POJ 3664

[NWPU][2014][TRN][1]水题堆

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A B C D E F G H I J K

D - Election Time

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

The cows are having their first election after overthrowing the tyrannical Farmer John, and Bessie is one of N cows (1 ≤ N ≤ 50,000) running for President. Before the election actually happens, however, Bessie wants to determine who has the best chance of winning.

The election consists of two rounds. In the first round, the K cows (1 ≤ K ≤ N) cows with the most votes advance to the second round. In the second round, the cow with the most votes becomes President.

Given that cow i expects to get A_i votes (1 ≤ A_i ≤ 1,000,000,000) in the first round and B_i votes (1 ≤ B_i ≤ 1,000,000,000) in the second round (if he or she makes it), determine which cow is expected to win the election. Happily for you, no vote count appears twice in the A_i list; likewise, no vote count appears twice in the B_i list.

Input

* Line 1: Two space-separated integers: N and K
* Lines 2..N+1: Line i+1 contains two space-separated integers: A_i and B_i

Output

* Line 1: The index of the cow that is expected to win the election.

Sample Input

5 33 109 25 68 46 5

Sample Output

5

---

//@auther Yang Zongjun#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>using namespace std;#define PI acos(-1.0)#define EPS 1e-8const int MAXN = 50005;const int INF = 2100000000;pair<int, int> p[MAXN];pair<int, int> a[MAXN];int n, k;int main(){    freopen("C:/Users/Administrator/Desktop/input.txt", "r", stdin);    scanf("%d%d", &n, &k);    for(int i = 1; i <= n; i++)    {        scanf("%d%d", &p[i].first, &p[i].second);        a[i].first = p[i].first;        a[i].second = p[i].second;    }    sort(p + 1, p + n + 1);    //由于sort排序之后，p的顺序，即牛的编号变了，所以我用    //了两个数组p和a    //for(int i = 1; i <= n; i++)        //printf("%d %d\n", p[i].first, p[i].second);    int maxa = p[n].second;    for(int i = n; i > n-k; i--)    {        maxa = max(maxa, p[i].second);    }    int j;    for(int i = 1; i <= n; i++)    {        if(a[i].second == maxa)        {            j = i;break;        }    }    printf("%d\n", j);    return 0;}

由于sort排序之后，p的顺序，即牛的编号变了，要开两个数组，而下面的代码用结构体，将牛的编号也设置成牛的属性，这样即使排序后每个牛的属性不变

//@auther Yang Zongjun#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>using namespace std;#define PI acos(-1.0)#define EPS 1e-8const int MAXN = 50005;const int INF = 2100000000;int n, k;struct Cow{    int f,s,num;}cow[MAXN];bool cmp1(const Cow& a, const Cow& b){    return a.f > b.f;}bool cmp2(const Cow& a, const Cow& b){    return a.s > b.s;}int main(){//    freopen("C:/Users/Administrator/Desktop/input.txt", "r", stdin);    scanf("%d%d", &n, &k);    for(int i = 0; i < n; i++)    {        scanf("%d%d", &cow[i].f, &cow[i].s);        cow[i].num = i + 1;    }    sort(cow, cow + n, cmp1);    sort(cow, cow + k, cmp2);    printf("%d\n", cow[0].num);    return 0;}