hdu3642Get The Treasury

链接

刚开始看n挺小，以为是二维的线段树，想了一会也没想到怎么解，之后看到z值非常小，想到可以直接枚举z，确定一个坐标，然后把三维转化为二维，把体积转化为面。

枚举z从-500到500，然后用面积并的解法求出单位z坐标上满足题意的面积。

把1写成了L，查错查了好久。其余还好，1A。

求覆盖超过两次的面积，up更新上的写法如下：

void up(int w,int l,int r)
{
    if(fs[w]>)
    {
        s[w][] = s[w][] = s[w][] = val[r+] - val[l];
    }
    else if(fs[w]>)
    {
        s[w][]  = s[w][] = val[r+]-val[l];
        if(l==r)
        s[w][] = ;
        else s[w][] = s[w<<][]+s[w<<|][];
    }
    else if(fs[w])
    {
        s[w][] = val[r+]-val[l];
        if(l==r)
        s[w][] = s[w][] = ;
        else
        {
            s[w][] = s[w<<][]+s[w<<|][];
            s[w][] =s[w<<][]+s[w<<|][];
        }
    }
    else
    {
        if(l==r)
        s[w][] = s[w][] = s[w][] = ;
        else
        {
            s[w][] = s[w<<][]+s[w<<|][];
            s[w][] = s[w<<][]+s[w<<|][];
            s[w][] = s[w<<][]+s[w<<|][];
        }
    }
}

代码：

 #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 2010
 #define LL __int64
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 map<int,int>f;
 struct node
 {
     int x1,x2,y,f;
     int z1,z2;
     node(){}
     node(int x1,int x2,int y,int f,int z1,int z2):x1(x1),x2(x2),y(y),f(f),z1(z1),z2(z2){}
     bool operator < (const node &S) const
     {
         return y<S.y;
     }
 }p[N],q[N];
 int a[N],val[N];
 int s[N<<][],fs[N<<];
 void up(int w,int l,int r)
 {
     if(fs[w]>)
     {
         s[w][] = s[w][] = s[w][] = val[r+] - val[l];
     }
     else if(fs[w]>)
     {
         s[w][]  = s[w][] = val[r+]-val[l];
         if(l==r)
         s[w][] = ;
         else s[w][] = s[w<<][]+s[w<<|][];
     }
     else if(fs[w])
     {
         s[w][] = val[r+]-val[l];
         if(l==r)
         s[w][] = s[w][] = ;
         else
         {
             s[w][] = s[w<<][]+s[w<<|][];
             s[w][] =s[w<<][]+s[w<<|][];
         }
     }
     else
     {
         if(l==r)
         s[w][] = s[w][] = s[w][] = ;
         else
         {
             s[w][] = s[w<<][]+s[w<<|][];
             s[w][] = s[w<<][]+s[w<<|][];
             s[w][] = s[w<<][]+s[w<<|][];
         }
     }
 }
 void build(int l,int r,int w)
 {
     s[w][] = s[w][] = s[w][] = ;
     fs[w] = ;
     if(l==r)
     return ;
     int m = (l+r)>>;
     build(l,m,w<<);
     build(m+,r,w<<|);
     up(w,l,r);
 }
 void update(int a,int b,int d,int l,int r,int w)
 {
    // cout<<l<<" "<<r<<" "<<w<<endl;
     if(a<=l&&b>=r)
     {
         fs[w]+=d;
         //cout<<l<<" "<<r<<" "<<fs[w]<<" "<<w<<endl;
         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()
 {
     int n,i,j;
     int t,kk=;
     cin>>t;
     while(t--)
     {
         scanf("%d",&n);
         f.clear();
         int g = ;
         for(i =  ;i <= n; i++)
         {
             int x1,x2,y1,y2,z1,z2;
             scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
             p[++g] = node(x1,x2,y1,,z1,z2);
             a[g] = x1;
             p[++g] = node(x1,x2,y2,-,z1,z2);
             a[g] = x2;
         }
         sort(a+,a+g+);
         sort(p+,p+g+);
         int o = ;
         f[a[]] = ++o;
         val[] = a[];
         for(i = ; i <= g; i++)
         if(a[i]!=a[i-])
         {
             f[a[i]] = ++o;
             val[o] = a[i];
         }
         LL ans = ;
         for(i = - ; i <  ; i++)
         {
             int e = ;
             build(,o-,);
             for(j = ; j <= g ;j++)
             {
                 if(p[j].z1>i||p[j].z1>i+||i>p[j].z2||p[j].z2<i+)
                 continue;
                 q[++e] = p[j];
             }
             for(j = ; j < e;  j++)
             {
                 int l = f[q[j].x1];
                 int r = f[q[j].x2]-;
                // cout<<q[i].f<<" "<<i<<endl;
                 if(l<=r)
                 {
                     update(l,r,q[j].f,,o-,);
                 }
                 LL sum = (LL)(q[j+].y-q[j].y)*s[][];
                 //cout<<sum<<" ."<<i<<" "<<s[1][2]<<endl;
                 ans+=sum;
             }
         }
         printf("Case %d: %I64d\n",++kk,ans);
     }
     return ;
 }