寒假集训日志（五）——期中测验

　　10道题目，A了4道，5道会做，5道不会，果然渣渣。。。

　　这次测验感觉无限段错误。。

　　最近写题段错误出现的比较多，问题主要在于：

　　1.命名一个变量未初始化，而之后的语句又恰好未赋值，并且使用

　　2.开超大数组一定不要开在函数里面！！！！

　　3.无限循环

　　4.scanf("%d",&x)少了＆有时候也会出现

一个是初始化的问题，还一个建树的过程出错，但还不知道错在哪里。先附上几道没A的题， 以后慢慢解决。

B - 拓扑

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 510C

Description

Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order.

After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted in lexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical!

She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order.

Lexicographical order is defined in following way. When we compare s and t, first we find the leftmost position with differing characters: s_i ≠ t_i. If there is no such position (i. e. s is a prefix of t or vice versa) the shortest string is less. Otherwise, we compare characters s_i and t_i according to their order in alphabet.

Input

The first line contains an integer n (1 ≤ n ≤ 100): number of names.

Each of the following n lines contain one string name_i (1 ≤ |name_i| ≤ 100), the i-th name. Each name contains only lowercase Latin letters. All names are different.

Output

If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on).

Otherwise output a single word "Impossible" (without quotes).

Sample Input

Input

3
rivest
shamir
adleman

Output

bcdefghijklmnopqrsatuvwxyz

Input

10
tourist
petr
wjmzbmr
yeputons
vepifanov
scottwu
oooooooooooooooo
subscriber
rowdark
tankengineer

Output

Impossible

Input

10
petr
egor
endagorion
feferivan
ilovetanyaromanova
kostka
dmitriyh
maratsnowbear
bredorjaguarturnik
cgyforever

Output

aghjlnopefikdmbcqrstuvwxyz

Input

7
car
care
careful
carefully
becarefuldontforgetsomething
otherwiseyouwillbehacked
goodluck

Output

acbdefhijklmnogpqrstuvwxyz

 

D - KMP循环节

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3746

Description

CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the entrepreneurial spirit of "HDU CakeMan", he wants to sell some little things to make money. Of course, this is not an easy task.

As Christmas is around the corner, Boys are busy in choosing christmas presents to send to their girlfriends. It is believed that chain bracelet is a good choice. However, Things are not always so simple, as is known to everyone, girl's fond of the colorful decoration to make bracelet appears vivid and lively, meanwhile they want to display their mature side as college students. after CC understands the girls demands, he intends to sell the chain bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls to show girls' lively, and the most important thing is that it must be connected by a cyclic chain which means the color of pearls are cyclic connected from the left to right. And the cyclic count must be more than one. If you connect the leftmost pearl and the rightmost pearl of such chain, you can make a CharmBracelet. Just like the pictrue below, this CharmBracelet's cycle is 9 and its cyclic count is 2:

Now CC has brought in some ordinary bracelet chains, he wants to buy minimum number of pearls to make CharmBracelets so that he can save more money. but when remaking the bracelet, he can only add color pearls to the left end and right end of the chain, that is to say, adding to the middle is forbidden.
CC is satisfied with his ideas and ask you for help.

 

Input

The first line of the input is a single integer T ( 0 < T <= 100 ) which means the number of test cases.
Each test case contains only one line describe the original ordinary chain to be remade. Each character in the string stands for one pearl and there are 26 kinds of pearls being described by 'a' ~'z' characters. The length of the string Len: ( 3 <= Len <= 100000 ).

 

Output

For each case, you are required to output the minimum count of pearls added to make a CharmBracelet.

 

Sample Input

3 aaa abca abcde

 

Sample Output

0 2 5

 

E - 组合

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1496

Description

Encoding schemes are often used in situations requiring encryption or information storage/transmission economy. Here, we develop a simple encoding scheme that encodes particular types of words with five or fewer (lower case) letters as integers.

Consider the English alphabet {a,b,c,...,z}. Using this alphabet, a set of valid words are to be formed that are in a strict lexicographic order. In this set of valid words, the successive letters of a word are in a strictly ascending order; that is, later letters in a valid word are always after previous letters with respect to their positions in the alphabet list {a,b,c,...,z}. For example,

abc aep gwz

are all valid three-letter words, whereas

aab are cat

are not.

For each valid word associate an integer which gives the position of the word in the alphabetized list of words. That is:

 a -> 1 
 b -> 2 
    .

    .

 z -> 26 
 ab -> 27 
 ac -> 28 
    .

    .

 az -> 51 
 bc -> 52 
    .

    .

 vwxyz -> 83681

Your program is to read a series of input lines. Each input line will have a single word on it, that will be from one to five letters long. For each word read, if the word is invalid give the number 0. If the word read is valid, give the word's position index in the above alphabetical list.

Input

The input consists of a series of single words, one per line. The words are at least one letter long and no more that five letters. Only the lower case alphabetic {a,b,...,z} characters will be used as input. The first letter of a word will appear as the first character on an input line.

The input will be terminated by end-of-file.

Output

The output is a single integer, greater than or equal to zero (0) and less than or equal 83681. The first digit of an output value should be the first character on a line. There is one line of output for each input line.

