题目链接

https://atcoder.jp/contests/agc034/tasks/agc034_d

题解

降智了没想出来建图……(不过这场是真的DE放反了)
注意到曼哈顿距离有一个重要的性质: 合法的最大，也就是对于任意维数两个点的曼哈顿距离而言，假设我们把每个绝对值符号任意地改变成正号或者负号，那么合法的(实际的)曼哈顿距离是这所有值里最大的。证明显然。
那么既然我们求的是最大距离和，就可以忽略合不合法的问题，四种拆绝对值的方法不对连接的两个点的坐标有要求。
考虑费用流建图，建两排点分别表示两种球，然后在中间建四个点代表四种匹配方式，两边的每个点和中间的每个点连对应匹配方式的权值，求最大费用最大流即可。
时间复杂度\(O(MFMC(n,10n))\), 似乎费用流算法的复杂度可以被估计为\(O(mC\log m)\)其中\(C\)为最大流量，而这里最大流量是\(O(S)\)的，故复杂度\(O(nS\log n)\).
题解里说可以模拟费用流做到\(O(S\log n)\)... 瑟瑟发抖

代码

#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 llong INF = 1e12;namespace NetFlow
{const int N = 2006;const int M = 10000;struct AEdge{int u,v,wl,wr; llong c;} ae[M+3];struct Edge{int u,v,nxt,w; llong c;} e[(M<<1)+3];int fe[N+3];llong dis[N+3];int que[N+5];bool inq[N+3];int lst[N+3];int n,m,en,s,t; llong mf,mc;void addedge(int u,int v,int w,llong c){
//      printf("addedge %d %d %d %lld\n",u,v,w,c);en++; e[en].u = u,e[en].v = v,e[en].w = w,e[en].c = c;e[en].nxt = fe[u]; fe[u] = en;en++; e[en].u = v,e[en].v = u,e[en].w = 0,e[en].c = -c;e[en].nxt = fe[v]; fe[v] = en;}bool spfa(){for(int i=1; i<=n; i++) dis[i] = -INF;int hd = 1,tl = 2; que[1] = s; dis[1] = 0;while(hd!=tl){int u = que[hd]; hd++; if(hd>n+1) hd-=n+1;for(int i=fe[u]; i; i=e[i].nxt){int v = e[i].v;if(e[i].w>0&&dis[e[i].v]<dis[u]+e[i].c){dis[e[i].v] = dis[u]+e[i].c; lst[e[i].v] = i;if(!inq[e[i].v]){inq[e[i].v] = true;que[tl] = e[i].v; tl++; if(tl>n+1) tl-=n+1;}}}inq[u] = false;}return dis[t]!=-INF;}void calcflow(){int flow = 1e5;for(int u=t; u!=s; u=e[lst[u]].u){flow = min(flow,e[lst[u]].w);}for(int u=t; u!=s; u=e[lst[u]].u){e[lst[u]].w -= flow; e[lst[u]^1].w += flow;}mf += flow; mc += 1ll*flow*dis[t];}llong mfmc(int _n,int _s,int _t){n = _n,s = _s,t = _t; mf = 0,mc = 0ll;while(spfa()) {calcflow();} return mc;}
}
using NetFlow::addedge;const int N = 1000;
struct Point
{int x,y,cnt;
} a[N+3],b[N+3];
int n;int main()
{NetFlow::en = 1;scanf("%d",&n);for(int i=1; i<=n; i++) scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].cnt);for(int i=1; i<=n; i++) scanf("%d%d%d",&b[i].x,&b[i].y,&b[i].cnt);for(int i=1; i<=n; i++){addedge(1,i+6,a[i].cnt,0);addedge(i+6,3,a[i].cnt,-a[i].x-a[i].y);addedge(i+6,4,a[i].cnt,-a[i].x+a[i].y);addedge(i+6,5,a[i].cnt,a[i].x-a[i].y);addedge(i+6,6,a[i].cnt,a[i].x+a[i].y);}for(int i=1; i<=n; i++){addedge(i+n+6,2,b[i].cnt,0);addedge(3,i+n+6,b[i].cnt,b[i].x+b[i].y);addedge(4,i+n+6,b[i].cnt,b[i].x-b[i].y);addedge(5,i+n+6,b[i].cnt,-b[i].x+b[i].y);addedge(6,i+n+6,b[i].cnt,-b[i].x-b[i].y);}llong ans = NetFlow::mfmc(n+n+6,1,2);printf("%lld\n",ans);return 0;
}