POJ1151 HDU1542 CODEVS3044 Atlantis 题解&代码

矩形面积求并
思路：离散化之后按x轴or按y轴刷扫描线，用线段树来表示区间范围。

**被POJ的%.2f输出坑了一个多小时的我并没有心情仔细写题解，有机会再补吧

#include
#include
#include
#include
#define lson (o<<1)
#define rson ((o<<1)|1)
using namespace std;
map<double,int> hx,hy;
map<int,int> vis;
const int maxn=105;
int T,n,p[maxn*2],val[maxn*2],tot,cnt,flag[maxn*2*4];
double x1,x2,y1,y2,ys[maxn*2],ans,st;
struct line{
    int v;
    double x,l,r;
};
line d[maxn*2],temp;
void build(int o,int l,int r)
{
    //cout<
    flag[o]=0;
    if(l==r)
    {
        val[l]=0;
        return;
    }
    int mid=(l+r)/2;
    build(lson,l,mid);
    build(rson,mid+1,r);
}
void pushdown(int o,int l,int r)
{
    if(l==r)val[l]+=flag[o];
    else
    {
        flag[lson]+=flag[o];
        flag[rson]+=flag[o];
    }
    flag[o]=0;
}
void addval(int o,int l,int r,int tx)
{
    pushdown(o,l,r);
    if(l>=hy[d[tx].r] || rreturn;
    if(l>=hy[d[tx].l] && rreturn;
    }
    int mid=(l+r)/2;
    addval(lson,l,mid,tx);
    addval(rson,mid+1,r,tx);
}
double getsum(int o,int l,int r)
{
    pushdown(o,l,r);
    if(l==r)
        return val[l]?ys[l+1]-ys[l]:0.0;
    int mid=(l+r)/2;
    return getsum(lson,l,mid)+getsum(rson,mid+1,r);
}
bool cmp1(int a,int b)
{
    return d[a].xbool cmp2(double a,double b)
{
    return aint main(void)
{
    while(scanf("%d",&n))
    {
        if(!n)break;
        build(1,1,2*maxn-1);
        hx.clear();
        hy.clear();
        vis.clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            temp.l=y1;
            temp.r=y2;
            temp.x=x1;
            temp.v=1;
            d[i]=temp;
            temp.x=x2;
            temp.v=-1;
            d[i+n]=temp;
            p[i]=i;
            p[i+n]=i+n;
            ys[i]=y1;
            ys[i+n]=y2;
        }
        tot=0;
        sort(p+1,p+n*2+1,cmp1);
        for(int i=1;i<=2*n;i++)
            if(!hx[d[p[i]].x])hx[d[p[i]].x]=++tot;
        cnt=0;
        sort(ys+1,ys+2*n+1,cmp2);
        for(int i=1;i<=2*n;i++)
            if(!hy[ys[i]])hy[ys[i]]=++cnt,ys[cnt]=ys[i];
        st=d[p[1]].x;ans=0;
        for(int i=1;i<=2*n;i++)
        {
            if(!vis[p[i]])
            {
                vis[p[i]]=1;
                ans+=(d[p[i]].x-st)*getsum(1,1,cnt-1);
                //cout<
                st=d[p[i]].x;
            }
            addval(1,1,cnt-1,p[i]);
        }
        printf("Test case #%d\n",++T);
        printf("Total explored area: %.2f\n\n",ans);
    }
}