Worried School
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 112 Accepted Submission(s): 34
There are totally G slots for EC sub-region. X slots will be distributed among five China regional sites and Y slots will be distributed to the EC-Final. Of course X and Y are non-negative integers and X + Y = G.
Here is how the X slots be distributed:
1. Slots are assigned to the Asia Regional sites from the first place, the second place, · · · , last place.
2. For schools having the same place across the sites, the slots will be given in the order of the number of “effective
teams” in the sites.
3. No school could be assigned a slot 2 times, which means the schools will be skipped if they already got a slot.
After X slots are distributed, the EC-Final ranklist from highest rank will be assigned Y slots for those schools that haven’t got a slot yet.
Now here comes a sad story, as X and Y are not announced until the end of the last regional contest of that year, even later!!!
Teachers from a school are worried about the whether they can advance to WF whatever the X and Y is. Let’s help them find out the results before the announcement of X and Y .
Each test case starts with a line consisting of 1 integer and 1 string, G representing the sum of X and Y and S representing the name of the worried school.
Next 5 lines each consists of 20 string representing the names of top 20 schools in each site. The sites are given in the order of the number of “effective teams” which means the first site has the largest number of “effective teams” and the last site has the
smallest numebr of “effective teams”.
The last line consists of 20 strings representing the names of top 20 schools in EC-Final site. No school can appear more than once in each ranklist
∙ 1
≤ T ≤ 200.
∙ School
names only consist of upper case characters ‘A’ - ‘Z’ and the length is at most 5.
∙ 1
≤ G ≤ 20.
10 IJU
UIV GEV LJTV UKV QLV TZTV AKOV TKUV
GAV DVIL TDBV ILVTU AKV VTUD IJU IEV
HVDBT YKUV ATUV TDOV
TKUV UIV GEV AKV AKOV GAV DOV TZTV
AVDD IEV LJTV CVQU HVDBT AKVU XIV TDVU
OVEU OVBB KMV OFV
QLV OCV TDVU COV EMVU TEV XIV
VFTUD OVBB OFV DVHC ISCTU VTUD OVEU DTV
HEVU TEOV TDV TDBV CKVU
CVBB IJU QLV LDDLQ TZTV GEV GAV KMV
OFV AVGF TXVTU VFTUD IEV OVEU OKV DVIL
TEV XIV TDVU TKUV
UIV DVIL VFTUD GEV ATUV AKV TZTV QLV
TIV OVEU TKUV UKV IEV OKV CVQU COV
OFOV CVBB TDVU IOV
UIV TKUV CVBB AKV TZTV VFTUD UKV GEV
QLV OVEU OVQU AKOV TDBV ATUV LDDLQ AKVU
GAV SVD TDVU UPOHK
For the first test case, the optimal solution is X = 6 and Y = 4, at this time the advanced schools were [UIV, TKUV, QLV, CVBB, GEV, OCV, AKV, TZTV, VFTUD, UKV].
题目是难读懂了点，但是本身并不是很难做，可以说直接进行模拟就行了。
说一下题目大意吧：
现在WF分配总共g个名额分为两块(x+y=g)：
首先，分配x个名额给五个中国区域站。
这五个中国区域站每个站都有20个学校，输入时这20个学校已经按照分数从大到小排好序。
分配这x个名额的顺序：
一号区域站第一名 -> 二号区域站第一名 -> 三号区域站第一名 -> 四号区域站第一名 -> 五号区域站第一名 -> 一号区域站第二名 -> 二号区域站第二名 -> …………
但是如果有重复就跳过这个学校，因为一个学校最多只能占一个名额。
然后，分配完x个名额后，剩下的y个名额分配给EC-Final中的20个学校，这20个学校输入时也按排名从高到低的顺序，这20个学校去除已经获得名额的学校后的第一名到第y名获得这y个名额。
输入：
每个test的第一行是总名额数g和要查询是否能进WF的那个学校——worried school；
接下来的1到5行分别是一到五号区域站的20个学校。
在接下来的一行是EC-Final的20个学校。
现在，我们要帮worried school看看它能不能进WF，如果对于任意的(x,y)都能使得worried school进WF，就输出“ADVANCED!”。
而如果对于一些(x,y)能使得worried school进WF，另一些(x,y)不能，就输出最小的y，这个时候(x,y)恰好使得worried school不能进WF。
#include<cstdio>
#include<cmath>
#include<set>
#include<iostream>
using namespace std;
int g;
string site[][],EC_site[],worried_school;
int main()
{
int t;
scanf("%d",&t);
for(int kase=;kase<=t;kase++)
{
cin>>g>>worried_school;
for(int i=;i<=;i++)
{
for(int j=;j<=;j++) cin>>site[i][j];
}
for(int j=;j<=;j++) cin>>EC_site[j]; int i,j,x,y=-;
set<string> adv_school;
for(int x=;x<=g;x++)
{
adv_school.clear();
bool can_adv_in_x=,can_adv_in_y=;
if(x>)//给5个site分配名额
{
for(int r=;r<=;r++)
{
i=r%; if(i==) i+=;
j=(int)ceil(r/5.0);
if(site[i][j]==worried_school) can_adv_in_x=;//目标学校可以进WF
adv_school.insert(site[i][j]);
if(adv_school.size()>=x) break;
}
}
if(g-x>)//给EC-Final分配名额
{
for(int j=;j<=;j++)
{
if(EC_site[j]==worried_school) can_adv_in_y=;
adv_school.insert(EC_site[j]);
if(adv_school.size()>=g) break;
}
}
if(!can_adv_in_x && !can_adv_in_y) y=g-x;
}
printf("Case #%d: ",kase);
if(y==-) printf("ADVANCED!\n");
else printf("%d\n",y);
}
}