思路：
这道题要把给定的每个坐标利用切比雪夫坐标表示，这样两个点的距离就是max（dx，dy），而不是一开始的dx + dy，有利于线段树的维护，又由于询问的是区间的最大差值，限制是两个点是属于不同门派的，所以我们可以维护每个区间的最大，次大值，最小次小值。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << " "; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; typedef pair<ll,int>pli; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ' ' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 9999973; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 5e5+9; int n,m; struct node{ ll x,y,c; }a[maxn]; struct Tree{ ///p[0] x轴 1,2 第一第二大,3,4第一第二小 ///p[1] y轴 pli p[2][5]; } t[maxn<<2]; void pushup34(int rt,int id){ pli tmp[5]; tmp[1] = t[rt<<1].p[id][3]; tmp[2] = t[rt<<1].p[id][4]; tmp[3] = t[rt<<1|1].p[id][3]; tmp[4] = t[rt<<1|1].p[id][4]; t[rt].p[id][3] = {inff,0}; t[rt].p[id][4] = {inff,0}; for(int i=1; i<=4; i++){ if(t[rt].p[id][3].fi > tmp[i].fi && tmp[i].se){ t[rt].p[id][3] = tmp[i]; } } for(int i=1; i<=4; i++){ if(t[rt].p[id][4].fi > tmp[i].fi && tmp[i].se != t[rt].p[id][3].se){ t[rt].p[id][4] = tmp[i]; } } } void pushup12(int rt,int id){ pli tmp[5]; tmp[1] = t[rt<<1].p[id][1]; tmp[2] = t[rt<<1].p[id][2]; tmp[3] = t[rt<<1|1].p[id][1]; tmp[4] = t[rt<<1|1].p[id][2]; t[rt].p[id][1] = {-inff,0}; t[rt].p[id][2] = {-inff,0}; for(int i=1; i<=4; i++){ if(t[rt].p[id][1].fi < tmp[i].fi && tmp[i].se){ t[rt].p[id][1] = tmp[i]; } } for(int i=1; i<=4; i++){ if(t[rt].p[id][2].fi < tmp[i].fi && tmp[i].se != t[rt].p[id][1].se){ t[rt].p[id][2] = tmp[i]; } } } void build(int l,int r,int rt){ if(l == r){ t[rt].p[0][1].fi = a[l].x; t[rt].p[0][1].se = a[l].c; t[rt].p[0][2] = {-inff,0}; t[rt].p[0][3].fi = a[l].x; t[rt].p[0][3].se = a[l].c; t[rt].p[0][4] = {inff,0}; /*-----------------------*/ t[rt].p[1][1].fi = a[l].y; t[rt].p[1][1].se = a[l].c; t[rt].p[1][2] = {-inff,0}; t[rt].p[1][3].fi = a[l].y; t[rt].p[1][3].se = a[l].c; t[rt].p[1][4] = {inff,0}; return; } int mid = (l + r) >> 1; build(l,mid,rt<<1); build(mid+1,r,rt<<1|1); pushup12(rt, 0); pushup34(rt, 0); pushup12(rt, 1); pushup34(rt, 1); } void update(int k,int c,int dx,int dy,int l,int r,int rt){ if(l == r){ a[l].x += 1ll*dx; a[l].y += 1ll*dy; a[l].c = c; t[rt].p[0][1].fi = a[l].x; t[rt].p[0][1].se = a[l].c; t[rt].p[0][2] = {-inff,0}; t[rt].p[0][3].fi = a[l].x; t[rt].p[0][3].se = a[l].c; t[rt].p[0][4] = {inff,0}; //-----------------------------------// t[rt].p[1][1].fi = a[l].y; t[rt].p[1][1].se = a[l].c; t[rt].p[1][2] = {-inff,0}; t[rt].p[1][3].fi = a[l].y; t[rt].p[1][3].se = a[l].c; t[rt].p[1][4] = {inff,0}; return; } int mid = (l + r) >> 1; if(k <= mid) update(k,c,dx,dy,l,mid,rt<<1); else update(k,c,dx,dy,mid+1,r,rt<<1|1); pushup12(rt, 0); pushup34(rt, 0); pushup12(rt, 1); pushup34(rt, 1); } void query(int L,int R,int l,int r,int rt,pli &b1,pli &b2,pli &s1, pli &s2,int id){ if( L <= l && r <= R){ for(int i=1; i<=2; i++) { if(b1.fi < t[rt].p[id][i].fi) { if(b1.se != t[rt].p[id][i].se) b2 = b1; b1 = t[rt].p[id][i]; } } for(int i=1; i<=2; i++){ if(b2.fi < t[rt].p[id][i].fi && b1.se != t[rt].p[id][i].se){ b2 = t[rt].p[id][i]; } } for(int i=3; i<=4; i++) { if(s1.fi > t[rt].p[id][i].fi){ if(s1.se != t[rt].p[id][i].se) s2 = s1; s1 = t[rt].p[id][i]; } } for(int i=3; i<=4; i++){ if(s2.fi > t[rt].p[id][i].fi && s1.se != t[rt].p[id][i].se){ s2 = t[rt].p[id][i]; } } return ; } int mid = (l + r) >> 1; if(mid >= L) query(L, R, l, mid, rt<<1, b1,b2,s1,s2,id); if(mid < R) query(L, R, mid+1, r,rt<<1|1, b1,b2,s1,s2,id); } int main(){ int T; scanf("%d", &T); for(int cas=1; cas<=T; cas++){ printf("Case #%d: ", cas); scanf("%d%d", &n, &m); for(int i=1; i<=n; i++){ int x,y,c; scanf("%d%d%d", &x, &y, &c); a[i].x = x + y, a[i].y = x - y; a[i].c = c; } build(1,n,1); while(m --) { int op; scanf("%d", &op); if(op == 1){ int k,x,y; scanf("%d%d%d", &k, &x, &y); update(k,a[k].c,x+y, x-y,1,n,1); } else if(op == 2){ int k,c; scanf("%d%d", &k, &c); update(k,c,0,0,1,n,1); } else { int l,r; scanf("%d%d", &l, &r); pli q[5]; q[1] = q[2] = {-inff, 0}; q[3] = q[4] = {inff, 0}; query(l,r,1,n,1,q[1],q[2],q[3],q[4],0); ll res = -inff; for(int i=1; i<=4; i++){ for(int j=1; j<=4; j++){ if(q[i].se != q[j].se && q[i].se && q[j].se){ res = max(res, abs(q[i].fi - q[j].fi)); } } } pli e[5]; e[1] = e[2] = {-inff, 0}; e[3] = e[4] = {inff, 0}; query(l,r,1,n,1,e[1],e[2],e[3],e[4],1); for(int i=1; i<=4; i++){ for(int j=1; j<=4; j++){ if(e[i].se != e[j].se && e[i].se && e[j].se){ res = max(res, abs(e[i].fi - e[j].fi)); } } } if(res <= -inff) res = 0; printf("%lld ", res); } } } return 0; }