Codeforces Round #129 (Div. 1) B. Little Elephant and Cards

D. Little Elephant and Cards

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Little Elephant loves to play with color cards.

He has n cards, each has exactly two colors (the color of the front side and the color of the back side). Initially, all the cards lay on the table with the front side up. In one move the Little Elephant can turn any card to the other side. The Little Elephant thinks that a set of cards on the table is funny if at least half of the cards have the same color (for each card the color of the upper side is considered).

Help the Little Elephant to find the minimum number of moves needed to make the set of n cards funny.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of the cards. The following n lines contain the description of all cards, one card per line. The cards are described by a pair of positive integers not exceeding 109 — colors of both sides. The first number in a line is the color of the front of the card, the second one — of the back. The color of the front of the card may coincide with the color of the back of the card.

The numbers in the lines are separated by single spaces.

Output

On a single line print a single integer — the sought minimum number of moves. If it is impossible to make the set funny, print -1.

Sample test(s)

input

34 74 77 4

output

0

input

54 77 42 119 71 1

output

2

Note

In the first sample there initially are three cards lying with colors 4, 4, 7. Since two of the three cards are of the same color 4, you do not need to change anything, so the answer is 0.

In the second sample, you can turn the first and the fourth cards. After that three of the five cards will be of color 7.

题目的大意是：给你n张卡片，每张卡片的正反面分别用一个数组代表一种颜色，刚开始时，卡片正面朝上，求最少的翻转次数使得卡片有一半以上是同一种颜色，如果不能实现，输出-1；

在这里要注意当卡片只有一张时的特殊情况！

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;struct cc{    LL d;    int a;} p[200001];bool cmp(cc n,cc m) {   return n.d<m.d; }int main(){    LL n,i,j,min,k,nn,sum,z=0;    LL ss,ee;    scanf("%lld",&n);    j = 0;    for(i=0;i<=n-1;i++)    {        scanf("%lld%lld",&ss,&ee);        if(ss!=ee)        {            p[j].d=ss;            p[j].a=1;            j ++;            p[j].d=ee;            p[j].a=2;            j ++;        }        else        {            p[j].d=ss;            p[j].a=1;            j ++;        }    }    sum=j;    if(n%2==0)        nn=n/2;    else        nn=n/2+1;    sort(p,p+sum,cmp);    j=1;    min=1e9+1;    if(p[0].a==1)        k=1;    else        k=0;    for(i=1;i<=sum-1;i++)    {        if(p[i].d==p[i-1].d)        {            j ++;            if(p[i].a==1) k++;            if(k>=nn)            {                z=1;                break;            }        if(j>=nn)            {                if(min>nn-k)                  min=nn-k;            }        }        else        {            if(j>=nn)            {                if(min>nn-k)                    min=nn-k;            }            j=1;            if(p[i].a==1)                k=1;            else                k=0;        }        if(j+sum-1-i<nn)        break;    }     if(z||n==1)       printf("0\n");    else       if(min==1e9+1)        printf("-1\n");       else        printf("%d\n",min);  return 0;}