9.10 simulated match

CCT

最近学校又发了n本五三题霸，BBS看到后十分高兴。但是，当他把五三拿到手后才发现，他已经刷过这些书了！他又认真地看了一会儿，发现新发的这些五三是2017版的，而他刷的是2016版的。现在他想找出所有他没有刷过的题来刷。每本五三都有m道题，并且它的特征（即它和去年版本的五三的差距）可以用一个m位二进制数来代表，二进制位上的1代表该题不同，0代表该题相同。比如4（100）就代表题目3和去年的有不同、5（101）就代表题目1和题目3和去年的有不同。而BBS热衷于给自己找麻烦，他要选择连续一段的几本五三一起刷，并且要求，所有选择的五三的特征中的所有k位中每一位出现1的次数都相同。他又想去刷最多的书，请你告诉他，他最多能刷多少本书？

输入格式：

第一行为两个整数 n、m，接下来的n行为 n 个整数，表示每本五三的特征。

输出格式：

一个整数，表示BBS最多能刷几本书。

样例输入	样例输出
7 3 7 6 7 2 1 4 2	4

样例解释：

这7本五三的特征分别为111,110,111,010,001,100,010。选择第3本至第6本五三，这些五三的特征中每一位都出现了2次1。当然，选择第4本到第6本也是可以的，这些五三的特征中每一位都出现了1次1。只是这样子BBS刷的书的数量就少了，他就会不高兴。

数据范围：

对于 100%的数据：1<=n<=100000，1<=k<=30。

 well the code is the best language 

#include <cstdio>
#include <set>
//-------------------------------------------------------------------------------
#define init(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout); 
#define cint    register int 
#define fr(i,a,b)  for(cint i=a;i<=b;i++)
//--------------------efficiency and interface optimization----------------------
using namespace std;
//-------------------------------read optimization---------------------------------------
#define getchar() (SS==TT&&(TT=(SS=BB)+fread(BB,1,1<<15,stdin),SS==TT)?EOF:*SS++) 
char BB[1<<15],*SS=BB,*TT=BB; 
int read(){cint x=0,f=1;char ch=getchar(); 
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 
    while('0'<=ch&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}return x*f;} 
//--------------------------------------------------------------------------------------
int n,x,ans,sum[100001][31],k;
struct sth{int num,dif[30];
    bool operator <(const sth &b)const{fr(i,1,k-1)if(dif[i]!=b.dif[i])return dif[i]<b.dif[i];return 0;}
}t;//create the data structure as the set's type,and
//the operator function is used to initializate the sort_compare that inside
set<sth>s;//use sth to make a set 
set<sth>::iterator it;//make an iterator named it
//if you don't know much about iterator,you can search it on the Internet and have a study
int max(int a,int b){return a>b?a:b;}//make a find_max_function by hand
//this means if a>b,return a ;or return b(just return the bigger one) 
int main(){init("CCT");n=read(),k=read();//read total_number of books and the given_binary_digit
    fr(i,1,n){//make a loop from 1 to n
        x=read();//read each eigenvalue
        fr(j,1,k){//calculate out the binary of each eigenvalue
            sum[i][j]=sum[i-1][j];//this use the ideology of prefix sum
            if(x&(1<<(j-1)))sum[i][j]++;//1<<(j-1)means 2*(j-1),if the solution >0,just renew the sum
        }}ans=0;t.num=0;//initializate the answer and t
    fr(i,1,k)t.dif[i]=0;//initializatation t's differente_num
    s.insert(t);//insert t into s
    fr(i,1,n){//make a loop from 1 to n
        t.num=i;//set t's num as i,just initializate t's num
        fr(j,1,k-1)//make a loop from 1 to k-1
        t.dif[j]=sum[i][j]-sum[i][j+1];//calculate out t's different_num
        //in this way we can ...
        it=s.find(t);//set the iterator point to t's location
        if(it==s.end())s.insert(t);//this just means that have not find t,just t is not real in s
        //just insert t to s  
        else ans=max(ans,i-(*it).num);//or ,just means that this way is OK,just mark it and renew the answer for better
    }printf("%d",ans);return 0;//out the answer
}

MHM

  LGL今天一共要上n节课，这n节课由0标号至n。由于过度劳累，除了第0节课和第n节课，LGL还打算睡上m节课，所以他做了一个睡觉计划表。通过小道消息，LGL得知WQ今天会在学校中检查，所以他想少睡k节课。但是由于某些原因，他又想使相邻的两节睡觉的课之间上的课数量的最小值最大。由于他很困，所以他请你来帮他计算这个值。

