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hdu3642Get The Treasury

2024-07-07 08:40:58   223人阅读

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]>2)
    {
        s[w][0] = s[w][1] = s[w][2] = val[r+1] - val[l];
    }
    else if(fs[w]>1)
    {
        s[w][0]  = s[w][1] = val[r+1]-val[l];
        if(l==r)
        s[w][2] = 0;
        else s[w][2] = s[w<<1][0]+s[w<<1|1][0];
    }
    else if(fs[w])
    {
        s[w][0] = val[r+1]-val[l];
        if(l==r)
        s[w][1] = s[w][2] = 0;
        else
        {
            s[w][2] = s[w<<1][1]+s[w<<1|1][1];
            s[w][1] =s[w<<1][0]+s[w<<1|1][0];
        }
    }
    else
    {
        if(l==r)
        s[w][0] = s[w][1] = s[w][2] = 0;
        else
        {
            s[w][0] = s[w<<1][0]+s[w<<1|1][0];
            s[w][1] = s[w<<1][1]+s[w<<1|1][1];
            s[w][2] = s[w<<1][2]+s[w<<1|1][2];
        }
    }
}

 

代码：

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 #include<map>
 11 using namespace std;
 12 #define N 2010
 13 #define LL __int64
 14 #define INF 0xfffffff
 15 const double eps = 1e-8;
 16 const double pi = acos(-1.0);
 17 const double inf = ~0u>>2;
 18 map<int,int>f;
 19 struct node
 20 {
 21     int x1,x2,y,f;
 22     int z1,z2;
 23     node(){}
 24     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){}
 25     bool operator < (const node &S) const
 26     {
 27         return y<S.y;
 28     }
 29 }p[N],q[N];
 30 int a[N],val[N];
 31 int s[N<<2][3],fs[N<<2];
 32 void up(int w,int l,int r)
 33 {
 34     if(fs[w]>2)
 35     {
 36         s[w][0] = s[w][1] = s[w][2] = val[r+1] - val[l];
 37     }
 38     else if(fs[w]>1)
 39     {
 40         s[w][0]  = s[w][1] = val[r+1]-val[l];
 41         if(l==r)
 42         s[w][2] = 0;
 43         else s[w][2] = s[w<<1][0]+s[w<<1|1][0];
 44     }
 45     else if(fs[w])
 46     {
 47         s[w][0] = val[r+1]-val[l];
 48         if(l==r)
 49         s[w][1] = s[w][2] = 0;
 50         else
 51         {
 52             s[w][2] = s[w<<1][1]+s[w<<1|1][1];
 53             s[w][1] =s[w<<1][0]+s[w<<1|1][0];
 54         }
 55     }
 56     else
 57     {
 58         if(l==r)
 59         s[w][0] = s[w][1] = s[w][2] = 0;
 60         else
 61         {
 62             s[w][0] = s[w<<1][0]+s[w<<1|1][0];
 63             s[w][1] = s[w<<1][1]+s[w<<1|1][1];
 64             s[w][2] = s[w<<1][2]+s[w<<1|1][2];
 65         }
 66     }
 67 }
 68 void build(int l,int r,int w)
 69 {
 70     s[w][0] = s[w][1] = s[w][2] = 0;
 71     fs[w] = 0;
 72     if(l==r)
 73     return ;
 74     int m = (l+r)>>1;
 75     build(l,m,w<<1);
 76     build(m+1,r,w<<1|1);
 77     up(w,l,r);
 78 }
 79 void update(int a,int b,int d,int l,int r,int w)
 80 {
 81    // cout<<l<<" "<<r<<" "<<w<<endl;
 82     if(a<=l&&b>=r)
 83     {
 84         fs[w]+=d;
 85         //cout<<l<<" "<<r<<" "<<fs[w]<<" "<<w<<endl;
 86         up(w,l,r);
 87         return ;
 88     }
 89     int m = (l+r)>>1;
 90     if(a<=m) update(a,b,d,l,m,w<<1);
 91     if(b>m) update(a,b,d,m+1,r,w<<1|1);
 92     up(w,l,r);
 93 }
 94 int main()
 95 {
 96     int n,i,j;
 97     int t,kk=0;
 98     cin>>t;
 99     while(t--)
100     {
101         scanf("%d",&n);
102         f.clear();
103         int g = 0;
104         for(i = 1 ;i <= n; i++)
105         {
106             int x1,x2,y1,y2,z1,z2;
107             scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
108             p[++g] = node(x1,x2,y1,1,z1,z2);
109             a[g] = x1;
110             p[++g] = node(x1,x2,y2,-1,z1,z2);
111             a[g] = x2;
112         }
113         sort(a+1,a+g+1);
114         sort(p+1,p+g+1);
115         int o = 0;
116         f[a[1]] = ++o;
117         val[1] = a[1];
118         for(i = 2; i <= g; i++)
119         if(a[i]!=a[i-1])
120         {
121             f[a[i]] = ++o;
122             val[o] = a[i];
123         }
124         LL ans = 0;
125         for(i = -500 ; i < 500 ; i++)
126         {
127             int e = 0;
128             build(1,o-1,1);
129             for(j = 1; j <= g ;j++)
130             {
131                 if(p[j].z1>i||p[j].z1>i+1||i>p[j].z2||p[j].z2<i+1)
132                 continue;
133                 q[++e] = p[j];
134             }
135             for(j = 1; j < e;  j++)
136             {
137                 int l = f[q[j].x1];
138                 int r = f[q[j].x2]-1;
139                // cout<<q[i].f<<" "<<i<<endl;
140                 if(l<=r)
141                 {
142                     update(l,r,q[j].f,1,o-1,1);
143                 }
144                 LL sum = (LL)(q[j+1].y-q[j].y)*s[1][2];
145                 //cout<<sum<<" ."<<i<<" "<<s[1][2]<<endl;
146                 ans+=sum;
147             }
148         }
149         printf("Case %d: %I64d\n",++kk,ans);
150     }
151     return 0;
152 }
View Code

 


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