HDU 4052 Adding New Machine (线段树+离散化)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4052

初始给你w*h的矩阵，给你n个矩形(互不相交)，按这些矩形尺寸把初始的矩形扣掉，形成一个新的'矩形'。然后给你1*m大小的矩形，问这个矩形在新'矩形'中有多少种放法。

一开始没想法==，然后看了看题解，说是线段树做的。

要是m为1的话，那答案就是剩下的面积了。

不为1的话，可以把n个矩形的面积扩展一下(我是向右扩展)，比如x1 y1 x2 y2 的矩形扣掉，就相当于x1  y1  x2+m-1  y2的地方扣掉了，这种是x轴的情况。那么y轴的情况就是x1 y1 x2  y2+m-1。最左边也要扣掉，比如x轴就是0 0 m - 1 h。所以最后就是算剩下的面积了。(我的方法比较笨，就是x轴一个线段树扫一下，y轴一个线段树扫一下)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
const int MAXN = 1e5 + 10;
typedef long long LL;
LL fab(LL a) {
    return (a > 0 ? a : -a);
}
struct data {
    int ll , rr , flag , l , r , h;
    bool operator <(const data &cmp) const {
        return h < cmp.h;
    }
}line1[MAXN] , line2[MAXN];
struct segtree {
    LL val;
    int l , r , add;
}T1[MAXN * 3] , T2[MAXN * 3];
LL x[MAXN] , y[MAXN];
map <int , int> mp1 , mp2;
int f1 , f2;

inline void add1(LL num) {
    if(!mp1[num]) {
        x[++f1] = num;
        mp1[num] = 1;
    }
}

inline void add2(LL num) {
    if(!mp2[num]) {
        y[++f2] = num;
        mp2[num] = 1;
    }
}

void pushupx(int p) {
    if(T1[p].add) {
        T1[p].val = x[T1[p].r] - x[T1[p].l];
    }
    else if(T1[p].r - T1[p].l == 1) {
        T1[p].val = 0;
    }
    else {
        T1[p].val = T1[p << 1].val + T1[(p << 1)|1].val;
    }
}

void pushupy(int p) {
    if(T2[p].add) {
        T2[p].val = y[T2[p].r] - y[T2[p].l];
    }
    else if(T2[p].r - T2[p].l == 1) {
        T2[p].val = 0;
    }
    else {
        T2[p].val = T2[p << 1].val + T2[(p << 1)|1].val;
    }
}

void buildx(int p , int l , int r) {
    int mid = (l + r) >> 1;
    T1[p].l = l , T1[p].r = r , T1[p].val = T1[p].add = 0;
    if(r - l == 1) {
        return ;
    }
    buildx(p << 1 , l , mid);
    buildx((p << 1)|1 , mid , r);
}

void buildy(int p , int l , int r) {
    int mid = (l + r) >> 1;
    T2[p].l = l , T2[p].r = r , T2[p].val = T2[p].add = 0;
    if(r - l == 1) {
        return ;
    }
    buildy(p << 1 , l , mid);
    buildy((p << 1)|1 , mid , r);
}

void updatex(int p , int l , int r , int add) {
    int mid = (T1[p].l + T1[p].r) >> 1;
    if(T1[p].l == l && T1[p].r == r) {
        T1[p].add += add;
        pushupx(p);
        return ;
    }
    if(r <= mid) {
        updatex(p << 1 , l , r , add);
    }
    else if(l >= mid) {
        updatex((p << 1)|1 , l , r , add);
    }
    else {
        updatex(p << 1 , l , mid , add);
        updatex((p << 1)|1 , mid , r , add);
    }
    pushupx(p);
}

void updatey(int p , int l , int r , int add) {
    int mid = (T2[p].l + T2[p].r) >> 1;
    if(T2[p].l == l && T2[p].r == r) {
        T2[p].add += add;
        pushupy(p);
        return ;
    }
    if(r <= mid) {
        updatey(p << 1 , l , r , add);
    }
    else if(l >= mid) {
        updatey((p << 1)|1 , l , r , add);
    }
    else {
        updatey(p << 1 , l , mid , add);
        updatey((p << 1)|1 , mid , r , add);
    }
    pushupy(p);
}