 

输入格式：

第一行为三个整数 n、m、k，接下来的m行为m个整数a_i，表示睡觉计划表中LGL想要睡觉的课。

输出格式：

一个整数，表示题目所求的值。

样例输入	样例输出
25 5 2 14 11 17 2 21	3

样例解释：

选择第2节和第14节不睡觉，这样子相邻的两节睡觉的课之间上的课数量的最小值为3，即第17节和第21节之间和第21节到第25节之间。没有答案更大的删除方案。

数据范围：

对于100%的数据：1<=n<=10⁹，1<=k<=m<=50000，0<a_i<n。

 well,it's just a two_point_answer's problem

 it's really easy in fact,but don't forget to sort it 

I'm so kind that I'd like to give a detailed explaination.

#include <cstdio>
#include <algorithm>
//-------------------------------------------------------------------------------
#define init(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout); 
#define cint    register int 
#define fr(i,a,b)  for(cint i=a;i<=b;i++)
//--------------------efficiency and interface optimization----------------------
using namespace std;
//-------------------------------read optimization---------------------------------------
#define getchar() (SS==TT&&(TT=(SS=BB)+fread(BB,1,1<<15,stdin),SS==TT)?EOF:*SS++) 
char BB[1<<15],*SS=BB,*TT=BB; 
int read(){cint x=0,f=1;char ch=getchar(); 
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 
    while('0'<=ch&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}return x*f;} 
//--------------------------------------------------------------------------------------
int n,m,k,ans,d[50002];
bool check(int x){int i=1,num=0,now=0;
    fr(i,1,m)if(d[i]-now<=x)//if one distance is shorter than the smallest_num
        //that means we need to kill a class to keep the smallest_num is still smallest
        {num++;if(num>k)return 0;}//if the killed number is bigger than requirement
     //just means that this case is not allowed,just return false
       else now=d[i];//or ,that means the distance is larger,just means we don't need to cut this class
       //just means if we go on,the distance is still larger
       //so we just need to enumerate from this class to find need_killed_class
       return 1;//if not return before,just means this case is allowed ,just return true
}
void TPA(){int l=1,r=n,mid;//set three variable
    while(l<=r){mid=(l+r)>>1;//think the mid as the smallest amount
       if(check(mid))ans=mid,l=mid+1;else r=mid-1;}
    //if the mid satisfies the requirement just renew the answer,and then search in the right part
    //or ,search in the left part
    //why we do in this way?
    //because when you find it's a best solution,that means the left part must be worse than it
    //(if you don't understand ,just read the check_function.)
    //(if you still understand ,I think you can have a look at the jumping_rock)
    //then we just go to the right part to search for the better solution 
    //or if you find it's not the best solution,just means that the right part must be not allowed either.
    //then we just go to the left part to search for an allowed solution 
}//two_point_answer
int main(){init("mhm");n=read(),m=read(),k=read();
    //read total_num of class ,total_num of sleepy class and the killed_number of sleepy class
    fr(i,1,m)d[i]=read();//read each sleepy class
    //because these classes are not arranged in order,so we need to sort them
    sort(d+1,d+m+1);//sort the array from 1 to m
    d[++m]=n;ans=1;//push the final_class into the array and initializate the answer
    TPA();printf("%d",ans);return 0;//two_point_ans and then out the answer
}

AAFA

 YYH有n道题要做。每一道题都有一个截止日期t，只要在该日期之前做完，他的父亲LRB就会奖励他w元钱。令人惊讶的是，每一道题他都只需要1秒来做。请问他最多能从父亲那里拿到多少钱？

