题目链接
题解
真是一道大码题，，，肝了一个上午
老司机的部分是一个(dp)，观察点是按(y)分层的，而且按每层点的上限来看可以使用(O(nd))的(dp)，其中(d)是每层的点数
我们设(f[i])表示从(i)点进入该层，直到走完为止所经过的最多点的数量，我们把原点也看做一棵树，计算答案时减去即可
转移只需枚举出点(j)，假如(i)在(j)的左侧，那么(j)及其左侧的点都能被经过，只需从(i)出发先走到左端点，再一直往右走到(j)
我们还需预处理出每个点向三个方向能到达的下一个点，这个使用排序 + 离散化 + 桶即可
具体地，向上记录(y)，向左上记录(x + y)，向右上记录(y - x)
原理是利用一次方程(y = x + b)和(y = -x + b)
好了(dp)的部分解决了，稍微记录一下转移就可以输出方案
现在我们解决小园丁的部分
显而易见这是一个带上下界网络流
对于可能出现的每条路径，都是一条流量下界为(1)的边
建源汇点(S)和(T)连向所有点
然后建超级源汇点跑一遍最小可行流即可
直接跑比较慢，需要用满流减去不加(T->S)的流量最大流
口胡真容易
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (register int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 1000005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int n,x[maxn],y[maxn],val[maxn],val2[maxn],b[maxn],bi,tot;
int nxtl[maxn],nxt[maxn],nxtr[maxn];
int bac[maxn],id[maxn],sit[maxn];
int L[maxn],R[maxn],pos[maxn],cnt;
int f[maxn],Nxt[maxn],Out[maxn];
int ok[maxn],d[maxn],vis[maxn],used[maxn],q[maxn],cur[maxn],head,tail,now;
int df[maxn],h[maxn],ne = 1;
struct EDGE{int to,nxt,f;}ed[maxm];
inline void build(int u,int v,int w){
ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;
ed[++ne] = (EDGE){u,h[v],0}; h[v] = ne;
}
void add(int u,int v){
Redge(u) if ((to = ed[k].to) == v) return;
ed[++ne] = (EDGE){v,h[u],INF}; h[u] = ne;
ed[++ne] = (EDGE){u,h[v],0}; h[v] = ne;
df[u]--; df[v]++; ok[v] = true;
}
inline bool cmp(const int& a,const int& b){
return val[a] == val[b] ? val2[a] < val2[b] : val[a] < val[b];
}
void preset(){
REP(i,n) id[i] = i,val[i] = y[i],b[++bi] = x[i];
sort(id + 1,id + 1 + n,cmp);
sort(b + 1,b + 1 + bi); tot = 1;
for (int i = 2; i <= bi; i++) if (b[i] != b[tot]) b[++tot] = b[i];
for (int i = n; i; i--){
int u = id[i],X = lower_bound(b + 1,b + 1 + tot,x[u]) - b;
nxt[u] = bac[X];
bac[X] = u;
}
bi = 0; cls(bac);
REP(i,n) val[i] = x[i] + y[i],val2[i] = x[i],b[++bi] = x[i] + y[i];
sort(id + 1,id + 1 + n,cmp);
sort(b + 1,b + 1 + bi); tot = 1;
for (int i = 2; i <= bi; i++) if (b[i] != b[tot]) b[++tot] = b[i];
for (int i = 1; i <= n; i++){
int u = id[i],V = lower_bound(b + 1,b + 1 + tot,x[u] + y[u]) - b;
nxtl[u] = bac[V];
bac[V] = u;
}
bi = 0; cls(bac);
REP(i,n) val[i] = y[i] - x[i],val2[i] = x[i],b[++bi] = y[i] - x[i];
sort(id + 1,id + 1 + n,cmp);
sort(b + 1,b + 1 + bi); tot = 1;
for (int i = 2; i <= bi; i++) if (b[i] != b[tot]) b[++tot] = b[i];
for (int i = n; i; i--){
int u = id[i],V = lower_bound(b + 1,b + 1 + tot,y[u] - x[u]) - b;
nxtr[u] = bac[V];
bac[V] = u;
}
REP(i,n) val[i] = y[i];
sort(id + 1,id + 1 + n,cmp);
for (int i = 1; i <= n; i++){
int u = id[i]; sit[u] = i;
if (i == 1 || y[u] != y[id[i - 1]]){
L[++cnt] = i;
if (cnt > 1) R[cnt - 1] = i - 1;
}
pos[i] = cnt;
}
R[cnt] = n;
}
void work_dp(){
for (int i = L[cnt]; i <= n; i++) f[id[i]] = n - L[cnt] + 1;
for (int i = cnt - 1; i; i--){
int mx = 0,now,nowf,tmp,to,from; now = 0; from = 0;
for (int j = L[i]; j <= R[i]; j++){
int u = id[j];
tmp = 0; to = 0; from = 0;
if (nxt[u] && f[nxt[u]] > tmp) tmp = f[nxt[u]],to = nxt[u],from = u;
if (nxtl[u] && f[nxtl[u]] > tmp) tmp = f[nxtl[u]],to = nxtl[u],from = u;
if (nxtr[u] && f[nxtr[u]] > tmp) tmp = f[nxtr[u]],to = nxtr[u],from = u;
tmp++;
