[USACO5.5] 矩形周长Picture

https://www.luogu.org/problemnew/show/P1856

1.每个矩形由两条横向边和两条纵向边组成.

2.对于横向边，按纵坐标排序。设当前讨论的边为 A [s , t]

如果 A 是某个矩形的靠下的边，在树中查询[s,t]区间中被覆盖的长度为x，那么加上这条边后将增加(t-s-x);

如果 A 是某个矩形的靠上的边，先删除它的对应边，再在树中查询[s,t]区间中被覆盖的长度为x，那么加上这条边后将增加(t-s-x);

3、对于纵向边，按横坐标排序，讨论方法与横向边相同。

注意建树方式是以单位线段为叶子节点

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>

#define LL long long

using namespace std;
const int MAXN = 2e4 + 5;
const int INF = 1e9;

int n,m,Ans = 0;

#define Max 10001

inline void read(int &x){
    x=0; int f=1; register char c = getchar();
    while(c>‘9‘||c<‘0‘){ if(c==‘-‘)f=-1; c=getchar(); }
    while(c>=‘0‘&&c<=‘9‘){ x=x*10+c-‘0‘; c=getchar(); } x*=f;
}

struct Seg_Ment{
    int s,t,id,p; // s、t线段端点 p是排序依据  id--排序之前的编号
    Seg_Ment(int a,int b,int c,int d):s(a),t(b),p(c),id(d){}
    Seg_Ment(){ s=t=p=id=0; }
    bool operator < (const Seg_Ment a) const {
        return p < a.p;
    }
}x[MAXN],y[MAXN];

int lazy[MAXN << 3],cnt[MAXN << 3],l,r;

inline void Push_Down(int u){
    int ls = u << 1 ,rs = u << 1|1;
    lazy[ls] += lazy[u],cnt[ls] += lazy[u],cnt[rs] += lazy[u],lazy[rs] += lazy[u];
    lazy[u] = 0;
}

void UpDate(int u,int L,int R,int Del){
    if(lazy[u]) Push_Down(u);
    if(l <= L && R <= r){
        cnt[u] += Del,lazy[u] += Del;
    }
    else if(L + 1 < R){
        int Mid = (L+R) >>1,ls = u<<1,rs = u<<1|1;
        if(l <= Mid && L <= r) UpDate(ls,L,Mid,Del);
        if(r >= Mid && l <= R) UpDate(rs,Mid,R,Del);
    }
}

int query(int u,int L,int R){
    if(lazy[u]) Push_Down(u);
    if(l <= L && R <= r && cnt[u] > 0) return R - L;
    if(L + 1 < R){
        int Mid = L+R >>1,a = 0,b = 0;
        if(l <= Mid) a = query(u<<1,L,Mid);
        if(r >= Mid) b = query(u<<1|1,Mid,R);
        return a + b;
    }
    return 0;
}

int main(int argc,char *argv[]){
    freopen("gg.out", "w", stdout);
    cin >> n;
    for(int i = 1; i <= n; i ++) {
        cout << i << ":" << " " << tan(i) << endl;
    }
    return 0;
    int x1,y1,xx,yy;
    read(n);
    for(int i=1; i<=n; ++i){
        read(x1),read(y1),read(xx),read(yy);
        x1 += Max,xx += Max,y1 += Max, yy += Max;
        x[i] = Seg_Ment(x1,xx,y1,i),y[i] = Seg_Ment(y1,yy,x1,i);
        x[i + n] = Seg_Ment(x1,xx,yy,i + n),y[i + n] = Seg_Ment(y1,yy,xx,i + n);
    }

    return 0;
    sort(x + 1,x + 2 * n + 1);
    sort(y + 1,y + 2 * n + 1);
    memset(lazy,0,sizeof lazy ); memset(cnt,0,sizeof cnt );
    for(int i=1; i<=n * 2; ++i){
        l = x[i].s, r = x[i].t;
        if(x[i].id <= n){
            Ans += abs(r - l - query(1,1,MAXN));
            UpDate(1,1,MAXN,1);
        }
        else {
            UpDate(1,1,MAXN,-1);
            Ans += abs(r - l - query(1,1,MAXN));
        }
    }
    memset(lazy,0,sizeof lazy ); memset(cnt,0,sizeof cnt );
    for(int i=1; i<=n * 2; ++i){
        l = y[i].s,r = y[i].t;
        if(y[i].id <= n){
            Ans += abs(r - l - query(1,1,MAXN));
            UpDate(1,1,MAXN,1);
        }
        else{
            UpDate(1,1,MAXN,-1);
            Ans += abs(r - l - query(1,1,MAXN));
        }
    }
    cout << Ans << endl;
    return 0;
}

原文地址：https://www.cnblogs.com/shandongs1/p/8406563.html