CF727D：D. T-shirts Distribution（贪心）

最新推荐文章于 2022-05-07 15:49:00 发布

原创最新推荐文章于 2022-05-07 15:49:00 发布 · 535 阅读

0 ·

CC 4.0 BY-SA版权

贪心专栏收录该内容

23 篇文章

订阅专栏

本文介绍了一种竞赛场景下的T恤分配算法，旨在解决不同尺码T恤如何有效分配给参与者的问题，确保每位参与者都能获得合适的T恤。

The organizers of a programming contest have decided to present t-shirts to participants. There are six different t-shirts sizes in this problem: S, M, L, XL, XXL, XXXL (sizes are listed in increasing order). The t-shirts are already prepared. For each size from S to XXXLyou are given the number of t-shirts of this size.

During the registration, the organizers asked each of the n participants about the t-shirt size he wants. If a participant hesitated between two sizes, he could specify two neighboring sizes — this means that any of these two sizes suits him.

Write a program that will determine whether it is possible to present a t-shirt to each participant of the competition, or not. Of course, each participant should get a t-shirt of proper size:

the size he wanted, if he specified one size;
any of the two neibouring sizes, if he specified two sizes.

If it is possible, the program should find any valid distribution of the t-shirts.

Input

The first line of the input contains six non-negative integers — the number of t-shirts of each size. The numbers are given for the sizes S, M, L, XL, XXL, XXXL, respectively. The total number of t-shirts doesn't exceed 100 000.

The second line contains positive integer n (1 ≤ n ≤ 100 000) — the number of participants.

The following n lines contain the sizes specified by the participants, one line per participant. The i-th line contains information provided by the i-th participant: single size or two sizes separated by comma (without any spaces). If there are two sizes, the sizes are written in increasing order. It is guaranteed that two sizes separated by comma are neighboring.

Output

If it is not possible to present a t-shirt to each participant, print «NO» (without quotes).

Otherwise, print n + 1 lines. In the first line print «YES» (without quotes). In the following n lines print the t-shirt sizes the orginizers should give to participants, one per line. The order of the participants should be the same as in the input.

If there are multiple solutions, print any of them.

Examples

input
0 1 0 1 1 0
3
XL
S,M
XL,XXL

output
YES
XL
M
XXL

input
1 1 2 0 1 1
5
S
M
S,M
XXL,XXXL
XL,XXL

output
NO

题意：6种码数的衣服，每件有一定数量，N个人买衣服，要么买指定码数的，要么可以买指定相邻码数二选一，问衣服能否分配够。

思路：将二选一的人挑出来，从小到大处理，优先买小码。

# include <bits/stdc++.h>
using namespace std;
string s[8] = {"S","M","L","XL","XXL","XXXL"};
int a[8], b[8], c[1<<17][2];
int buf[8][2];
int cal(string st)
{
    for(int i=0; i<6; ++i)
        if(s[i] == st) return i;
}
int main()
{
    int n;
    string t;
    memset(c, -1, sizeof(c));
    for(int i=0; i<6; ++i) scanf("%d",&a[i]);
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
    {
        cin >> t;
        int pos = t.find(",",0);
        if(pos == string::npos)
        {
            int id = cal(t);
            if(--a[id] < 0) return 0*puts("NO");
            else c[i][0] = id;
        }
        else
        {
            int id = cal(t.substr(0, pos));
            ++b[id];
            c[i][1] = id;
        }
    }
    for(int i=0; i<5; ++i)
    {
        if(b[i])
        {
            int imin = min(b[i], a[i]);
            a[i] -= imin, b[i] -= imin;
            buf[i][0] = imin;
            imin = min(b[i], a[i+1]);
            a[i+1] -= imin;
            b[i] -= imin;
            buf[i][1] = imin;
            if(b[i]) return 0*puts("NO");
        }
    }
    puts("YES");
    for(int i=0; i<n; ++i)
    {
        if(c[i][0] != -1)
            cout << s[c[i][0]] << endl;
        else
        {
            if(buf[c[i][1]][0] > 0)
            {
                --buf[c[i][1]][0];
                cout << s[c[i][1]] << endl;
            }
            else
            {
                --buf[c[i][1]][1];
                cout << s[c[i][1]+1] << endl;
            }
        }
    }
    return 0;
}