Professor GukiZ and Two Arrays CodeForces
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传送门:CodeForces - 620D
题意:给定两个序列,可以交换两个序列中的任意一个元素,但是最多交换两次,问交换后两个序列的和的差最小是多少。
思路:根据数据量显然我们可以直接暴力求出来交换零次和一次的最小差值,两次的话就不能直接暴力了,考虑优化,假设未交换时两序列差值为s,那么交换两次后(假设交换了a[i],a[j],b[u],b[v])差值为:
s-2*a[i]-2*a[j]+2*b[u]+2*b[v],要让这个式子尽可能地小,我们可以将s-2*a[i]-2*a[j]移项,即让2*b[u]+2*b[v]尽可能地等于2*a[i]+2*a[j]-s,显然我们可以预处理出所有2*b[u]+2*b[v],然后我们就可以愉快的二分优化了。
代码:
#include<bits/stdc++.h>#define ll long long#define pi acos(-1)#define inf 0x3f3f3f3f#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define rep(i,x,n) for(int i=x;i<n;i++)#define per(i,n,x) for(int i=n;i>=x;i--)using namespace std;typedef pair<int,int>P;const int MAXN=2010;int gcd(int a,int b){return b?gcd(b,a%b):a;}int n,m;ll a[MAXN],b[MAXN];pair<ll,P>p[MAXN*MAXN];int ans[5][2];int main(){std::ios::sync_with_stdio(0);ll sum=0,SUM=0;int tot=0;cin>>n;for(int i=1;i<=n;i++)cin>>a[i],sum+=a[i];cin>>m;for(int i=1;i<=m;i++)cin>>b[i],sum-=b[i];SUM=sum;sum=abs(sum);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(abs(SUM+2*b[j]-2*a[i])<sum)sum=abs(SUM+2*b[j]-2*a[i]),ans[0][0]=i,ans[0][1]=j,tot=1;int cnt=0;for(int i=1;i<=m;i++)for(int j=i+1;j<=m;j++)p[cnt++]=make_pair(2*b[i]+2*b[j],P(i,j));sort(p,p+cnt);for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){ll tmp=2*a[i]+2*a[j]-SUM;int x=lower_bound(p,p+cnt,make_pair(tmp,P(0,0)))-p;if(x!=cnt){tmp=abs(SUM-2*a[i]-2*a[j]+p[x].first);if(tmp<sum){sum=tmp;tot=2;ans[0][0]=i;ans[0][1]=p[x].second.first;ans[1][0]=j;ans[1][1]=p[x].second.second;}}if(x>0){tmp=abs(SUM-2*a[i]-2*a[j]+p[x-1].first);if(tmp<sum){sum=tmp;tot=2;ans[0][0]=i;ans[0][1]=p[x-1].second.first;ans[1][0]=j;ans[1][1]=p[x-1].second.second;}}}}cout<<abs(sum)<<endl<<tot<<endl;for(int i=0;i<tot;i++)cout<<ans[i][0]<<" "<<ans[i][1]<<endl; return 0;}
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