BNU Training 2015 07 27 Problem Solving

Last Update:2015-07-31 Source: Internet

Author: User

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Ultraviolet A 12435 C. consistent Verdicts.

[Solution]: O (n ^ 2) brute force enumeration + unique function deduplication adjacent elements. It only ran 3 ms ,~~

Code:

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                                                using namespace std;#define rep(i,j,k) for(int i=(int)j;i<(int)k;++i)#define per(i,j,k) for(int i=(int)j;i>(int)k;--i)#define lowbit(a) a&-a#define Max(a,b) a>b?a:b#define Min(a,b) a>b?b:a#define mem(a,b) memset(a,b,sizeof(a))typedef long long LL;typedef unsigned long long LLU;typedef double db;const int N=1e6+10;const int inf=0x3f3f3f3f;char str[N];bool vis[N];int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};int movv[5][2]= {{1,0},{0,1},{0,0},{-1,0},{0,-1}};inline LL read(){ int c=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') { if(ch=='-')f=-1; ch=getchar(); } while(ch>='0'&&ch<='9') { c=c*10+ch-'0'; ch=getchar(); } return c*f;}char mon1[N],mon2[N];LL day1,year1;LL day2,year2;LL row,line,x,t,y,i,res;struct node{ LL codrx; LL codry;};node num[N];LL mum[N];int main(){ LL tot=1; t=read(); while(t--){ x=read(); int len=0; for(i=0; i

The number of leap data between two dates: ps: the conventional method determines the TE, so another idea is to determine the leap year using the Division code:

// C#ifndef _GLIBCXX_NO_ASSERT#include 
 
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                                                using namespace std;#define rep(i,j,k) for(int i=(int)j;i<(int)k;++i)#define per(i,j,k) for(int i=(int)j;i>(int)k;--i)#define lowbit(a) a&-a#define Max(a,b) a>b?a:b#define Min(a,b) a>b?b:a#define mem(a,b) memset(a,b,sizeof(a))typedef long long LL;typedef unsigned long long LLU;typedef double db;const int N=1e5;const int inf=0x3f3f3f3f;char str[N];bool vis[N];int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};int movv[5][2]= {{1,0},{0,1},{0,0},{-1,0},{0,-1}};bool leap_year(int y){ if(y%4==0&&y%100!=0||y%400==0) return 1; return 0;}inline LL read(){ int c=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') { if(ch=='-')f=-1; ch=getchar(); } while(ch>='0'&&ch<='9') { c=c*10+ch-'0'; ch=getchar(); } return c*f;}char mon1[N],mon2[N];LL day1,year1;LL day2,year2;LL row,line,x,t,y,i,res;int main(){ scanf(%lld,&t); for(i=1; i<=t; ++i) { int res=0; scanf(%s %lld,%lld,mon1,&day1,&year1); scanf(%s %lld,%lld,mon2,&day2,&year2); if((mon1[0]=='J'&&mon1[1]=='a')||mon1[0]=='F') year1=year1; else year1++; if((mon2[0]=='J'&&mon2[1]=='a')||mon2[0]=='F'&&day2<=28) year2--; res=(year2/4)-((year1-1)/4); res=res-(year2/100)+((year1-1)/100); res=res+(year2/400)-((year1-1)/400); printf(Case %lld: %lld,i,res); } return 0;}

Ultraviolet A 12442 J. forwarding Emails, find the number of the person who received the message for the first time in the case of the longest chain of people that the message can be delivered. [solution]: a more intuitive thought process is displayed on the Internet. Copy the process to make it easy to understand: Solution Description:

DFS problem
Read this line carefully in problem description- They each pick one other person they know to email those things to every time-exactly one, no less, no more (and never themselves)I. e Each person send email only one. for each person has only one adjacency person. the input can not be (1 to 3, 2 to 3, 1 to 2), because for person 1 here 2 adjacency person 3 and 2. so, you can represent this graph using one dimen=array adj [N]. do not need to stack, you can use recursion.
Example:

1 2

2 1

4 3

3 2

The adjacency list, adj[1]=2, adj[2]=1, adj[3]=2, and adj[4]=3.

Use a boolean array visit[N]

Visit [1]	Visit [2]	Visit [3]	Visit [4]
False	False	False	False

At first start from 1

If visit[1]==false then run DFS in this time count the visited node and update the visit[] array (Set visit[i]=True here i is a visited node).  Remember dot not use the array visit[] for cycle finding you can use another boolean array visit2[] for that purpose. After run DFS the visit[] array is

Visit [1]	Visit [2]	Visit [3]	Visit [4]
True	True	False	False

And count_visited_node = 2 (1-> 2)

Now 2

visit[2]==true so, do not need to do anything.

Now 3

visit[3]==false so, run the DFS. After that the visit[] array is

Visit [1]	Visit [2]	Visit [3]	Visit [4]
True	True	True	False

And count_visited_node = 3 (3-> 2-> 1)

Now 4

visit[4]==false so, run the DFS. After that the visit[] array is

Visit [1]	Visit [2]	Visit [3]	Visit [4]
True	True	True	True

And count_visited_node = 4 (4-> 3-> 2-> 1)

For 4, the count_visited_node is maximum. Ans is 4.

Code:

#include 
 
  #define MAX 50005int T,N;int vis[MAX], f[MAX], c[MAX];int ans, flag;typedef long long LL;inline LL read(){    int c=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();}    return c*f;}int dfs(int u){    int v = f[u];/// 2=f[1];1=f[2];2=f[3];    int r = 0;   ///    vis[u] = 1;  /// vis[1]=1;vis[2]=1;vis[3]=1;    if(!vis[v]) r = dfs(v) + 1;///vis[2]=1,r=0+1=1;    vis[u] = 0;    c[u] = r;   ///c[1]=1,c[2]=0,c[3]=2;    return r;}int main(){    int u,v;    T=read();    for(int t=1; t<=T; t++){        N=read();        for(int i=1; i<=N; i++){            u=read(),v=read();            f[u] = v;            vis[u] = 0;            c[u] = -1;        }        ans = -1;        for(int i=1; i<=N; i++){            if(c[i]==-1) dfs(i);            if(c[i]>m)            {                m=c[i];                flag=i;            }          /*  printf(vertex %d children %d,i,c[i]);*/        }        printf(Case %d: %d,t,flag);    }    return 0;}

Several groups of data:

731 22 33 141 22 14 33 251 22 15 33 44 521 22 131 22 33 144 22 14 33 2101 22 33 44 55 66 77 88 99 1010 1

Case 1: 1Case 2: 4Case 3: 3Case 4: 1Case 5: 1Case 6: 4Case 7: 1

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BNU Training 2015 07 27 Problem Solving

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