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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#472539#6414. Classical Maximization ProblemUESTC_xxx#WA 110ms8036kbC++201.9kb2024-07-11 17:03:062024-07-11 17:03:06

Judging History

你现在查看的是最新测评结果

  • [2024-07-11 17:03:06]
  • 评测
  • 测评结果:WA
  • 用时:110ms
  • 内存:8036kb
  • [2024-07-11 17:03:06]
  • 提交

answer

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
#include<stack>
#include<vector>
#include<map>
#define ll long long
#define lowbit(x) x&(-x)
using namespace std;
struct node{
	int x,y,d,id;
}p[500005];
map<ll,ll>mp;
vector<int>e[500005];
int a[500005],n,tt,g[500005],cx[50005],cy[50005];
bool use[500005];
bool cmp(int ca,int cb){
	return p[ca].d<p[cb].d;
}
bool cmp2(node a,node b){
	return a.d<b.d;
}
int main(){
	scanf("%d",&tt);
	while(tt--){
		scanf("%d",&n);
		int tot=0;
		for(int i=1;i<=2*n;++i){
			p[i].id=i;
			scanf("%d%d",&p[i].x,&p[i].y);
			if(!mp[p[i].x]) tot++,mp[p[i].x]=tot;
			if(!mp[p[i].y+1e10]) tot++,mp[p[i].y+1e10]=tot;
			e[mp[p[i].x]].push_back(i);
			e[mp[p[i].y+1e10]].push_back(i);
		}
		for(int i=1;i<=2*n;++i){
			p[i].d=e[mp[p[i].x]].size()+e[mp[p[i].y+1e10]].size();
		}
		for(int i=1;i<=tot;++i){
			for(int j=0;j<e[i].size();++j) a[j+1]=e[i][j];
			sort(a+1,a+e[i].size()+1,cmp);
			for(int j=0;j<e[i].size();++j) e[i][j]=a[j+1];
		}
		sort(p+1,p+n+1,cmp2);
		int cnt=0;
		for(int i=1;i<=2*n;++i){
			if(use[i]) continue;
			int u=mp[p[i].x],v=mp[p[i].y+1e10],p1=0,p2=0;
			for(int j=g[u];j<e[u].size();++j){
				g[u]=j;
				int k=e[u][j];
				if(use[k]||k==i) continue;
				p1=k;
				break;
			}
			for(int j=g[v];j<e[v].size();++j){
				g[v]=j;
				int k=e[v][j];
				if(use[k]||k==i) continue;
				p2=k;
				break;
			}
			if(p1==0&&p2==0) continue;
			cnt++;
			if(!p1) p1=p2;
			else if(p1&&p2) p1=(p[p1].d<=p[p2].d? p1 :p2);
			cx[cnt]=p[i].id,cy[cnt]=p[p1].id;
			use[p1]=1,use[i]=1;
		}
		printf("%d\n",cnt);
		int ls=0;
		for(int i=1;i<=2*n;++i){
			if(!use[i]){
				if(!ls) ls=i;
				else{
					cnt++;
					cx[cnt]=p[ls].id,cy[cnt]=p[i].id;
					ls=0;
				}
			}
			use[i]=0;
		}
		for(int i=1;i<=n;++i) printf("%d %d\n",cx[i],cy[i]);
		mp.clear();
		for(int i=1;i<=tot;++i) e[i].clear();
	}
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 7976kb

input:

3
2
0 0
0 1
1 0
1 1
2
0 0
0 1
0 2
0 3
2
0 0
1 1
2 2
3 3

output:

2
1 2
3 4
2
1 2
3 4
0
1 2
3 4

result:

ok ok (3 test cases)

Test #2:

score: -100
Wrong Answer
time: 110ms
memory: 8036kb

input:

10000
2
-107276936 -310501829
419434212 585811870
-65754386 -491212232
381152038 897148193
3
-474045168 493506332
299114415 540203303
165808153 983551
-506936261 -694189769
766718170 -725540031
975267148 -593051087
1
-818952276 -762387923
584023914 -612401389
6
-77701228 -266484128
659434465 6322062...

output:

0
1 2
3 4
0
1 2
3 4
5 6
0
1 2
0
1 2
3 4
5 6
7 8
9 10
11 12
0
1 2
3 4
5 6
7 8
9 10
11 12
13 14
0
1 2
16
26 9
18 1
19 16
20 15
21 14
22 13
23 12
24 11
25 10
17 33
27 8
28 7
29 6
30 5
31 4
32 3
2 34
35 36
37 38
39 40
41 42
43 44
45 46
47 48
49 50
51 52
53 54
55 56
57 58
59 60
61 62
63 64
65 66
0
1 2
3 ...

result:

wrong answer wrong number of friendly pairs: printed 16, but it is actually 0 (test case 7)