Sample Input

z
a
cat
vwxyz

Sample Output

26
1
0
83681

F - 最短路

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1062

Description

年轻的探险家来到了一个印第安部落里。在那里他和酋长的女儿相爱了，于是便向酋长去求亲。酋长要他用10000个金币作为聘礼才答应把女儿嫁给他。探险家 拿不出这么多金币，便请求酋长降低要求。酋长说："嗯，如果你能够替我弄到大祭司的皮袄，我可以只要8000金币。如果你能够弄来他的水晶球，那么只要 5000金币就行了。"探险家就跑到大祭司那里，向他要求皮袄或水晶球，大祭司要他用金币来换，或者替他弄来其他的东西，他可以降低价格。探险家于是又跑 到其他地方，其他人也提出了类似的要求，或者直接用金币换，或者找到其他东西就可以降低价格。不过探险家没必要用多样东西去换一样东西，因为不会得到更低 的价格。探险家现在很需要你的帮忙，让他用最少的金币娶到自己的心上人。另外他要告诉你的是，在这个部落里，等级观念十分森严。地位差距超过一定限制的两 个人之间不会进行任何形式的直接接触，包括交易。他是一个外来人，所以可以不受这些限制。但是如果他和某个地位较低的人进行了交易，地位较高的的人不会再 和他交易，他们认为这样等于是间接接触，反过来也一样。因此你需要在考虑所有的情况以后给他提供一个最好的方案。
为了方便起见，我们把所有的物品从1开始进行编号，酋长的允诺也看作一个物品，并且编号总是1。每个物品都有对应的价格P，主人的 地位等级L，以及一系列的替代品Ti和该替代品所对应的"优惠"Vi。如果两人地位等级差距超过了M，就不能"间接交易"。你必须根据这些数据来计算出探 险家最少需要多少金币才能娶到酋长的女儿。

Input

输入第一行是两个整数M，N（1 <= N <= 100），依次表示地位等级差距限制和物品的总数。接下来按照编号从小到大依次给出了N个物品的描述。每个物品的描述开头是三个非负整数P、L、X（X < N），依次表示该物品的价格、主人的地位等级和替代品总数。接下来X行每行包括两个整数T和V，分别表示替代品的编号和"优惠价格"。

Output

输出最少需要的金币数。

Sample Input

1 4
10000 3 2
2 8000
3 5000
1000 2 1
4 200
3000 2 1
4 200
50 2 0

Sample Output

5250

H - 最短路难

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4370

Description

Given a n*n matrix C _ij (1<=i,j<=n),We want to find a n*n matrix X _ij (1<=i,j<=n),which is 0 or 1.

Besides,X _ij meets the following conditions:

1.X ₁₂+X ₁₃+...X _1n=1
2.X _1n+X _2n+...X _n-1n=1
3.for each i (1<i<n), satisfies ∑X _ki (1<=k<=n)=∑X _ij (1<=j<=n).

For example, if n=4,we can get the following equality:

X ₁₂+X ₁₃+X ₁₄=1
X ₁₄+X ₂₄+X ₃₄=1
X ₁₂+X ₂₂+X ₃₂+X ₄₂=X ₂₁+X ₂₂+X ₂₃+X ₂₄
X ₁₃+X ₂₃+X ₃₃+X ₄₃=X ₃₁+X ₃₂+X ₃₃+X ₃₄

Now ,we want to know the minimum of ∑C _ij*X _ij(1<=i,j<=n) you can get.

Hint

For sample, X ₁₂=X ₂₄=1,all other X _ij is 0.

 

Input

The input consists of multiple test cases (less than 35 case).
For each test case ,the first line contains one integer n (1<n<=300).
The next n lines, for each lines, each of which contains n integers, illustrating the matrix C, The j-th integer on i-th line is C _ij(0<=C _ij<=100000).

 

Output

For each case, output the minimum of ∑C _ij*X _ij you can get.

 

Sample Input

4 1 2 4 10 2 0 1 1 2 2 0 5 6 3 1 2

 

Sample Output

3

 

Hint

For sample, X 

₁₂

=X 

₂₄

=1,all other X 

_ij

 is 0. 

I - Manacher+DP

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4513

Description

　　吉哥又想出了一个新的完美队形游戏！
　　假设有n个人按顺序站在他的面前，他们的身高分别是h[1], h[2] ... h[n]，吉哥希望从中挑出一些人，让这些人形成一个新的队形，新的队形若满足以下三点要求，则就是新的完美队形：

　　1、挑出的人保持原队形的相对顺序不变，且必须都是在原队形中连续的；
　　2、左右对称，假设有m个人形成新的队形，则第1个人和第m个人身高相同，第2个人和第m-1个人身高相同，依此类推，当然如果m是奇数，中间那个人可以任意；
　　3、从左到中间那个人，身高需保证不下降，如果用H表示新队形的高度，则H[1] <= H[2] <= H[3] .... <= H[mid]。

　　现在吉哥想知道：最多能选出多少人组成新的完美队形呢？

 

Input

　　输入数据第一行包含一个整数T，表示总共有T组测试数据(T <= 20)；
　　每组数据首先是一个整数n(1 <= n <= 100000)，表示原先队形的人数，接下来一行输入n个整数，表示原队形从左到右站的人的身高（50 <= h <= 250，不排除特别矮小和高大的）。

 

Output

　　请输出能组成完美队形的最多人数，每组输出占一行。

 

Sample Input

2 3 51 52 51 4 51 52 52 51

 

Sample Output

3 4

//此题可以直接用MARACHER算法来做，但不知道为什么老是超时，还可以将原数列翻转过来，再求LCIS
#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=220;

int n,a[N],b[N],dp[N];

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            b[n+1-i]=a[i];
        }
        memset(dp,0,sizeof(dp));
        int len,ans=1;
        for(int i=1;i<=n;i++){
            len=0;
            for(int j=1;j<=(n-i+1);j++){        //j<=n-i+1 保证最多中间只重复一人
                if(a[i]>b[j])
                    len=max(len,dp[j]);
                else if(a[i]==b[j])
                    dp[j]=max(dp[j],len+1);
                if(i<(n+1-j))       //判断是否重叠
                    ans=max(ans,dp[j]*2);
                else
                    ans=max(ans,dp[j]*2-1);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}