Codeforces 493C - Vasya and Basketball (二分)

题意

划定一个得分线,在这个线上算三分,以下算两分,要求A队和B队的分差尽量大

思路

一开始用了二分,但是没考虑到如果分数相等的时候怎么划分。这种思路应该是错的。

其实直接暴力就行了,检验一下每一个得分点。。。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <ctime>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-6;
const int MAXN = 2e5 + 10;
const int MOD = 1000007;
const int dir[][2] = { {1, 0}, {-1, 0}, {0, -1}, {0, 1} };
 
int fir[MAXN], sec[MAXN], ans, m, n, pivot = -INF;
 
 
int GetAns(int *arr, int n)
{
    int cnt = upper_bound(arr, arr + n, ans) - arr;
    return 2ll * cnt + (n - cnt) * 3;
}
 
vector<int> tot;
set<int> mp;
 
int main()
{
    //ROP;
    int i, j;
    scanf("%d", &m);
    for (int i = 0; i < m; i++)
    {
        scanf("%d", &fir[i]);
        tot.PB(fir[i]);
    }
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &sec[i]);
        tot.PB(sec[i]);
    }
    tot.PB(0);
    sort(fir, fir + m); sort(sec, sec + n);
    LL firAns, secAns, rec;
    for (int i = 0; i < SZ(tot); i++)
    {
        if (mp.count(tot[i])) continue;
        mp.insert(tot[i]);
        int firCnt = upper_bound(fir, fir + m, tot[i]) - fir, secCnt = upper_bound(sec, sec + n, tot[i]) - sec;
        firAns = 2ll * firCnt + (m - firCnt) * 3, secAns = 2ll * secCnt + (n - secCnt) * 3;
        if (firAns - secAns > pivot)
        {
            pivot = firAns - secAns;
            rec = firAns;
            ans = tot[i];
        }
        else if (firAns - secAns == pivot && firAns > rec)
        {
            pivot = firAns - secAns;
            ans = tot[i];
        }

        
    }
    firAns = GetAns(fir, m), secAns = GetAns(sec, n);
    printf("%lld:%lld\n", firAns, secAns);
    return 0;
}

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