POJ 1177

线段树+离散化+扫描线

可以只扫一遍竖线段（下面代码），也可以横竖都扫一遍（见http://archive.cnblogs.com/a/2197109/）

#include<cstdio>#include<algorithm>#define N 10010using namespace std;struct Tree{    int l,r,num,len,cover;   //num表示区间【l,r】内线段条数，len表示区间【l,r】被占用的总长度     bool lb,rb;}tree[3*N];struct Line{    int x,y1,y2,flag;}l[N];int y[N];bool cmp(struct Line a,struct Line b){    //这里表示当2条线段x坐标重合时优先处理是入边的矩形的线段    if(a.x==b.x)return a.flag>b.flag;    return a.x<b.x;}void build(int s,int t,int id){    tree[id].l=s,tree[id].r=t;    tree[id].cover=0,tree[id].lb=0,tree[id].rb=0;    tree[id].len=0,tree[id].num=0;    if(s!=t-1){        int mid=(s+t)>>1;        build(s,mid,id<<1);        build(mid,t,id<<1|1);    }}void update(int id){    if(tree[id].cover>0){        tree[id].num=tree[id].lb=tree[id].rb=1;        tree[id].len=y[tree[id].r-1]-y[tree[id].l-1];   //注意这里都要减一    }    else if(tree[id].l==tree[id].r-1){        tree[id].num=tree[id].lb=tree[id].rb=tree[id].len=0;    }    else{        tree[id].lb=tree[id<<1].lb;        tree[id].rb=tree[id<<1|1].rb;        tree[id].len=tree[id<<1].len+tree[id<<1|1].len;        tree[id].num=tree[id<<1].num+tree[id<<1|1].num-tree[id<<1|1].lb*tree[id<<1].rb; //这里合并时要注意    }}void query(int s,int t,int flag,int id){    if(tree[id].l==s && tree[id].r==t){        tree[id].cover+=flag;        update(id);        return;    }    int mid=(tree[id].l+tree[id].r)>>1;    if(t<=mid)query(s,t,flag,id<<1);    else if(s>=mid)query(s,t,flag,id<<1|1);    else{        query(s,mid,flag,id<<1);        query(mid,t,flag,id<<1|1);    }    update(id);}int main(){    int i,n,x1,x2,y1,y2;    scanf("%d",&n);    int cnt=0;    for(i=1;i<=n;i++){        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);        l[cnt].x=x1,l[cnt].y1=y1,l[cnt].y2=y2,l[cnt].flag=1,y[cnt++]=y1;        l[cnt].x=x2,l[cnt].y1=y1,l[cnt].y2=y2,l[cnt].flag=-1,y[cnt++]=y2;    }    sort(y,y+cnt);    sort(l,l+cnt,cmp);    int t=unique(y,y+cnt)-y;    build(1,t,1);    int ans=0,last=0,lines=0;    for(i=0;i<cnt;i++){        int ll=lower_bound(y,y+t,l[i].y1)-y+1;        int rr=lower_bound(y,y+t,l[i].y2)-y+1;        query(ll,rr,l[i].flag,1);        if(i)ans+=2*lines*(l[i].x-l[i-1].x);   //计算平行于x轴的长度        ans+=abs(tree[1].len-last);            //计算平行于y轴的长度        last=tree[1].len;        lines=tree[1].num;    }    printf("%d\n",ans);    return 0;}