• Luogu1856 [USACO5.5]矩形周长Picture (线段树扫描线)


    对于横轴,加上与上一次扫描的差值;对于竖轴,加上高度差与区间内不相交线段(*2)的积;
    难点在pushdown,注意维护覆盖关系。再就注意负数

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
    #define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
    #define Max(a,b) ((a) > (b) ? (a) : (b))
    #define Min(a,b) ((a) < (b) ? (a) : (b))
    #define Fill(a,b) memset(a, b, sizeof(a))
    #define Abs(a) ((a) < 0 ? -(a) : (a))
    #define Swap(a,b) a^=b^=a^=b
    #define ll long long
    
    #define ON_DEBUG
    
    #ifdef ON_DEBUG
    
    #define D_e_Line printf("
    
    ----------
    
    ")
    #define D_e(x)  cout << #x << " = " << x << endl
    #define Pause() system("pause")
    
    #else
    
    #define D_e_Line ;
    
    #endif
    
    struct ios{
        template<typename ATP>ios& operator >> (ATP &x){
            x = 0; int f = 1; char c;
            for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
            while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
            x*= f;
            return *this;
        }
    }io;
    using namespace std;
    
    #define lson rt << 1, l, mid
    #define rson rt << 1 | 1, mid + 1, r
    
    const int N = 20007;
    
    struct Line{
        int l, r, h, tag;
        
        bool operator < (const Line &com)const{
    		if(h != com.h) return h < com.h;
    		return tag > com.tag;
    	}
    }a[N];
    struct Tree{
        int sum, num, len, ltag, rtag;
    }t[N << 2];
    
    inline void Pushdown(int rt, int l, int r){
    	if(t[rt].sum){
    		t[rt].num = 1;
    		t[rt].len = r - l + 1;
    		t[rt].ltag = t[rt].rtag = 1;
    		return;
    	}
    	if(l == r){
    		t[rt].len = 0;
    		t[rt].num = 0;
    		t[rt].ltag = t[rt].rtag = 0;
    		return;
    	}
    	t[rt].len = t[rt << 1].len + t[rt << 1 | 1].len;
    	t[rt].num = t[rt << 1].num + t[rt << 1 | 1].num;
    	if(t[rt << 1].rtag && t[rt << 1 | 1].ltag) --t[rt].num;
    	t[rt].ltag = t[rt << 1].ltag;
    	t[rt].rtag = t[rt << 1 | 1].rtag;
    }
    inline void Updata(int rt, int l, int r, int L, int R, int w){
    	if(L <= l && r <= R){
    		t[rt].sum += w;
    		Pushdown(rt, l, r);
    		return;
    	}
    	int mid = (l + r) >> 1;
    	if(L <= mid) Updata(lson, L, R, w);
    	if(R > mid)  Updata(rson, L, R, w);
    	Pushdown(rt, l, r);
    }
    int main(){
    	int n, m = 0;
        io >> n;
        int maxx = 0xcfcfcfcf, minn = 0x7fffffff;
        R(i,1,n){
        	int X1, Y1, X2, Y2;
        	io >> X1 >> Y1 >> X2 >> Y2;
        	maxx = Max(maxx, X2);
        	minn = Min(minn, X1);
            a[++m] = (Line){X1, X2, Y1, 1};
    		a[++m] = (Line){X1, X2, Y2, -1};
        }
        if(minn <= 0){
        	R(i,1,m){
        		a[i].l -= minn - 1;
        		a[i].r -= minn - 1;
        	}
        	maxx -= minn - 1;
        }
        sort(a + 1, a + m + 1);
        int ans = 0, last = 0;
        R(i,1,m){
        	Updata(1, 1, maxx, a[i].l, a[i].r - 1, a[i].tag);
        	while(a[i].h == a[i + 1].h && a[i].tag == a[i + 1].tag){
        		++i;
        		Updata(1, 1, maxx, a[i].l, a[i].r - 1, a[i].tag);
        	}
        	ans += Abs(t[1].len - last) + (t[1].num * (a[i + 1].h - a[i].h) << 1);
        	
        	last = t[1].len;    	
        }
        printf("%d
    ",ans);
        return 0;
    }
    

  • 相关阅读:
    JavaScript获取查询字符串
    Struts2 验证码图片实例
    js函数重载
    js面向对象基础
    js上下文
    java克隆入门和深入
    js类型检查
    js闭包
    Phonegap移动开发:布局总结(一) 全局
    Python发送多附件邮件的方法
  • 原文地址:https://www.cnblogs.com/bingoyes/p/11220405.html
Copyright © 2020-2023  润新知