sgu Kalevich Strikes Back

这道题就是求一个大矩形被n个矩形划分成n+1个部分的面积，这些矩形之间不会相交，可能包含。。

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <vector>

  4 #include <algorithm>

  5 #define maxn 120100

  6 using namespace std;

  7 

  8 long long s[maxn];

  9 vector<int>g[maxn];

 10 int pre[maxn];

 11 int X[maxn];

 12 int n,m,w,h,x1,y1,x2,y2;

 13 struct node

 14 {

 15     int l,r;

 16     int ll,rr;

 17     int cover;

 18 }tree[maxn*4];

 19 

 20 struct line

 21 {

 22     int y,x1,x2,lr;

 23     bool operator <(const line &a)const

 24     {

 25         return y<a.y;

 26     }

 27 }p[maxn];

 28 

 29 void build(int i,int l,int r)

 30 {

 31     tree[i].l=l;

 32     tree[i].r=r;

 33     tree[i].ll=X[l];

 34     tree[i].rr=X[r];

 35     tree[i].cover=0;

 36     if(r-l==1) return ;

 37     int mid=(l+r)>>1;

 38     build(i<<1,l,mid);

 39     build(i<<1|1,mid,r);

 40 }

 41 void down(int i)

 42 {

 43     if(tree[i].l+1==tree[i].r) return;

 44     if(tree[i].cover!=-1)

 45     {

 46         tree[i<<1].cover=tree[i<<1|1].cover=tree[i].cover;

 47         tree[i].cover=-1;

 48     }

 49 }

 50 void update(int i, line a)

 51 {

 52     if(tree[i].ll==a.x1&&tree[i].rr==a.x2)

 53     {

 54         if(a.lr>0)

 55             tree[i].cover=a.lr;

 56         else

 57             tree[i].cover=pre[-a.lr];

 58         return ;

 59     }

 60     down(i);

 61     if(a.x2<=tree[i<<1].rr) update(i<<1,a);

 62     else if(a.x1>=tree[i<<1|1].ll) update(i<<1|1,a);

 63     else

 64     {

 65         line tmp=a;

 66         tmp.x2=tree[i<<1].rr;

 67         update(i<<1,tmp);

 68         tmp=a;

 69         tmp.x1=tree[i<<1|1].ll;

 70         update(i<<1|1,tmp);

 71     }

 72 }

 73 

 74 int search1(int i,line m)

 75 {

 76     if(tree[i].cover!=-1)

 77     {

 78         return tree[i].cover;

 79     }

 80     if(m.x2<=tree[i<<1].rr) return search1(i<<1,m);

 81     else if(m.x1>=tree[i<<1|1].ll) return search1(i<<1|1,m);

 82     else

 83     {

 84         line tmp;

 85         tmp=m;

 86         tmp.x2=tree[i<<1].rr;

 87         return search1(i<<1,tmp);

 88     }

 89 }

 90 void dfs(int u)

 91 {

 92     for(int i=0; i<(int)g[u].size(); i++)

 93     {

 94         int v=g[u][i];

 95         s[u]-=s[v];

 96         dfs(v);

 97     }

 98 }

 99 

100 int main()

101 {

102     while(scanf("%d",&n)!=EOF)

103     {

104         for(int i=0; i<=n; i++)

105         {

106             g[i].clear();

107         }

108         scanf("%d%d",&w,&h);

109         s[0]=(long long)w*h;

110         int t1=0;

111         for(int i=1; i<=n; i++)

112         {

113             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);

114             if(x1>x2)swap(x1,x2);

115             if(y1>y2)swap(y1,y2);

116             s[i]=(long long)(x2-x1)*(y2-y1);

117             p[t1].y=y1;

118             p[t1].x1=x1;

119             p[t1].x2=x2;

120             p[t1].lr=i;

121             X[t1++]=x1;

122             p[t1].y=y2;

123             p[t1].x1=x1;

124             p[t1].x2=x2;

125             p[t1].lr=-i;

126             X[t1++]=x2;

127         }

128         sort(X,X+t1);

129         t1=unique(X,X+t1)-X;

130         build(1,0,t1-1);

131         sort(p,p+2*n);

132         for(int i=0; i<2*n; i++)

133         {

134             if(p[i].lr>0)

135             {

136                 pre[p[i].lr]=search1(1,p[i]);

137                 g[pre[p[i].lr]].push_back(p[i].lr);

138             }

139             update(1,p[i]);

140         }

141         dfs(0);

142         sort(s,s+n+1);

143         for(int i=0; i<=n; i++)

144         {

145             if(i==n) printf("%I64d\n",s[i]);

146             else printf("%I64d ",s[i]);

147         }

148     }

149     return 0;

150 }

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