BNU Training 2015 07 27

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UVA 12435 C. Consistent verdicts"to the main idea": give you two-dimensional plane some people's coordinates, each person has a gun, ask all people at the same time after the shooting of the number of the number of shots can be heard.

"Problem Solving idea": O (n^2) Violent enumeration +unique function to re-weigh adjacent elements. I just ran 3ms,~~.

Code:

<span style= "FONT-SIZE:14PX;" >//c#ifndef _glibcxx_no_assert#include <cassert> #endif # include <cctype> #include <cerrno># Include <cfloat> #include <ciso646> #include <climits> #include <clocale> #include <cmath > #include <csetjmp> #include <csignal> #include <cstdarg> #include <cstddef> #include < cstdio> #include <cstdlib> #include <cstring> #include <ctime>//C + + #include <algorithm># Include <bitset> #include <complex> #include <deque> #include <exception> #include <fstream > #include <functional> #include <iomanip> #include <ios> #include <iosfwd> #include < iostream> #include <istream> #include <iterator> #include <limits> #include <list> #include <locale> #include <map> #include <memory> #include <new> #include <numeric> #include < ostream> #include <queue> #include <set> #include <sstream> #include <stack> #include <stdexcept> #include <streambuf> #include <string># Include <typeinfo> #include <utility> #include <valarray> #include <vector>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<x; ++i) {num[i].codrx=read ();        Num[i].codry=read (); } for (int i=0, i<x; ++i) {for (int j=i+1; j<x; ++j) mum[len++]= (Num[i].codrx-num[j].        CODRX) * (NUM[I].CODRX-NUM[J].CODRX) + (num[i].codry-num[j].codry) * (Num[i].codry-num[j].codry);        } sort (Mum,mum+len);        LL res = unique (mum,mum+len)-mum;    printf ("Case%LLD:%lld\n", tot++,res+1); } return 0;} </span>

UVA 12439 G. February"Main Idea" the number of leap data between two dates: PS: The general method of judging will te, so a change of mind, to judge leap years can be used divisionCode:

<span style= "FONT-SIZE:14PX;" >//c#ifndef _glibcxx_no_assert#include <cassert> #endif # include <cctype> #include <cerrno># Include <cfloat> #include <ciso646> #include <climits> #include <clocale> #include <cmath > #include <csetjmp> #include <csignal> #include <cstdarg> #include <cstddef> #include < cstdio> #include <cstdlib> #include <cstring> #include <ctime>//C + + #include <algorithm># Include <bitset> #include <complex> #include <deque> #include <exception> #include <fstream > #include <functional> #include <iomanip> #include <ios> #include <iosfwd> #include < iostream> #include <istream> #include <iterator> #include <limits> #include <list> #include <locale> #include <map> #include <memory> #include <new> #include <numeric> #include < ostream> #include <queue> #include <set> #include <sstream> #include <stack> #include <stdexcept> #include <streambuf> #include <string># Include <typeinfo> #include <utility> #include <valarray> #include <vector>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 VI S[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,&AMP;DAY1,&AMP;YEAR1);        scanf ("%s%lld,%lld", MON2,&AMP;DAY2,&AMP;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\n", i,res); } return 0;} </span>

UVA 12442 J. Forwarding Emails"The main idea": The chief's long hair a letter to the tribe of people, give the tribe of people's relations, no one can only be received once and once, for the letter can be transmitted the longest number of people in the case of the first letter received the number of the person"problem-solving ideas":Online to see a more intuitive process of thinking, copied over, easy to understand: Solution Description:

DFS problem
Read This line carefully in problem description-"They all 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 is only one . For all person have only one adjacency person. The input can not is (1 to 3, 2 to 3, 1 to 2), because for person 1 here 2 adjacency person 3 and 2. So, you can represent the this graph using one dimensional array adj[n].Do not need to the stack, you can use recursion.
Example:

4

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 a 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, does not need to do anything.

Now 3

Visit[3]==false So, run the DFS. After, 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, 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:

<span style= "FONT-SIZE:14PX;" > #include <stdio.h> #define MAX 50005int t,n;int Vis[max], F[max], c[max];int ans, flag;typedef long long Ll;inlin    e 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\n", i,c[i]); */} printf ("Case%d:%d\n", T,flag); } return 0;} </span>

several sets 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

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

BNU Training 2015 07 27