洛谷P3887 [GDOI2014]世界杯 题解 队列模拟

题目链接:https://www.luogu.com.cn/problem/P3887

解题思路:
一开始给4组人从大到小排序然后放入4个队列,然后每次从队列中选出来人,用4个队列模拟即可。

实现代码如下:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int K, D, M, F,
    k[maxn], d[maxn], m[maxn], f[maxn],
    Q, a, b, c;
queue<int> que_k, que_d, que_m, que_f;
bool cmp(int a, int b) {
    return a > b;
}
int main() {
    cin >> K >> D >> M >> F;
    for (int i = 0; i < K; i ++) cin >> k[i];
    sort(k, k+K, cmp);
    for (int i = 0; i < K; i ++) que_k.push(k[i]);
    for (int i = 0; i < D; i ++) cin >> d[i];
    sort(d, d+D, cmp);
    for (int i = 0; i < D; i ++) que_d.push(d[i]);
    for (int i = 0; i < M; i ++) cin >> m[i];
    sort(m, m+M, cmp);
    for (int i = 0; i < M; i ++) que_m.push(m[i]);
    for (int i = 0; i < F; i ++) cin >> f[i];
    sort(f, f+F, cmp);
    for (int i = 0; i < F; i ++) que_f.push(f[i]);
    cin >> Q;
    while (Q --) {
        cin >> a >> b >> c;
        long long sum = que_k.front();
        long long num = a + b + c + 1;
        que_k.pop();
        while (a --) {
            sum += que_d.front();
            que_d.pop();
        }
        while (b --) {
            sum += que_m.front();
            que_m.pop();
        }
        while (c --) {
            sum += que_f.front();
            que_f.pop();
        }
        printf("%.2lf\n", (double) sum / (double) num);
    }
    return 0;
}
posted @ 2020-05-15 15:22  quanjun  阅读(252)  评论(0编辑  收藏  举报