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]>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