if (tmp > f[u]) f[u] = tmp,Nxt[u] = to,Out[u] = from;
tmp += R[i] - j;
if (mx > f[u]) f[u] = mx,Nxt[u] = now,Out[u] = nowf;
if (tmp > mx) mx = tmp,now = to,nowf = u;
}
mx = 0; now = 0; from = 0;
for (int j = R[i]; j >= L[i]; j--){
int u = id[j];
tmp = 0; to = 0; from = 0;
if (nxt[u] && f[nxt[u]] > tmp) tmp = f[nxt[u]],to = nxt[u],from = u;
if (nxtl[u] && f[nxtl[u]] > tmp) tmp = f[nxtl[u]],to = nxtl[u],from = u;
if (nxtr[u] && f[nxtr[u]] > tmp) tmp = f[nxtr[u]],to = nxtr[u],from = u;
tmp++;
if (tmp > f[u]) f[u] = tmp,Nxt[u] = to,Out[u] = nowf;
tmp += j - L[i];
if (mx > f[u]) f[u] = mx,Nxt[u] = now,Out[u] = nowf;
if (tmp > mx) mx = tmp,now = to,nowf = u;
}
}
}
void print(){
printf("%d
",f[n] - 1);
int u = Nxt[n],now,num = f[n] - 1;
while (u){
now = pos[sit[u]];
if (now == cnt){
printf("%d ",u),num--; if (!num) break;
for (int i = sit[u] - 1; num && i >= L[cnt]; i--){
printf("%d ",id[i]),num--; if (!num) break;
}
for (int i = sit[u] + 1; num && i <= R[cnt]; i++){
printf("%d ",id[i]),num--; if (!num) break;
}
break;
}
if (x[Out[u]] == x[u]){
printf("%d ",u),num--;
}
else if (x[Out[u]] <= x[u]){
for (int i = sit[u]; num && i <= R[now]; i++){
printf("%d ",id[i]),num--; if (!num) break;
}
for (int i = sit[u] - 1; num && i >= sit[Out[u]]; i--){
printf("%d ",id[i]),num--; if (!num) break;
}
}
else {
for (int i = sit[u]; num && i >= L[now]; i--){
printf("%d ",id[i]),num--; if (!num) break;
}
for (int i = sit[u] + 1; num && i <= sit[Out[u]]; i++){
printf("%d ",id[i]),num--; if (!num) break;
}
}
if (!num) break;
u = Nxt[u];
}
puts("");
}
void solve1(){
preset();
work_dp();
print();
}
bool bfs(int S,int T){
q[head = tail = 0] = S; now++;
int u;
while (head <= tail){
u = q[head++];
Redge(u) if (ed[k].f && vis[to = ed[k].to] != now){
d[to] = d[u] + 1;
vis[to] = now;
q[++tail] = to;
if (to == T) return true;
}
}
return false;
}
int dfs(int u,int minf,int T){
if (u == T || !minf) return minf;
int f,flow = 0,to;
if (used[u] != now) used[u] = now,cur[u] = h[u];
for (int& k = cur[u]; k; k = ed[k].nxt)
if (vis[to = ed[k].to] == now && d[to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f),T))){
ed[k].f -= f; ed[k ^ 1].f += f;
flow += f; minf -= f;
if (!minf) break;
}
return flow;
}
void work_sol(int u){
int S = pos[sit[u]],l = L[S],r = R[S],a,b,c,v,sum;
for (int i = l; i <= r; i++){
v = id[i]; a = nxtl[v]; b = nxt[v]; c = nxtr[v];
if (v == u){
if (a && f[a] + 1 == f[u]) add(v,a);
if (b && f[b] + 1 == f[u]) add(v,b);
if (c && f[c] + 1 == f[u]) add(v,c);
continue;
}
if (sit[v] < sit[u]) sum = r - sit[v] + 1;
else sum = sit[v] - l + 1;
if (a && f[a] + sum == f[u]) add(v,a);
if (b && f[b] + sum == f[u]) add(v,b);
if (c && f[c] + sum == f[u]) add(v,c);
}
}
void setG(){
ok[n] = true;
for (int i = 1; i < cnt; i++){
for (int j = L[i]; j <= R[i]; j++)
if (ok[id[j]]) work_sol(id[j]);
}
}
void solve2(){
setG();
int S = n + 1,T = n + 2,ans = 0;
for (int i = 1; i <= n; i++){
if (!df[i]) continue;
if (df[i] > 0) ans += df[i],build(S,i,df[i]);
else build(i,T,-df[i]);
}
while (bfs(S,T)) ans -= dfs(S,INF,T);
printf("%d
",ans);
}
int main(){
n = read();
REP(i,n) x[i] = read(),y[i] = read();
n++;
solve1();
solve2();
return 0;
}