題目鏈接

https://codeforces.com/contest/685/problem/C

題解

我怎么又還差最后一步的時(shí)候放棄了然后往別的方向上想了一小時(shí)才發(fā)現(xiàn)這個(gè)思路能做……

首先二分答案，轉(zhuǎn)化為所有點(diǎn)半徑為\(mid\)的曼哈頓距離區(qū)域內(nèi)是否有交。
考慮二維問(wèn)題，顯然可以用曼哈頓轉(zhuǎn)切比雪夫來(lái)做，把\((x,y)\)變成\((x+y,x-y)\)那么原來(lái)兩點(diǎn)的曼哈頓距離就等于后來(lái)兩點(diǎn)的切比雪夫距離。考慮一下這個(gè)做法的證明: \[|x_1-x_2|+|y_1-y_2|=\max(x_1-x_2+y_1-y_2,x_1-x_2-y_1+y_2,-x_1+x_2+y_1-y_2,-x_1+x_2-y_1+y_2)=\max(|(x_1+y_1)-(x_2+y_2)|,|(x_1-y_1)-(x_2-y_2)|)\]
三維的問(wèn)題也可以用類似的思路。\[|x_1-x_2|+|y_1-y_2|+|z_1-z_2|=\max(|(x_1+y_1+z_1)-(x_2+y_2+z_2)|,|(x_1-y_1+z_1)-(x_2-y_2+z_2)|,|(x_1+y_1-z_1)-(x_2+y_2-z_2)|,|(x_1-y_1-z_1)-(x_2-y_2-z_2)|)\]于是我們可以把\((x,y,z)\)變成一個(gè)四維的坐標(biāo)\((x+y+z,x-y+z,x+y-z,x-y-z)\), 曼哈頓距離就變成了切比雪夫距離，求一下區(qū)間交就行了。但是有以下兩個(gè)附加限制：

(1) 新的坐標(biāo)\((x,y,z,w)\)必須滿足\(w=y+z-x\).

(2) 由于題目里要求所有坐標(biāo)都是整數(shù)，新的坐標(biāo)\((x,y,z,w)\)必須滿足\(y-x,z-x\)都是\(2\)的倍數(shù)。
那么可以枚舉\(x,y,z,w\)都是奇數(shù)或偶數(shù)的情況，分別求一下在\(x,y,z\)取值范圍內(nèi)\(y+z-x\)的最大最小值，然后和\(w\)的取值范圍求交，調(diào)整一下即可求解。
時(shí)間復(fù)雜度\(O(n\log C)\), \(C\)為值域。

代碼

#include<bits/stdc++.h> #define llong long long #define mkpr make_pair #define riterator reverse_iterator using namespace std;inline int read() {int x = 0,f = 1; char ch = getchar();for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}return x*f; }const int N = 1e5; const llong INF = 6e18+1; struct Point {llong x,y,z,w; } a[N+3]; int n;llong fun(llong x,llong y,llong z) {return x&1?x+z:x+y;}bool check(llong mid,bool typ=false) {llong lx = -INF,ly = -INF,lz = -INF,lw = -INF,rx = INF,ry = INF,rz = INF,rw = INF;for(int i=1; i<=n; i++){lx = max(lx,a[i].x-mid); rx = min(rx,a[i].x+mid);ly = max(ly,a[i].y-mid); ry = min(ry,a[i].y+mid);lz = max(lz,a[i].z-mid); rz = min(rz,a[i].z+mid);lw = max(lw,a[i].w-mid); rw = min(rw,a[i].w+mid);}if(lx>rx||ly>ry||lz>rz||lw>rw) return false;llong lx2 = fun(lx,0,1),ly2 = fun(ly,0,1),lz2 = fun(lz,0,1),lw2 = fun(lw,0,1),rx2 = fun(rx,0,-1),ry2 = fun(ry,0,-1),rz2 = fun(rz,0,-1),rw2 = fun(rw,0,-1);if(lx2<=rx2&&ly2<=ry2&&lz2<=rz2&&lw2<=rw2){llong lw3 = ly2+lz2-rx2,rw3 = ry2+rz2-lx2;if(max(lw2,lw3)<=min(rw2,rw3)){llong w = max(lw2,lw3),x = rx2,y = ly2,z = lz2;if(typ){if(y+z-x<w) {y += min(ry2-ly2,w-(y+z-x));}if(y+z-x<w) {z += min(rz2-lz2,w-(y+z-x));}if(y+z-x<w) {x -= min(rx2-lx2,w-(y+z-x));}llong xx = (y+z)/2ll,yy = (x-y)/2ll,zz = (x-z)/2ll;printf("%I64d %I64d %I64d\n",xx,yy,zz);}return true;}}lx2 = fun(lx,1,0),ly2 = fun(ly,1,0),lz2 = fun(lz,1,0),lw2 = fun(lw,1,0),rx2 = fun(rx,-1,0),ry2 = fun(ry,-1,0),rz2 = fun(rz,-1,0),rw2 = fun(rw,-1,0);if(lx2<=rx2&&ly2<=ry2&&lz2<=rz2&&lw2<=rw2){llong lw3 = ly2+lz2-rx2,rw3 = ry2+rz2-lx2;if(max(lw2,lw3)<=min(rw2,rw3)){if(typ){llong w = max(lw2,lw3),x = rx2,y = ly2,z = lz2;if(y+z-x<w) {y += min(ry2-ly2,w-(y+z-x));}if(y+z-x<w) {z += min(rz2-lz2,w-(y+z-x));}if(y+z-x<w) {x -= min(rx2-lx2,w-(y+z-x));}llong xx = (y+z)/2ll,yy = (x-y)/2ll,zz = (x-z)/2ll;printf("%I64d %I64d %I64d\n",xx,yy,zz);}return true;}}return false; }int main() {int T; scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=1; i<=n; i++){llong x,y,z; scanf("%I64d%I64d%I64d",&x,&y,&z);a[i].x = x+y+z,a[i].y = x-y+z,a[i].z = x+y-z,a[i].w = x-y-z;}llong left = 0ll,right = 3e18;while(left<right){llong mid = left+(right-left>>1);if(check(mid)) {right = mid;}else {left = mid+1;}} // printf("ans=%I64d\n",right);check(right,true);}return 0; }

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