Cinema(离散化 / map)

发布于 2020-06-30  0 次阅读


题面:

Cinema

题目大意:

n个人去电影院,每个人有自己喜欢的语言(用数字表示)——a[n]。
m个电影院,电影院有audio——b[m]和subtitles——c[m], 人与audio相同高兴,人与subtitles相同比较高兴。
求使最多人高兴的电影(b在a的体现),当b相等时求比较高兴的人多的电影(b相等情况下,c在a中的体现)。

思路:

因为a,b,c的数据范围都是[1,1e9],所以采取离散化或者用map存储
离散化后语言种数最多n + 2m
map直接存储后进行模拟

AC Code:

离散化:

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int MAXN = 2e5 + 9;
int n, m;
int a[MAXN], b[MAXN], c[MAXN];
int unite[3 * MAXN], unite_a[3 * MAXN];

int find(int len, int x)   //在离散化数组中寻找值
{
    return lower_bound(unite + 1, unite + len + 1, x) - unite;
}

int discrete()            //对数组进行离散化处理
{
    int len = 0;
    for (int i = 1; i <= n; i++)   //存a数组
        unite[++len] = a[i];
    for (int i = 1; i <= m; i++)   //存b,c数组
    {
        unite[++len] = b[i];
        unite[++len] = c[i];
    }
    sort(unite + 1, unite + len + 1);   //去重前先排序
    return unique(unite + 1, unite + len + 1) - (unite + 1);    //离散化数组去重,返回离散化数组长度
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    cin >> m;
    for (int i = 1; i <= m; i++)
        cin >> b[i];
    for (int i = 1; i <= m; i++)
        cin >> c[i];
    int unite_len = discrete();              //求得离散化数组的长度
    for (int i = 1; i <= n; i++)
        unite_a[find(unite_len, a[i])]++;    //a中每个值在离散化数组的个数
    int max_b = 0, max_c = 0;
    int index;
    for (int i = 1; i <= m; i++)
    {
        int find_b = unite_a[find(unite_len, b[i])];            //b在a中的体现
        int find_c = unite_a[find(unite_len, c[i])];            //c在a中的体现
        if (find_b > max_b || find_b == max_b && find_c > max_c)
        {
            index = i;
            max_b = find_b;
            max_c = find_c;
        }
    }
    cout << index << endl;
    return 0;
}

map:

#include<cstdio>
#include<iostream>
#include<map>
using namespace std;
const int MAXN = 2e5 + 9;
int n, m;
int a[MAXN], b[MAXN], c[MAXN];
map<int,int>mp;

int main(){
    scanf("%d", &n);
    for(int i=1; i<=n; i++){
        scanf("%d", &a[i]);
        mp[a[i]]++;
    }
    scanf("%d", &m);
    for(int i=1; i<=m; i++)
        scanf("%d", &b[i]);
    for(int i=1; i<=m; i++)
        scanf("%d", &c[i]);
    int max_b=0, max_c=0, index;
    for(int i=1; i<=m; i++){
        if(mp[b[i]]>max_b || mp[b[i]]==max_b&&mp[c[i]]>max_c){
            index=i;
            max_b=mp[b[i]];
            max_c=mp[c[i]];
        }
    }
    printf("%d\n", index);
    return 0;
}

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