[luogu3067 USACO12OPEN] 平衡的奶牛群

標籤：string   https   open   algo   type   ++   namespace   mem   cstring   

傳送門

Solution

折半搜尋模板題
考慮枚舉每個點在左集合和右集合或者不在集合中，然後排序合并即可

Code

//By Menteur_Hxy#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define Re register#define Ms(a,b) memset(a,(b),sizeof(a))#define Fo(i,a,b) for(Re int i=(a),_=(b);i<=_;i++)#define Ro(i,a,b) for(Re int i=(b),_=(a);i>=_;i--)#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<16,stdin)),p1==p2?EOF:*p1++)using namespace std;typedef long long LL;char buf[1<<16],*p1,*p2;inline int read() {    int x=0,f=1; char c=getchar();    while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}    while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();    return x*f;}inline void writ(int x) {    if(x>9) writ(x/10);    putchar(x%10+'0');}const int N=25,M=1e8+5;int n,ans,tt1,tt2;bool vis[1<<N];int da[N];struct Data{int val,cur;}T1[1<<N],T2[1<<N];void dfs(int pos,int ed,int sum,int S) {    if(pos>ed) {        if(ed==n/2) T1[++tt1].val=sum,T1[tt1].cur=S;        else T2[++tt2].val=sum,T2[tt2].cur=S;//1        return ;    }    dfs(pos+1,ed,sum,S);    dfs(pos+1,ed,sum+da[pos],S|(1<<(pos-1)));    dfs(pos+1,ed,sum-da[pos],S|(1<<(pos-1)));}bool cmp1(Data a,Data b) {return a.val<b.val;}bool cmp2(Data a,Data b) {return a.val>b.val;}int main() {    n=read();    Fo(i,1,n) da[i]=read();    dfs(1,n/2,0,0); dfs(n/2+1,n,0,0);    sort(T1+1,T1+1+tt1,cmp1);    sort(T2+1,T2+1+tt2,cmp2);    int l=1,r=1,las;    while(l<=tt1&&r<=tt2) {        while(r<=tt2&&T1[l].val+T2[r].val>0) r++;//2        las=r;        while(r<=tt2&&T1[l].val+T2[r].val==0) {            if(!vis[T1[l].cur|T2[r].cur])                vis[T1[l].cur|T2[r].cur]++,ans++;            r++;        }        l++;        if(T1[l-1].val==T1[l].val) r=las;//3    }    writ(ans-1);    return 0;}

[luogu3067 USACO12OPEN] 平衡的奶牛群