2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 F. Overlapping Rectangles（面积并）

题目链接：Overlapping Rectangles 面积并模板题，类似题还有HDU1542

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<cmath>#include<map>#include<set>#include<cstdlib>#define mem(a,b) memset(a,b,sizeof(a))#define INF 0x7ffffffftypedef long long ll;using namespace std;const int maxn = 2010;struct Seg{    double l;    double r;    double h;    int s;    Seg(){};    Seg(double a,double b,double c,int d) : l(a),r(b),h(c),s(d){};    bool operator < (const Seg &cmp) const{        return h < cmp.h;    }}ss[maxn<<2];double X[maxn<<2];double sum[maxn<<2];int cnt[maxn<<2];int bs(double key,int kind,double X[]){    int l = 0, r = kind-1, mid;    while(l <= r){        mid = (l+r)>>1;        if(X[mid] == key) return mid;        else if(X[mid] < key) l = mid+1;        else r = mid-1;    }    return -1;}void PushUp(int rt,int L,int R){    if(cnt[rt]) sum[rt] = X[R+1] - X[L];    else if(L == R) sum[rt] = 0;    else sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void update(int l,int r,int c,int L,int R,int rt){    if(l <= L && R <= r){        cnt[rt] += c;        PushUp(rt,L,R);        return;    }    int mid = (L+R)>>1;    if(l <= mid) update(l,r,c,L,mid,rt<<1);    if(mid < r) update(l,r,c,mid+1,R,rt<<1|1);    PushUp(rt,L,R);}int main(){    int n,m;    double x1,y1,x2,y2;    while(scanf("%d",&n)){        if(n == 0){            puts("*");            break;        }        m = 0;        for(int i = 0; i < n; i++){            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);            X[m] = x1;            ss[m++] = Seg(x1,x2,y1,1);            X[m] = x2;            ss[m++] = Seg(x1,x2,y2,-1);        }        sort(X,X+m);        sort(ss,ss+m);        int kind = 1;        for(int i = 1; i < m; i++){            if(X[i] != X[i-1]) X[kind++] = X[i];        }        memset(cnt,0,sizeof(cnt));        memset(sum,0,sizeof(sum));        double ans = 0;        for(int i = 0; i < m-1; i++){            int l = bs(ss[i].l,kind,X);            int r = bs(ss[i].r,kind,X) - 1;            if(l <= r){                update(l,r,ss[i].s,0,kind-1,1);            }            ans += sum[1]*(ss[i+1].h - ss[i].h);        }        printf("%.0lf\n",ans);    }    return 0;}