = 0;
for (ll j = 1; j < (1<<n1); j <<= 1)
if (i&j)
{
node1[i].v += a[j];
node1[i].l++;
}
}
for (ll i = 0; i < cnt2; i++)
{
node2[i].v = node2[i].l = 0;
for (ll j = 1; j <(1<<n2); j <<= 1)
if (i&j)
{
node2[i].v += a[n1+j];
node2[i].l++;
}
}
sort(node2 + 1,node2 +cnt2, cmp);
node2[0].l = node2[0].v = 0;
ll minn = inf, mlen = 40;
for (ll i = 0; i < cnt1; i++)
{
ll m = node1[i].v;
ll j = binary(-m,n2);
if (absx(m + node2[j].v)<minn)
{
minn = absx(m + node2[j].v);
mlen = node1[i].l + node2[j].l;
}
else if (minn = absx(m + node2[j].v))
if (mlen>node1[i].l + node2[j].l)
mlen = node2[i].l + node2[j].l;
j--;
if (absx(m + node2[j].v)<minn)
{
minn = absx(m + node2[j].v);
mlen = node1[i].l + node2[j].l;
}
else if (minn == absx(m + node2[j].v))
if (mlen>node1[i].l + node2[j].l)
mlen = node2[i].l + node2[j].l;
}
printf("%lld %lld\n", minn, mlen);
}
return 0;
}