瑞士轮
题意很简单,暴力点就是每次弄完都排序就OK了,但只有60分
考虑代码重复多余的运算在哪,超复杂度是因为N次排序,但事实上不需要这么多次排序,
或者说,每次对于N个队,每个队的胜者分数+1,依然大于排序后面队的胜者,败者+0也是如此。
组成两个长度为N的数组,归并排序O(n)合并得出排序结果。
第一步排序手写了个heapsort,实际上用sort就可以了,主要是暴力写完第一遍懒得改了。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<stack>
#define FOR(a,b) for(register int i=a;i<=b;i++)
#define LL long long
using namespace std;
struct node
{
int id;
int w;
int score;
}A[200000];
void sift(node a[], int low, int high)
{
int i = low, j = 2 * i;
node temp = a[i];
while (j <= high)
{
if (j < high&&((a[j].score > a[j + 1].score)||a[j].score==a[j+1].score&&a[j].id<a[j+1].id))
j++;
if((temp.score > a[j].score)||(temp.score==a[j].score&&temp.id<a[j].id))
{
a[i] = a[j];
i = j;
j = 2 * i;
}
else break;
}a[i] = temp;
}
void Heapsort(node a[], int n)
{
int i;
for (i = n / 2; i >= 1; i--)
{
sift(a, i, n);
}
node temp;
for (i = n; i >= 2; i--)
{
temp = a[i];
a[i] = a[1];
a[1] = temp;
sift(a, 1, i - 1);
}
}
node B[100010];
node C[100010];
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL); cout.tie(NULL);
int N, R, Q;
cin >> N >> R >> Q;
FOR(1,2*N)
{
A[i].id = i;
cin >> A[i].score;
}
FOR(1,2*N)
{
cin >> A[i].w;
}
Heapsort(A, 2*N);
int cntb = 0, cntc = 0;
int cnt = 0;
FOR(1, R)
{
cntb = 0;
cntc = 0;
/* for (int i = 1; i <= 2 * N; i++)
{
cout << A[i].id<<":"<<A[i].score << " ";
}*/
for (int j = 1; j <= N; j++)
{
if (A[2 * j - 1].w < A[2 * j].w)
{
A[2 * j].score++;
B[++cntb] = A[2 * j];
C[++cntc] = A[2 * j - 1];
}
else
{
A[2 * j - 1].score++;
B[++cntb] = A[2 * j - 1];
C[++cntc] = A[2 * j];
}
}
cnt = 0;
cntb = 1, cntc = 1;
while (cntb <= N && cntc <= N)
{
if ((B[cntb].score > C[cntc].score) || (B[cntb].score == C[cntc].score&&B[cntb].id < C[cntc].id))
{
A[++cnt] = B[cntb++];
}
else A[++cnt] = C[cntc++];
}
while (cntb <= N)
{
A[++cnt] = B[cntb++];
}
while (cntc <= N)
{
A[++cnt] = C[cntc++];
}
}
cout << A[Q].id << endl;
system("pause");
}