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文章目錄

• 題意：
• 思路：

題意：

給你三個數組$a,b,c$，讓你從每個數組中選擇一個數$x,y,z$，使得$(x?y{)}^2+(x?z{)}^2+(y?z{)}^2$最小，求這個最小值。

思路：

由于答案一定是$x\le y\le z$的形式(當然這里的$x,y,z$與題意的$x,y,z$不同)，所以我們就可以跑一個$3!$排列，讓他們分別為$x,y,z$，讓后枚舉其中一個二分另外倆就行了。
具體實現可以寫個函數就比較方便了。
我這里寫的是枚舉$x$，讓后分兩種即找$y$和找$z$，如果找$y$的話，假設現在已經有$x\le y$了，那么找的$z$根據平方的性質，最好是$x+(y?x)/2$附近的位置，二分找即可，對于$z$同理。
總結：這個題瞎搞都能過。

// Problem: D. Xenia and Colorful Gems // Contest: Codeforces - Codeforces Round #635 (Div. 2) // URL: https://codeforces.com/contest/1337/problem/D // Memory Limit: 256 MB // Time Limit: 3000 ms // // Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native") //#pragma GCC optimize(2) #include<cstdio> #include<iostream> #include<string> #include<cstring> #include<map> #include<cmath> #include<cctype> #include<vector> #include<set> #include<queue> #include<algorithm> #include<sstream> #include<ctime> #include<cstdlib> #define X first #define Y second #define L (u<<1) #define R (u<<1|1) #define pb push_back #define mk make_pair #define Mid (tr[u].l+tr[u].r>>1) #define Len(u) (tr[u].r-tr[u].l+1) #define random(a,b) ((a)+rand()%((b)-(a)+1)) #define db puts("---") using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); } //void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); } //void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f; const LL inf=0x3f3f3f3f3f3f3f3f; const double eps=1e-6;int n1,n2,n3; LL a[N],b[N],c[N];LL get(int i,int j,int k) {//cout<<i<<' '<<j<<' '<<k<<endl;return (a[i]-b[j])*(a[i]-b[j])+(a[i]-c[k])*(a[i]-c[k])+(b[j]-c[k])*(b[j]-c[k]); }LL solve() {LL ans=inf;for(int i=1;i<=n1;i++) {int val1=a[i];int pos1=lower_bound(b+1,b+1+n2,val1)-b;if(pos1<=n2) {int pos2=lower_bound(c+1,c+1+n3,a[i]+(b[pos1]-a[i])/2)-c;if(pos2<=n3) ans=min(ans,get(i,pos1,pos2)); pos2--;if(pos2>=1) ans=min(ans,get(i,pos1,pos2));}pos1++;if(pos1<=n2) {int pos2=lower_bound(c+1,c+1+n3,a[i]+(b[pos1]-a[i])/2)-c;if(pos2<=n3) ans=min(ans,get(i,pos1,pos2)); pos2--;if(pos2>=1) ans=min(ans,get(i,pos1,pos2));}pos1-=2;if(pos1>=1) {int pos2=lower_bound(c+1,c+1+n3,b[pos1]+(a[i]-b[pos1])/2)-c;if(pos2<=n3) ans=min(ans,get(i,pos1,pos2)); pos2--;if(pos2>=1) ans=min(ans,get(i,pos1,pos2));}pos1=lower_bound(c+1,c+1+n3,val1)-c;if(pos1<=n3) {int pos2=lower_bound(b+1,b+1+n2,a[i]+(c[pos1]-a[i])/2)-b;if(pos2<=n2) ans=min(ans,get(i,pos2,pos1)); pos2--;if(pos2>=1) ans=min(ans,get(i,pos2,pos1));}pos1++;if(pos1<=n3) {int pos2=lower_bound(b+1,b+1+n2,a[i]+(c[pos1]-a[i])/2)-b;if(pos2<=n2) ans=min(ans,get(i,pos2,pos1)); pos2--;if(pos2>=1) ans=min(ans,get(i,pos2,pos1));}pos1-=2;//cout<<pos1<<' '<<c[pos1]<<' '<<a[i]<<endl;if(pos1>=1) {int pos2=lower_bound(b+1,b+1+n2,c[pos1]+(a[i]-c[pos1])/2)-b;if(pos2<=n2) ans=min(ans,get(i,pos2,pos1)); pos2--;if(pos2>=1) ans=min(ans,get(i,pos2,pos1));}}return ans; }int main() { // ios::sync_with_stdio(false); // cin.tie(0);int _; scanf("%d",&_);while(_--) {scanf("%d%d%d",&n1,&n2,&n3);for(int i=1;i<=n1;i++) scanf("%lld",&a[i]);for(int i=1;i<=n2;i++) scanf("%lld",&b[i]);for(int i=1;i<=n3;i++) scanf("%lld",&c[i]);sort(a+1,a+1+n1); sort(b+1,b+1+n2); sort(c+1,c+1+n3);n1=unique(a+1,a+1+n1)-a-1;n2=unique(b+1,b+1+n2)-b-1;n3=unique(c+1,c+1+n3)-c-1;printf("%lld\n",solve());}return 0; } /**/

總結

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