【codeforces 767D】Cartons of milk

【题目链接】:http://codeforces.com/problemset/problem/767/D

【题意】



每个牛奶都有最晚可以喝的时间;

每天喝K瓶牛奶;

你有n瓶牛奶->已知每个牛奶的期限;

然后商店里面有m瓶牛奶；->已知每个牛奶的期限;

然后问你最多能再从商店里买几瓶牛奶;

使得你买的这些牛奶都不会因为过期而被扔掉;

或者你原本的n瓶牛奶都不能做到则输出-1

【题解】



首先肯定喝的顺序是先喝马上就要过期的

->所以原有的n瓶升序排;

然后买的话,肯定是买晚过期的;

所以商店里的m瓶降序排;

然后就是二分答案;

二分x表示买几瓶;->买最晚的x瓶;

然后O(N+X)判断可不可以喝完.

判断的时候没必要把新加的X个再加进去;

直接用类似归并排序的方式看看哪个小就先喝哪个就好;

(两个数组嘛)



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define ps push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)

typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 2e6 + 100;

struct abc
{
    int day, id;
};

int n, m, k, num, f[N];
map <int, int> extra;
map <int, int> dic;
abc s[N];

bool cmp1(abc a, abc b)
{
    return a.day > b.day;
}

bool check(int x)
{
    int R = x,now = k,day = 0,i = 1;
    while (i <= n)
    {
        if (R == 0 || f[i] < s[R].day)
        {
            if (day > f[i]) return false;
            now--;
            i++;
        }
        else
        {
            if (day > s[R].day) return false;
            R--;
            now--;
        }

        if (now == 0) now = k,day++;
    }
    rep2(i, R, 1)
    {
        if (day > s[i].day)
            return false;
        now--;
        if (now == 0) day++, now = k;
    }
    return true;
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    rei(n), rei(m), rei(k);
    rep1(i, 1, n)
        rei(f[i]);
    rep1(i, 1, m)
        rei(s[i].day), s[i].id = i;

    sort(f + 1, f + 1 + n);
    sort(s + 1, s + 1 + m, cmp1);

    if (!check(0))
    {
        return puts("-1"), 0;
    }

    int l = 1, r = m, ans = 0;
    while (l <= r)
    {
        int mid = (l + r) >> 1;
        if (check(mid))
        {
            ans = mid, l = mid + 1;
        }
        else
        {
            r = mid - 1;
        }
    }
    printf("%d\n", ans);
    rep1(i, 1, ans)
        printf("%d ", s[i].id);
    puts("");
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}