hdu4419Colourful Rectangle

链接

分别求出7种颜色覆盖的面积。

做法：每种颜色设定一个标号，以二进制表示R：100 G：010 B：001 。这样很明显可以知道RG：110 GB：011 以此类推。

求解时，需要开一个二维标记数组，标记了这一段的某种颜色被标记了几次，然后类似状压的方式求解。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 #include<map>
 using namespace std;
 #define N 41010
 #define LL __int64
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 map<int,int>f;
 int s[N<<][];
 int fs[N<<],a[N],ff[N<<][];
 int val[N];
 int id[];
 LL num[N];
 struct node
 {
     int x1,x2,y;
     int f;
     node(){}
     node(int x1,int x2,int y,int f):x1(x1),x2(x2),y(y),f(f){}
     bool operator <(const node &S)const
     {
         return y<S.y;
     }
 }p[N];
 void up(int w,int l,int r)
 {
     int i;
     if(fs[w]==)
     {
         for(i = ; i <=  ; i++)
         {
             if(l==r)
             s[w][i] = ;
             else
             s[w][i] = s[w<<][i]+s[w<<|][i];
         }
     }
     else
     {
         int res = val[r+]-val[l];
         int cnt[] = {};
         for(i = ; i <= ; i++)
         {
             if(l==r)
             {
                 if(i==fs[w])
                 cnt[i] += val[r+]-val[l];
                 else
                 cnt[i] += ;
             }
             else
             {
                 cnt[i|fs[w]] += s[w<<][i]+s[w<<|][i];
             }
         }
         for(i = ; i <= ; i++)
         res-=cnt[i];
         cnt[fs[w]]+=res;
         for(i = ; i <=  ; i++)
         s[w][i] = cnt[i];
     }
 }
 void build(int l,int r,int w)
 {
     int i;
     for(i = ;i<= ; i++)
     {s[w][i] = ff[w][i] = ;}
     fs[w] = ;
     if(l==r)
     {
         return ;
     }
     int m = (l+r)>>;
     build(l,m,w<<);
     build(m+,r,w<<|);
 }
 void update(int a,int b,int d,int l,int r,int w)
 {
     if(a<=l&&b>=r)
     {
         ff[w][abs(d)]+=d/abs(d);
         if(d>)
         fs[w]|=d;
         else if(ff[w][-d]==) fs[w]+=d;
         up(w,l,r);
         return ;
     }
     int m = (l+r)>>;
     if(a<=m) update(a,b,d,l,m,w<<);
     if(b>m) update(a,b,d,m+,r,w<<|);
     up(w,l,r);
 }
 int main()
 {
     id['R'] = ;
     id['G'] = ;
     id['B'] = ;
     int i,j,t,n,cas = ;
     char sr[];
     cin>>t;
     while(t--)
     {
         f.clear();
         scanf("%d",&n);
         memset(num,,sizeof(num));
         int g=  ,dd[];
         for(i = ; i <= n; i++)
         {
             int x1,x2,y1,y2,k;
             scanf("%s%d%d%d%d",sr,&x1,&y1,&x2,&y2);
             if(sr[]=='R') k = id['R'];
             else if(sr[]=='G') k = id['G'];
             else k = id['B'];
             p[++g] = node(x1,x2,y1,k);
             a[g] = x1;
             p[++g] = node(x1,x2,y2,-k);
             a[g] = x2;
         }
         sort(a+,a+g+);
         sort(p+,p+g+);
         int o = ;
         f[a[]] = ++o;
         val[o] = a[];
         for(i = ; i <= g ; i++)
         if(a[i]!=a[i-])
         {
             f[a[i]] = ++o;
             val[o] = a[i];
         }
         build(,o-,);

         for(i = ; i < g; i++)
         {
             int l = f[p[i].x1];
             int r = f[p[i].x2]-;        //cout<<l<<" "<<r<<endl;
             if(l<=r)
             {
                 update(l,r,p[i].f,,o-,);
             }
             for(j = ; j <= ;  j++)
             num[j] += (LL)(p[i+].y-p[i].y)*s[][j];
         }
         dd[] = ,dd[] = ,dd[] = ,dd[] = ,dd[] = ,dd[] = ,dd[] = ;
         printf("Case %d:\n",cas++);
         for(i = ; i <= ; i++)
         printf("%I64d\n",num[dd[i]]);
     }
     return ;
 }