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  • hdu1542 Atlantis(扫描线+线段树+离散)矩形相交面积

    题目链接:点击打开链接

    题目描写叙述:给定一些矩形,求这些矩形的总面积。假设有重叠。仅仅算一次


    解题思路:扫描线+线段树+离散(代码从上往下扫描)

    代码:

    #include<cstdio>
    #include <algorithm>
    #define MAXN 110
    #define LL ((rt<<1)+1)
    #define RR ((rt<<1)+2)
    using namespace std;
    int n;
    struct segment{
        double l,r,h;
        int f;
        bool operator<(const segment& b)const{
            return h>b.h;
        }
    }sg[2*MAXN];
    double pos[2*MAXN];
    int id;
    void addSegment(double x1,double y1,double x2,double y2){
        sg[id].l=x1;sg[id].r=x2;
        sg[id].h=y1;sg[id].f=1;
        pos[id++]=x1;
        sg[id].l=x1;sg[id].r=x2;
        sg[id].h=y2;sg[id].f=-1;
        pos[id++]=x2;
    }
    int binary(double key,int low,int high){
        while(low<=high){
            int mid=(low+high)/2;
            if(pos[mid]==key)
                return mid;
            else if(key<pos[mid])
                high=mid-1;
            else
                low=mid+1;
        }
        return -1;
    }
    struct Tree{
        int l,r;
        int cover;
        double len;
    }tree[8*MAXN];
    void build(int rt,int l,int r){
        tree[rt].l=l;
        tree[rt].r=r;
        tree[rt].cover=0;
        tree[rt].len=0;
        if(l==r-1)
            return;
        int mid=(l+r)>>1;
        build(LL,l,mid);
        build(RR,mid,r);
    }
    void pushup(int rt){
        if(tree[rt].cover)
            tree[rt].len=pos[tree[rt].r]-pos[tree[rt].l];
        else if(tree[rt].l==tree[rt].r-1)
            tree[rt].len=0;
        else
            tree[rt].len=tree[LL].len+tree[RR].len;
    }
    void update(int rt,int l,int r,int f){
        if(tree[rt].l==l&&tree[rt].r==r){
            tree[rt].cover+=f;
            pushup(rt);
            return;
        }
        int mid=(tree[rt].l+tree[rt].r)>>1;
        if(r<=mid)
            update(LL,l,r,f);
        else if(l>=mid)
            update(RR,l,r,f);
        else{
            update(LL,l,mid,f);
            update(RR,mid,r,f);
        }
        pushup(rt);
    }
    int main(){
        int Case=0;
        while(scanf("%d",&n)!=EOF&&n!=0){
            id=0;
            double x1,y1,x2,y2;
            for(int i=0;i<n;++i){
                scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
                addSegment(x1,y1,x2,y2);
            }
            n=(n<<1);
            sort(sg,sg+n);
            sort(pos,pos+n);
            int m=1;
            for(int i=1;i<n;++i)
                if(pos[i]!=pos[i-1])
                    pos[m++]=pos[i];
            build(0,0,m-1);
            double ans=0;
            int l=binary(sg[0].l,0,m-1);
            int r=binary(sg[0].r,0,m-1);
            update(0,l,r,sg[0].f);
            for(int i=1;i<n;i++){
                ans+=(sg[i-1].h-sg[i].h)*tree[0].len;
                l=binary(sg[i].l,0,m-1);
                r=binary(sg[i].r,0,m-1);
                update(0,l,r,sg[i].f);
            }
            printf("Test case #%d
    ",++Case);
            printf("Total explored area: %.2f
    
    ",ans);
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/brucemengbm/p/7107406.html
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