CodeForces Round 693 E
技術標籤：Acm 貪心
連結:
https://codeforces.ml/contest/1472/problem/E
題目大意:
有T(1
≤
\leq
≤
T
T
T
≤
\leq
≤
1
0
4
10^4
104)組資料,你有N(1
≤
\leq
≤
N
N
N
≤
\leq
≤
2
∗
1
0
5
2*10^5
2∗105)張照片.每張照片給定
w
i
,
h
i
w_i,h_i
wi,hi(1
≤
\leq
≤
w
i
,
h
i
w_i,h_i
wi,hi
≤
\leq
≤
1
0
9
10^9
109)表示寬度和高度.定義兩張照片
i
,
j
i,j
i,j,j包含i的含義是:
w
i
<
w
j
,
h
i
<
h
j
w_i<w_j,h_i<h_j
分析:
每張照片
i
i
i我們可以根據大小把令
w
i
=
m
i
n
(
w
i
,
h
i
)
,
h
i
=
m
a
x
(
w
i
,
h
i
)
w_i=min(w_i,h_i),h_i=max(w_i,h_i)
#include<bits/stdc++.h>
#define MAXN 200005
using namespace std;
int n,Min[MAXN],Minval[MAXN];
struct Data{
int x,y,id,ans;
}a[MAXN];
bool Cmp_(Data Cmp1,Data Cmp2){
return(Cmp1.y<Cmp2.y)||(Cmp1.y==Cmp2.y&&Cmp1.x<Cmp2.x);
}
bool Cmp(Data Cmp1,Data Cmp2){
return(Cmp1.id<Cmp2.id);
}
int main(){
int Tmp;
cin>>Tmp;
while(Tmp--){
cin>>n;
for(int i=1;i<=n;++i){
int xx,yy;
cin>>xx>>yy;
a[i].x=min(xx,yy);
a[i].y=max(xx,yy);
a[i].id=i;
a[i].ans=-1;
}
sort(a+1,a+n+1,Cmp_);
Min[1]=a[1].x;
Minval[1]=a[1].id;
for(int i=2;i<=n;++i){
if(Min[i-1]<a[i].x){
Minval[i]=Minval[i-1];
Min[i]=Min[i-1];
}
else{
Minval[i]=a[i].id;
Min[i]=a[i].x;
}
}
int l=0;
Min[0]=INT_MAX;
for(int i=2;i<=n;++i){
if(a[i].y>a[i-1].y)l=i-1;
if(Min[l]<a[i].x)a[i].ans=Minval[l];
}
sort(a+1,a+n+1,Cmp);
for(int i=1;i<=n;++i)
cout<<a[i].ans<<" ";
puts("");
}
return 0;
}
時間複雜度O( ∑ i = 1 T N i l o g N i \displaystyle\sum_{i=1}^T N_ilogN_i i=1∑TNilogNi)