2017 ACM-ICPC 亞洲區（南寧賽區）網路賽 Overlapping Rectangles 矩形並面積和

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Overlapping Rectangles

題意：求 n 個矩形並面積和

tags：掃描線+線段樹，模板題

參考部落格

#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  secondtypedef long long ll;const int N=1000+5;int col[N<<3], cnt, res;double X[N<<3], Sum[N<<3], Sum2[N<<3];  //多開一倍，因為 x座標有兩個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])    //覆蓋一次        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) // 覆蓋兩次以上        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]當前區間，[L,R]目標區間{    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座標過大的話，就離散化        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);      //二分左端點        int r = binary_find(S[i].r)-1;    // 左閉右開,二分右端點        update(l, r, S[i].flag, 1, 1, res);        ans += Sum[1]*(S[i+1].h-S[i].h);    //矩陣並        //ans += Sum2[1]*(S[i+1].h-S[i].h); //矩陣交集    }    return ans;}int main(){    int n;    while(scanf("%d", &n), n)    {        printf("%.0f\n", solve(n));    }    puts("*");    return 0;}

2017 ACM-ICPC 亞洲區（南寧賽區）網路賽 Overlapping Rectangles 矩形並面積和