2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 Overlapping Rectangles 矩形并面积和

Overlapping Rectangles

题意：求 n 个矩形并面积和

tags：扫描线+线段树，模板题

参考博客

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;

const int N=1000+5;
int col[N<<3], cnt, res;
double X[N<<3], Sum[N<<3], Sum2[N<<3];  //多开一倍，因为 x坐标有两个
struct Seg {
    double l, r, h;
    int flag;
    Seg(){}
    Seg(double l,double r,double h,int flag):l(l),r(r),h(h),flag(flag){}
    bool operator <(const Seg & object ) const{
        return h < object.h;
    }
}S[N<<2];
void pushup(int ro, int l, int r)
{
    if(col[ro])    //覆盖一次
        Sum[ro]=X[r+1]-X[l];
    else if(l==r) Sum[ro]=0;
    else Sum[ro]=Sum[ro<<1]+Sum[ro<<1|1];

    if(col[ro]>=2) // 覆盖两次以上
        Sum2[ro]=X[r+1]-X[l];
    else if(l==r) Sum2[ro]=0;
    else if(col[ro]==1) Sum2[ro]=Sum[ro<<1] +Sum[ro<<1|1];
    else if(col[ro]==0) Sum2[ro]=Sum2[ro<<1] +Sum2[ro<<1|1];
}
void update(int L, int R, int c, int ro, int l, int r)  // [l,r]当前区间，[L,R]目标区间
{
    if(L<=l && r<=R)
    {
        col[ro] += c;
        pushup(ro, l, r);
        return ;
    }
    int mid = (l+r)>>1;
    if(L<=mid) update(L, R, c, ro<<1, l, mid);
    if(R>mid) update(L, R, c, ro<<1|1, mid+1, r);
    pushup(ro, l, r);
}
int binary_find(double x)
{
    int l=1, r=res, ans, mid;
    while(l<=r)
    {
        mid = (l+r)>>1;
        if(X[mid] >= x) ans=mid, r=mid-1;
        else l=mid+1;
    }
    return ans;
}
double solve(int n)
{
    cnt = res = 0;
    for(int i=1; i<=n; ++i)
    {
        double x1, y1, x2, y2;
        scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
        S[++cnt]=Seg(x1,x2,y1,1);
        X[cnt]=x1;
        S[++cnt]=Seg(x1,x2,y2,-1);
        X[cnt]=x2;
    }
    sort(X+1, X+1+cnt);
    sort(S+1, S+1+cnt);
    ++res;
    for(int i=2; i<=cnt; ++i) {  // 去重，x坐标过大的话，就离散化
        if(X[i]!=X[i-1])  X[++res]=X[i];
    }
    memset(Sum, 0, sizeof(Sum));
    memset(col, 0, sizeof(col));
    memset(Sum2, 0, sizeof(Sum2));
    double ans=0;
    for(int i=1; i<cnt; ++i)
    {
        int l = binary_find(S[i].l);      //二分左端点
        int r = binary_find(S[i].r)-1;    // 左闭右开,二分右端点
        update(l, r, S[i].flag, 1, 1, res);
        ans += Sum[1]*(S[i+1].h-S[i].h);    //矩阵并
        //ans += Sum2[1]*(S[i+1].h-S[i].h); //矩阵交集
    }
    return ans;
}

int main()
{
    int n;
    while(scanf("%d", &n), n)
    {
        printf("%.0f
", solve(n));
    }
    puts("*");

    return 0;
}