hdu 4419 Colourful Rectangle

http://acm.hdu.edu.cn/showproblem.php?pid=4419

题意：给出3种颜色，重叠会生成新的颜色，然后有一些矩形，求出每种颜色的面积。

转化为二进制表示颜色：001 R ，010G，100B，011RG,101RB,....111RGB;

在结构体里面加上一个len[8]和cover[8]表示每种颜色所占的长度和在区间的覆盖次数。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define maxn 100010
 #define ll __int64
 using namespace std;

 int t,n;
 char ch;
 ll Y[maxn];
 struct node1
 {
     ll x,y1,y2;
     int lr,c;
     bool operator <(const node1 &a)const
     {
         return (x<a.x)||(x==a.x&&lr>a.lr);
     }
 }p[maxn];
 struct node
 {
     int l,r;
     ll len[];
     int cover[];
 } tree[maxn*];

 void build(int i,int l,int r)
 {
     tree[i].l=l;
     tree[i].r=r;
     for(int j=; j<; j++)
     {
         tree[i].cover[j]=tree[i].len[j]=;
     }
     if(l==r-) return;
     int mid=(l+r)>>;
     build(i<<,l,mid);
     build(i<<|,mid,r);
 }

 void update(int i,int l,int r,int lr,int c)
 {
     if(tree[i].l==l&&tree[i].r==r)
     {
         tree[i].cover[c]+=lr;
         if(tree[i].cover[c])
             tree[i].len[c]=Y[tree[i].r]-Y[tree[i].l];
         else if(tree[i].r-==tree[i].l)
             tree[i].len[c]=;
         else
             tree[i].len[c]=tree[i<<].len[c]+tree[i<<|].len[c];
         return ;
     }
     int mid=(tree[i].r+tree[i].l)>>;
     if(r<=mid)
     {
         update(i<<,l,r,lr,c);
     }
     else if(l>=mid)
     {
         update(i<<|,l,r,lr,c);
     }
     else
     {
         update(i<<,l,mid,lr,c);
         update(i<<|,mid,r,lr,c);
     }
     if(tree[i].cover[c])
         tree[i].len[c]=Y[tree[i].r]-Y[tree[i].l];
     else if(tree[i].r-==tree[i].l)
         tree[i].len[c]=;
     else
         tree[i].len[c]=tree[i<<].len[c]+tree[i<<|].len[c];
 }

 int main()
 {
     scanf("%d",&t);
     int cas=;
     while(t--)
     {
         scanf("%d",&n);
         getchar();
         int t1=;
         for(int i=; i<=n; i++)
         {
             ll x1,y1,x2,y2;
             scanf("%c %I64d%I64d%I64d%I64d",&ch,&x1,&y1,&x2,&y2);
             if(ch=='R')
             {
                 p[t1].c=; p[t1+].c=;
             }
             else if(ch=='G')
             {
                 p[t1].c=; p[t1+].c=;
             }
             else if(ch=='B')
             {
                 p[t1].c=; p[t1+].c=;
             }
             p[t1].x=x1; p[t1].y1=y1; p[t1].y2=y2;p[t1].lr=;Y[t1++]=y1;
             p[t1].x=x2; p[t1].y1=y1; p[t1].y2=y2;p[t1].lr=-;Y[t1++]=y2;
             getchar();
         }
         sort(Y,Y+t1);
         int cnt=unique(Y,Y+t1)-Y;
         sort(p,p+t1);
         build(,,cnt-);
         ll s[maxn]={};
         for(int i=; i<t1; i++)
         {
             int l1=lower_bound(Y,Y+cnt,p[i].y1)-Y;
             int rr=lower_bound(Y,Y+cnt,p[i].y2)-Y;
             for(int j=; j<; j++)
             {
                 if(p[i].c&j) update(,l1,rr,p[i].lr,j);
                 if(i+<t1) s[j]+=tree[].len[j]*(p[i+].x-p[i].x);
             }
         }
         printf("Case %d:\n",cas);
         cas++;
         printf("%I64d\n",s[]-s[]);
         printf("%I64d\n",s[]-s[]);
         printf("%I64d\n",s[]-s[]);
         printf("%I64d\n",s[]+s[]-s[]-s[]);
         printf("%I64d\n",s[]+s[]-s[]-s[]);
         printf("%I64d\n",s[]+s[]-s[]-s[]);
         printf("%I64d\n",s[]+s[]+s[]-s[]-s[]-s[]+s[]);
     }
     return ;
 }