C - Oranges and Apples CodeForces - 23C
题目链接:http://codeforces.com/problemset/problem/23/C
In 2N - 1 boxes there are apples and oranges. Your task is to choose N boxes so, that they will contain not less than half of all the apples and not less than half of all the oranges.
Input
The first input line contains one number T — amount of tests. The description of each test starts with a natural number N — amount of boxes. Each of the following 2N - 1 lines contains numbers ai and oi — amount of apples and oranges in the i-th box (0 ≤ ai, oi ≤ 109). The sum of N in all the tests in the input doesn't exceed 105. All the input numbers are integer.
Output
For each test output two lines. In the first line output YES, if it's possible to choose N boxes, or NO otherwise. If the answer is positive output in the second line N numbers — indexes of the chosen boxes. Boxes are numbered from 1 in the input order. Otherwise leave the second line empty. Separate the numbers with one space.
Examples
2
2
10 15
5 7
20 18
1
0 0
YES
1 3
YES
1
题意:
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<stack> #include<map> #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<stack> #include<map> #include<queue> #include<cmath> using namespace std; typedef long long LL; typedef unsigned long long ull; #define sc1(a) scanf("%lld",&a) #define pf1(a) printf("%lld\n",a) #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 const LL INF=1e18; const ull base=2333; const int maxn=1e5+55; const int maxm=1e4+50; const int maxv=1e6+5; const int maxk=1e8+5; const int mod=1e9+7; struct Node { LL a,b,id; Node(){} Node(LL a1,LL b1,LL id1) { a=a1;b=b1;id=id1; } bool operator < (const Node x) const { return a<x.a; } }box[maxn*2]; int main() { LL T;cin>>T; LL N; while(T--) { cin>>N; for(LL i=1;i<=2*N-1;i++) { sc1(box[i].a);sc1(box[i].b); box[i].id=i; } sort(box+1,box+1+2*N-1); printf("YES\n"); for(LL i=1;i+1<=2*N-1;i+=2) { if(box[i].b<box[i+1].b) { printf("%lld ",box[i+1].id); } else printf("%lld ",box[i].id); } printf("%lld\n",box[2*N-1].id); } return 0; } /** */