2017冬季24集训模拟-3.耀西岛

Posted 2020-08-31 sigongzi

——————————————————————————题解

路径的长度是1-200000

然后路径的条数有n*(n+1)/2

根据鸽巢原理n*(n+1)/2 > 200000就一定是YES

所以复杂度只有n^2

 1 #include <iostream>
 2 #include <queue>
 3 #include <set>
 4 #include <cstdio>
 5 #include <cstring>
 6 #include <vector>
 7 #include <algorithm>
 8 #include <cmath>
 9 #define siji(i,x,y) for(int i=x;i<=y;++i)
10 #define gongzi(j,x,y) for(int j=x;j>=y;--j)
11 #define xiaosiji(i,x,y) for(int i=x;i<y;++i)
12 #define sigongzi(j,x,y) for(int j=x;j>y;--j)
13 #define ivorysi
14 #define inf 0x3f3f3f3f
15 #define mo 97797977
16 #define ha 974711
17 #define ba 47
18 #define fi first
19 #define se second
20 #define pii pair<int,int>
21 typedef long long ll;
22 using namespace std;
23 int T,n,m;
24 int x[100005],y[100005];
25 bool check[200005],flag;
26 int dist(int a,int b) {
27     return abs(x[a]-x[b])+abs(y[a]-y[b]);
28 }
29 void init() {
30     scanf("%d%d",&n,&m);
31     siji(i,1,n) {
32         scanf("%d%d",&x[i],&y[i]);
33     }
34     flag=0;
35     memset(check,0,sizeof(check));
36 }
37 void solve() {
38     scanf("%d",&T);
39     while(T--){
40         init();
41         if(n>=30000) puts("YES");
42         siji(i,1,n) {
43             siji(j,i+1,n) {
44                 int x=dist(i,j);
45                 if(check[x]) {flag=1;goto suc;}
46                 else check[x]=1;
47             }
48         }
49         suc:
50         if(flag) puts("YES");
51         else puts("N0");
52     }
53 }
54 int main(int argc, char const *argv[])
55 {
56 #ifdef ivorysi
57     freopen("island.in","r",stdin);
58     freopen("island.out","w",stdout);
59 #else
60     freopen("f1.in","r",stdin);
61 #endif    
62     solve();
63     return 0;
64 }