2018-2019 ACM-ICPC, Asia Seoul Regional Contest——A - Circuits

A - Circuits

不難發現x座標根本沒用，只需要儲存y座標。

題目所求的兩條直線$y_1=a$，$y_2=b(a<b)$
我們列舉$y_2=b$這條線，這條線一定可以是矩形的邊界，於是我們掃描矩形邊界差分計算當前這條線覆蓋的矩形個數，對於這條線沒有覆蓋的矩形把它丟到線段樹中（維護區間+和區間max即可）然後區間查詢$y_1=a$覆蓋的最大矩形即可。兩種相加即是當前的情況的最大數量。

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const ll mod=998244353;
const int N=200010;
int n,m;
struct line
{
    int id,v,op;
    bool operator<(const line &o) const
    {
        if(v==o.v) return op<o.op;
        return v<o.v;
    }
}q[N];
int y[N];
pii a[N];
int find(int x)
{
    return lower_bound(y+1,y+1+m,x)-y;
}
struct node
{
    int l,r;
    int v,lazy;
    
}tree[N*4];
void pushup(int u)
{
    tree[u].v=max(tree[u<<1].v,tree[u<<1|1].v);
}
void build(int u,int l,int r)
{
    tree[u]={l,r};
    if(l==r) 
    {
        tree[u].v=0;
        return;
    }
    int mid=l+r>>1;
    build(u<<1,l,mid),build(u<<1|1,mid+1,r);
    pushup(u);
}
void pushdown(int u)
{
    if(!tree[u].lazy) return;
    
    tree[u<<1].lazy+=tree[u].lazy;
    tree[u<<1|1].lazy+=tree[u].lazy;
    tree[u<<1].v+=tree[u].lazy;
    tree[u<<1|1].v+=tree[u].lazy;
    tree[u].lazy=0;
    
}
void modify(int u,int l,int r,int v)
{
    if(l>r) return;
    if(tree[u].l>=l&&tree[u].r<=r)
    {
        tree[u].lazy+=v;
        tree[u].v+=v;
        return;
    }
    pushdown(u);
    int mid=tree[u].l+tree[u].r>>1;                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       l+tree[u].r>>1;
    if(l<=mid) modify(u<<1,l,r,v);
    if(r>mid) modify(u<<1|1,l,r,v);
    pushup(u);
}
int main()
{
    IO;
    int T=1;
    //cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
           int t1,t2;
           cin>>t1>>y[i];
           cin>>t2>>y[i+n];
           a[i]={y[i+n],y[i]};
           
           y[i]++;
           q[i]={i,y[i],-1};
           q[i+n]={i,y[i+n],+1};
           
        }
        sort(y+1,y+1+2*n);
        m=unique(y+1,y+1+2*n)-y-1;
        
        build(1,1,m);
        sort(q+1,q+1+2*n);
        int pre=0;
        int res=0;
        for(int i=1;i<=2*n;i++)
        {
            pre+=q[i].op;
            res=max(res,pre+tree[1].v);
            
            if(q[i].op==-1)
            {
                int id=q[i].id;
                modify(1,find(a[id].first),find(a[id].second+1)-1,1);
            }
        }
        cout<<res<<'\n';
    }
    return 0;
}

這題有一個錯誤的貪心思路即求兩邊最大值。
先讓第一條線覆蓋最多的矩形，然後把這些矩形刪除，然後再求出第二條線覆蓋最多的矩形。

我也不知道這個貪心思路錯在哪裡（還沒舉出反例
先貼一個隊友寫的錯誤程式碼

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#define N 20000005
using namespace std;
const int m = 10000000,inf = 20000000;
int n;
struct operation{
	int c1,c2,r1,r2;
}op[100005];
int x[N],y[N],cans1,cans2;
int read(){
	char ch = getchar();
	int re = 0,fl = 1;
	while(ch<'0'||ch>'9') {if(fl == '-')fl = -1; ch = getchar();}
	while(ch>='0'&&ch<='9') {re = (re<<1)+(re<<3)+ch-'0'; ch = getchar();}
	return re*fl;
}
int main(){
	//freopen("1.in","r",stdin); 
	int r1,r2,c1,c2,rf = inf,rl = 0,cf = inf,cl = 0;
	int pre = 0,fd = 0; 
	n = read();
	for(int i=1;i<=n;++i){
	  r1 = read()+m; c1 = read()+m;//r1<r2  c1>c2
	  r2 = read()+m; c2 = read()+m;
	  y[c1+1]--; y[c2]++;
	  op[i].c1 = c1; op[i].c2 = c2; op[i].r1 = r1; op[i].r2 = r2;
	}
	pre = 0; fd = -1;
	for(int i=0;i<=2*m;++i){
	  pre += y[i];
	  if(pre >= cans1){
	  	cans1 = pre; fd = i;
	  }
	}
	for(int i=1;i<=n;++i)
	  if(op[i].c1 >= fd && op[i].c2 <= fd){
	  	y[op[i].c1+1]++; y[op[i].c2]--;
	  }
	  
	pre = 0;
	for(int i=0;i<=2*m;++i){
	  pre += y[i];
	  cans2 = max(cans2,pre);
	}
	printf("%d\n",cans1+cans2);
	return 0;
}