题意：给定一个n*m的平面，有k条垂直或平行的直线，问将平面分成了几个互不联通的部分
n，m<=1e9，k<=1e5
思路：
刻在DNA里的二维数点
1 #include<bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 typedef unsigned int uint;
5 typedef unsigned long long ull;
6 typedef pair<int,int> PII;
7 typedef pair<ll,ll> Pll;
8 typedef vector<int> VI;
9 typedef vector<PII> VII;
10 #define N 410000
11 #define M 4100000
12 #define fi first
13 #define se second
14 #define MP make_pair
15 #define pi acos(-1)
16 #define mem(a,b) memset(a,b,sizeof(a))
17 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
18 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
19 #define lowbit(x) x&(-x)
20 #define Rand (rand()*(1<<16)+rand())
21 #define id(x) ((x)<=B?(x):m-n/(x)+1)
22 #define ls p<<1
23 #define rs p<<1|1
24
25 const ll MOD=1e9+7,inv2=(MOD+1)/2;
26 double eps=1e-6;
27 int INF=1e9;
28 int da[4]={-1,1,0,0};
29 int db[4]={0,0,-1,1};
30
31 char ch[N][2];
32 int t[N<<2],x[N],y[N],c[N],p;
33
34 struct arr1
35 {
36 int t,x,y;
37 }a[N];
38
39 bool cmp1(arr1 a,arr1 b)
40 {
41 return a.t<b.t;
42 }
43
44 struct arr2
45 {
46 int x1,x2,y;
47 }b[N];
48
49 bool cmp2(arr2 a,arr2 b)
50 {
51 return a.y<b.y;
52 }
53
54 int read()
55 {
56 int v=0,f=1;
57 char c=getchar();
58 while(c<48||57<c) {if(c==‘-‘) f=-1; c=getchar();}
59 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
60 return v*f;
61 }
62
63 int lisan(int x)
64 {
65 int l=1,r=p,last=1;
66 while(l<=r)
67 {
68 int mid=(l+r)>>1;
69 if(c[mid]>x) r=mid-1;
70 if(c[mid]==x){last=mid; r=mid-1;}
71 if(c[mid]<x) l=mid+1;
72 }
73 return last;
74 }
75
76 void build(int l,int r,int p)
77 {
78 t[p]=0;
79 if(l==r) return;
80 int mid=(l+r)>>1;
81 build(l,mid,ls);
82 build(mid+1,r,rs);
83 }
84
85 int query(int l,int r,int x,int y,int p)
86 {
87 if(x<=l&&r<=y) return t[p];
88 int mid=(l+r)>>1;
89 int s=0;
90 if(x<=mid) s+=query(l,mid,x,y,ls);
91 if(y>mid) s+=query(mid+1,r,x,y,rs);
92 return s;
93 }
94
95 void update(int l,int r,int x,int v,int p)
96 {
97 if(l==r)
98 {
99 t[p]+=v;
100 return;
101 }
102 int mid=(l+r)>>1;
103 if(x<=mid) update(l,mid,x,v,ls);
104 else update(mid+1,r,x,v,rs);
105 t[p]=t[ls]+t[rs];
106 }
107
108 int main()
109 {
110 //freopen("1.in","r",stdin);
111 //freopen("1.out","w",stdout);
112
113 int cas;
114 scanf("%d",&cas);
115
116 while(cas--)
117 {
118 int n=read(),m=read(),K=read();
119 ll ans=0;
120 int m1=0,m2=0;
121 p=0;
122 rep(i,1,K)
123 {
124 x[i]=read(),y[i]=read();
125 scanf("%s",ch[i]+1);
126 //if(ch[i][1]==‘U‘&&y[i]==) ans++;
127 //if(ch[i][1]==‘R‘&&x[i]==1) ans++;
128 c[++p]=x[i];
129 c[++p]=y[i];
130 }
131 //printf("ans=%I64d\n",ans);
132 c[++p]=n;
133 c[++p]=m;
134 c[++p]=1;
135 sort(c+1,c+p+1);
136 rep(i,1,K)
137 {
138 x[i]=lisan(x[i]);
139 y[i]=lisan(y[i]);
140 }
141 n=lisan(n),m=lisan(m);
142 rep(i,1,K)
143 {
144 if(ch[i][1]==‘U‘)
145 {
146 m1++;
147 a[m1].t=y[i];
148 a[m1].x=x[i];
149 a[m1].y=1;
150 //a[m1].x=x[i];
151 //a[m1].y1=y[i];
152 //a[m1].y2=m;
153 }
154 if(ch[i][1]==‘D‘)
155 {
156 m1++;
157 a[m1].t=1;
158 a[m1].x=x[i];
159 a[m1].y=1;
160
161 m1++;
162 a[m1].t=y[i]+1;
163 a[m1].x=x[i];
164 a[m1].y=-1;
165 //a[m1].x=x[i];
166 //a[m1].y1=1;
167 //a[m1].y2=y[i];
168 }
169 if(ch[i][1]==‘L‘)
170 {
171 m2++;
172 b[m2].x1=1;
173 b[m2].x2=x[i];
174 b[m2].y=y[i];
175 }
176 if(ch[i][1]==‘R‘)
177 {
178 m2++;
179 b[m2].x1=x[i];
180 b[m2].x2=n;
181 b[m2].y=y[i];
182 }
183 }
184 sort(a+1,a+m1+1,cmp1);
185 sort(b+1,b+m2+1,cmp2);
186 build(1,p,1);
187 int j1=1,j2=1;
188 rep(i,1,p)
189 {
190 while(j1<=m1&&a[j1].t==i)
191 {
192 update(1,p,a[j1].x,a[j1].y,1);
193 j1++;
194 }
195 while(j2<=m2&&b[j2].y==i)
196 {
197 ans+=query(1,p,b[j2].x1,b[j2].x2,1);
198 j2++;
199 }
200 }
201 printf("%I64d\n",ans+1);
202
203
204 }
205
206 return 0;
207 }
原文地址：https://www.cnblogs.com/myx12345/p/11666961.html
时间： 2024-10-22 15:00:42