int main()
{
    LL w , h , n , m , x1 , x2 , y1 , y2;
    while(~scanf("%lld %lld %lld %lld" , &w , &h , &n , &m)) {
        mp1.clear();
        mp2.clear();
        f1 = f2 = 0;
        if(m == 1) {
            LL sum = 0;
            while(n--) {
                scanf("%lld %lld %lld %lld" , &x1 , &y1 , &x2 , &y2);
                sum += (fab(x1 - x2) + 1) * (fab(y1 - y2) + 1);
            }
            printf("%lld\n" , h * w - sum);
            continue;
        }
        for(int i = 0 ; i < n ; i++) {
            scanf("%lld %lld %lld %lld" , &x1 , &y1 , &x2 , &y2);
            x1-- , y1--;
            int ls = i << 1 , rs = (i << 1)|1;
            line1[ls].l = x1 , line1[ls].r = min(w , x2 + m - 1) , line1[ls].h = y1 , line1[ls].flag = 1;
            line1[rs].l = x1 , line1[rs].r = line1[ls].r , line1[rs].h = y2 , line1[rs].flag = -1;
            line2[ls].l = y1 , line2[ls].r = min(h , y2 + m - 1) , line2[ls].h = x1 , line2[ls].flag = 1;
            line2[rs].l = y1 , line2[rs].r = line2[ls].r , line2[rs].h = x2 , line2[rs].flag = -1;
            add1(line1[ls].l);
            add1(line1[ls].r);
            add1(line1[rs].l);
            add1(line1[rs].r);
            add2(line2[ls].l);
            add2(line2[ls].r);
            add2(line2[rs].l);
            add2(line2[rs].r);
        }
        int ls = n << 1 , rs = (n << 1)|1 , f = ((n << 1)|1) + 1;
        line1[ls].l = 0 , line1[ls].r = m - 1 , line1[ls].h = 0 , line1[ls].flag = 1;
        line1[rs].l = 0 , line1[rs].r = m - 1 , line1[rs].h = h , line1[rs].flag = -1;
        line2[ls].l = 0 , line2[ls].r = m - 1 , line2[ls].h = 0 , line2[ls].flag = 1;
        line2[rs].l = 0 , line2[rs].r = m - 1 , line2[rs].h = w , line2[rs].flag = -1;
        add1(line1[ls].l);
        add1(line1[ls].r);
        add1(line1[rs].l);
        add1(line1[rs].r);
        add2(line2[ls].l);
        add2(line2[ls].r);
        add2(line2[rs].l);
        add2(line2[rs].r);
        sort(line1 , line1 + f);
        sort(line2 , line2 + f);
        sort(x + 1 , x + f1 + 1);
        sort(y + 1 , y + f2 + 1);
        for(int i = 0 ; i < f ; i++) {
            line1[i].ll = lower_bound(x + 1 , x + f1 + 1 , line1[i].l) - x;
            line1[i].rr = lower_bound(x + 1 , x + f1 + 1 , line1[i].r) - x;
            line2[i].ll = lower_bound(y + 1 , y + f2 + 1 , line2[i].l) - y;
            line2[i].rr = lower_bound(y + 1 , y + f2 + 1 , line2[i].r) - y;
        }
        LL res1 = 0 , res2 = 0;
        buildx(1 , 1 , f1);
        buildy(1 , 1 , f2);
        updatex(1 , line1[0].ll , line1[0].rr , line1[0].flag);
        updatey(1 , line2[0].ll , line2[0].rr , line2[0].flag);
        for(int i = 1 ; i < f ; i++) {
            res1 += (line1[i].h - line1[i - 1].h) * T1[1].val;
            res2 += (line2[i].h - line2[i - 1].h) * T2[1].val;
            updatex(1 , line1[i].ll , line1[i].rr , line1[i].flag);
            updatey(1 , line2[i].ll , line2[i].rr , line2[i].flag);
        }
        printf("%lld\n" , h * w * 2 - res1 - res2);
    }
}