CF 670C Cinema(算竞进阶习题)

离散化+排序

离散化统计人数就好,本来不难,但是测试点太丧心病狂了。。。CF还是大哥啊

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C yql){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;
    return ans;
}
const int N = 200005;
int a[N], b[N<<2], tot, num[N<<2];
struct Film{
    int x, y, cnt1, cnt2, id;
    bool operator < (const Film &rhs) const {
        if(cnt1 != rhs.cnt1) return cnt1 > rhs.cnt1;
        if(cnt2 != rhs.cnt2) return cnt2 > rhs.cnt2;
        return id < rhs.id;
    }
}film[N];
int main(){

    int n = read();
    for(int i = 0; i < n; i ++) a[i] = read(), b[++tot] = a[i];
    int m = read();
    for(int i = 0; i < m; i ++) film[i].x = read(), b[++tot] = film[i].x, film[i].id = i;
    for(int i = 0; i < m; i ++) film[i].y = read(), b[++tot] = film[i].y;
    sort(b + 1, b + tot + 1);
    tot = unique(b + 1, b + tot + 1) - b - 1;
    for(int i = 0; i < n; i ++){
        num[lower_bound(b + 1, b + tot + 1, a[i]) - b] ++;
    }
    for(int i = 0; i < m; i ++){
        film[i].cnt1 = num[lower_bound(b + 1, b + tot + 1, film[i].x) - b];
        film[i].cnt2 = num[lower_bound(b + 1, b + tot + 1, film[i].y) - b];
    }
    sort(film, film + m);
    printf("%d\n", film[0].id + 1);
    return 0;
}

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转载自www.cnblogs.com/onionQAQ/p/10533850.html