输入格式：

第一行为一个整数 n，接下来的n行每一行都有两个数t_i和w_i，分别表示第i题的截止日期和奖励。

输出格式：

一个整数，表示YYH的最大获利。

样例输入	样例输出
3 2 10 1 5 1 7	17

样例解释：

第1秒做第3道题，第2秒做第1道题。

数据范围：

对于 100%的数据：1<=n、t_i 、w_i <=100000。

#include <cstdio>
#include <algorithm>
#include <queue>
//-------------------------------------------------------------------------------
#define init(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout); 
#define cint    register int 
#define fr(i,a,b)  for(cint i=a;i<=b;i++)
//--------------------efficiency and interface optimization----------------------
using namespace std;
//-------------------------------read optimization---------------------------------------
#define getchar() (SS==TT&&(TT=(SS=BB)+fread(BB,1,1<<15,stdin),SS==TT)?EOF:*SS++) 
char BB[1<<15],*SS=BB,*TT=BB; 
int read(){cint x=0,f=1;char ch=getchar(); 
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 
    while('0'<=ch&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}return x*f;} 
//--------------------------------------------------------------------------------------
typedef long long ll;
struct work{ll d,p;}w[100001];
//d:killed time
//p:reward
priority_queue<int,vector<int>,greater<int> >q;
int n;ll sz,ans;
bool comp(work a,work b){return a.d<b.d;}
int main(){init("aafa");n=read();//read the total_num of problems
    fr(i,1,n)w[i].d=read(),w[i].p=read();//read each killed time and reward
    sort(w+1,w+1+n,comp);ans=0;//sort the array to make it in order,and initializate the answer
    fr(i,1,n){sz=q.size();//use sz to mark the size of priority_queue
        if(sz<w[i].d)q.push(w[i].p);//if has not arrived this killed time,
        //just arrive it and push the reward into the queue
        else if(sz==w[i].d&&q.top()<w[i].p)q.pop(),q.push(w[i].p);}
        //or ,if has arrived this killed time,and can make the smallest one better,just renew the solution
    while(!q.empty())ans+=q.top(),q.pop();//calculate the answer 
    printf("%I64d",ans);return 0;//out the answer
}

ZZI

  YYH拿到了父亲给的钱欣喜若狂，把这些钱拿来造了n栋房子。现在他要给这些房子通电。他有两种方法：第一种是在房间里搭核电发电机发电，对于不同的房子，他需要花不同的代价V_i；，第二种是将有电的房子i的电通过电线通到没电的房子j中，这样子他需要花的代价为a_ij。他现在请你帮他算出他最少要花多少钱才能让所有的房子通上电。

输入格式：

第一行为一个整数 n。接下来的n行为 n 个整数v_i，再接下来的n行每行n个数，第i行第j列的数表示a_ij。

输出格式：

一个整数，表示最小代价。

样例输入	样例输出
4 5 4 4 3 0 2 2 2 2 0 3 3 2 3 0 4 2 3 4 0	9

样例解释：

在第4栋房子造核电发电机，再将其他三栋房子通过电线连向它。

数据范围：

对于 100%的数据：1<=n<=300，1<=v_i，a_ij<=100000，保证a_ii=0，a_ij=a_ji。

#include <cstdio>
#define INF 0x3f3f3f3f
//-------------------------------------------------------------------------------
#define init(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout); 
#define cint    register int 
#define fr(i,a,b)  for(cint i=a;i<=b;i++)
//--------------------efficiency and interface optimization----------------------
using namespace std;
//-------------------------------read optimization---------------------------------------
#define getchar() (SS==TT&&(TT=(SS=BB)+fread(BB,1,1<<15,stdin),SS==TT)?EOF:*SS++) 
char BB[1<<15],*SS=BB,*TT=BB; 
int read(){cint x=0,f=1;char ch=getchar(); 
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 
    while('0'<=ch&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}return x*f;} 
//--------------------------------------------------------------------------------------
int n,dist[301],g[301][301];
bool used[301];
int min(int a,int b){return a<b?a:b;}
//make a find_min_function by hand
//this means if a<b,return a ;or return b(just return the smaller one) 
int prim(){int next,maxn,res=0;//set three variables
    while(true){next=-1;maxn=INF;//initializate the next and the maxn
        fr(i,1,n)//make a loop from 1 to n
        if((!used[i])&&dist[i]<maxn)maxn=dist[i],next=i;//it this case can make better and it has not been used
        //just renew the solution ,and make next point to it
        if(next==-1)break;//this means that all case have been used or all case can not make better,just break
        res+=dist[next],used[next]=1,dist[next]=0;//or,just renew the answer  and mark it has been used
        //and set the distance(cost) to 0,just means it has added into the ans and can not choose it again
        fr(i,1,n)//make a loop from 1 to n
        if(!used[i])//if this case has not been used
        dist[i]=min(dist[i],g[next][i]);//if the distance can make better(by connecting to the house had electricity)
        //just improve it
    }return res;}//minimal spanning tree
int main(){init("zzi");n=read();//read the total_num of houses
    fr(i,1,n)dist[i]=read();//read the make_electricity_by_hand's cost
    fr(i,1,n)fr(j,1,n)g[i][j]=read();//read the make_electricity_by_connecting's cost
    printf("%d",prim());return 0;//set up a minimal spanning tree and out the ans
}