HDU 4277

最新推荐文章于 2017-09-07 23:57:42 发布

Benzema67

最新推荐文章于 2017-09-07 23:57:42 发布

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分类专栏： STL 文章标签： struct ini im

本文链接：https://blog.youkuaiyun.com/waitfor_/article/details/7968287

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这个乱搞题

用hash+set搞

或着DFS即可

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<set>
#define N 32768
using namespace std;
int a[20];
int num[N+100];
struct Point{
    int sum,id;
}p[N+100];
set<unsigned long long>s;
bool cmp(struct Point a,struct Point b){
    return a.sum<b.sum;
}
int main(){
    int t,T,i,j,n,m;
    int sum;
    scanf("%d",&T);
    for(t=1;t<=T;t++){
        scanf("%d",&n);
        sum=0;
        for(i=0;i<n;i++){
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        memset(p,0,sizeof(p));
        m=1<<n;
        for(i=1;i<m;i++){
            for(j=0;j<n;j++)
                if(i&(1<<j))
                    p[i].sum+=a[j];
            p[i].id=i;
        }
        sort(p+1,p+m,cmp);
        int ans=0;
        s.clear();
        int p1=0,p2=0;
        for(i=1;i<m && p[i].sum<=sum/3;i++){
            if(p[i].sum!=p1){
                ans+=s.size();
                p1=p[i].sum;
                s.clear();
            }
            for(j=i+1;j<m && p[j].sum<=sum/2;j++){
                if((p[i].id&p[j].id)!=0)continue;
                if(sum-(p[i].sum+p[j].sum)<p[j].sum)continue;
                if(p[i].sum+p[j].sum>sum-(p[i].sum+p[j].sum)){
                    unsigned long long tem=(unsigned long long)p[j].sum+(unsigned long long)160000*p[j].sum+(unsigned long long)160000*160000*(sum-(p[i].sum+p[j].sum));
                    s.insert(tem);
                }
            }
        }
        ans+=s.size();
        printf("%d\n",ans);
    }
    return 0;
}

但是直接枚举1到3^15是不行的

#include"iostream"
#include"vector"
#include"string"
#include"cstdio"
#include"cstdlib"
#include"cmath"
#include"algorithm"
#include"queue"
#include"cstring"
#include"map"
#include"set"
#include"fstream"
#include"sstream"
#include"numeric"
#include"stack"
#include"iomanip"
#include"bitset"
#include"list"
#include"functional"
#include"utility"
#include"ctime"
typedef long long ll;
#define sz(a) sizeof(a)
#define mp make_pair
#define pb push_back
#define ms(a,i) memset((a),(i),sz(a))
const double eps=1e-5;
const int inf=0x7fffffff;
const int mod=1000000007;

using namespace std;

set<pair<int,pair<int,int> > > S;
int n;
int dat[16];
int sum[3];
int three[16];
int main()
{
    int t;
    scanf("%d",&t);
    three[0]=1;
    for(int i=1;i<=15;i++)
        three[i]=3*three[i-1];
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&dat[i]);
        S.clear();
        ms(sum,0);
        for(int i=0;i<three[n];i++)
        {
            ms(sum,0);
            int ii=i;
            for(int j=0;j<n;j++)
            {
                sum[ii%3]+=dat[j];
                ii/=3;
            }
            sort(sum,sum+3);
            if(sum[0]+sum[1]>sum[2])
                S.insert(mp(sum[0], mp(sum[1],sum[2])));
        }
        printf("%d\n",S.size());
    }
}

写成DFS形式复杂度也是3^15，但是就不用每个状态乘以常数15了

#include<cstdio>
#include<cstring>
#include<set>
#include<algorithm>
using namespace std;

int s[20],res,cas,sum[3];
int n;
set<unsigned long long>S;
void dfs(int x){
    int i;
    if(x==n){
        if(sum[0]>sum[1] || sum[1]>sum[2]) return ;
        if(sum[0]+sum[1]>sum[2]){
            unsigned long long tem=(unsigned long long)sum[0]+(unsigned long long)160000*sum[1]+(unsigned long long)160000*160000*sum[2];
            S.insert(tem);
        }
        return ;
    }
    for(i=0;i<3;i++){
        sum[i]+=s[x];
        dfs(x+1);
        sum[i]-=s[x];
    }
}

int main(){
    int i,t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        memset(has,0,sizeof(has));
        for(i=0;i<n;i++)
            scanf("%d",&s[i]);
        res=0;
        sum[0]=sum[1]=sum[2]=0;
        S.clear();
        dfs(0);
        printf("%d\n",S.size());

    }
    return 0;
}