2008年5月28日
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
struct NODE
{
int p_x,p_y,step,c_x,c_y,d;
};
queue<NODE> que;
const int MAXN = 100;
bool p[MAXN][MAXN][4],c[MAXN][MAXN];
int map[MAXN][MAXN];
int mv[4][2]={-1,0,0,1,1,0,0,-1},p_sm,p_sn,c_sm,c_sn,m,n;
void init ( )
{
memset(p,false,sizeof(p));
for ( int i=0 ; i<m ; i++ )
for ( int j=0 ; j<n ; j++ )
{
scanf("%d",&map[i][j]);
if ( map[i][j]==4 )
p_sm=i,p_sn=j;
else
if ( map[i][j]==2 )
c_sm=i,c_sn=j;
}
}
int bfs ( )
{
NODE temp,go;
int i;
temp.p_x = p_sm;
temp.p_y = p_sn;
temp.c_x = c_sm;
temp.c_y = c_sn;
temp.step = 0;
temp.d=0;
for ( i=0 ; i<4 ; i++ )
{
p[p_sm][p_sn][i]=true;
temp.d=i;
que.push(temp);
}
c[c_sm][c_sn]=true;
while ( !que.empty () )
{
NODE head=que.front();
que.pop() ;
for ( i=0 ; i<4 ; i++ )
{
int tm=head.p_x+mv[i][0] , tn=head.p_y+mv[i][1] ;
if ( tm>=0 && tm<m && tn>=0 && tn<n && !p[tm][tn][i] && i!=(head.d+2)%4 && map[tm][tn]!=1 )
{
double t=sqrt( (double)( (tm-head.c_x)*(tm-head.c_x) ) +(double)( (tn-head.c_y)*(tn-head.c_y) ));
if ( t <= sqrt(2.0) && t>0 )
{
go=head;
go.p_x = tm;
go.p_y = tn;
go.step ++ ;
go.d=i;
p[tm][tn][i]=true;
que.push(go);
}
else
{
int tmm=tm+mv[i][0] , tnn=tn+mv[i][1];
if ( t==0 && tmm>=0 && tmm<m && tnn>=0 && tnn<n && map[tmm][tnn]!=1 && !c[tmm][tnn] )
{
go.c_x = tmm ;
go.c_y = tnn ;
go.p_x = tm;
go.d=i;
go.p_y = tn;
go.step = head.step+1;
if ( map[tmm][tnn]==3 )
return go.step ;
c[tmm][tnn]=true;
p[tm][tn][i]=true;
que.push(go);
}
}
}
}
}
return -1;
}
int main ( )
{
while ( scanf("%d%d",&n,&m)!=EOF )
{
init();
printf("%d\n",bfs());
// que.
}
}
2008年5月7日
#include<iostream>
#include<string>
using namespace std;
class FIELDDiagrams
{
public:
void dfs ( int k, int max, int n , long long& sum , int a )
{
int i;
if ( max==n )
{
sum++;
return;
}
if (( k-1)*a+1<max-n )
return ;
for ( i=a ; i>=1 ; i-- )
dfs(k-1,max,n+i ,sum , i );
}
long long countDiagrams ( int t )
{
long long map[40][40],sum=0,i;
memset(map,0,sizeof(map));
for ( i=1 ; i<=t*(t-1)/2 ; i++ )
dfs( t, i,0,sum, i );
return sum;
}
};
int main ( )
{
FIELDDiagrams a;
int n;
while (cin>>n)
cout<<a.countDiagrams (n)<<endl;
}
2008年4月16日
2008年4月3日
2008年4月2日
#include<iostream>
#include<sstream>
#include<map>
#include<string>
#include<vector>
#include<iomanip>
#include<fstream>
#include<algorithm>
using namespace std;
typedef map< string , vector<int> > WORD;
void init ( WORD & m )
{
ifstream fin;
fin.open("keywords.txt");
string keyword;
while ( fin>>keyword )
m[keyword];
fin.close();
}
void count ( WORD & m )
{
int line=0;
ifstream fin;
fin.open("text.txt");
string str;
while ( getline(fin,str) )
{
line++;
for ( WORD::iterator iter = m.begin () ; iter!=m.end( ); iter++ )
if ( str.find ( iter->first )!=string::npos )
m[iter->first].push_back(line);
}
fin.close();
}
void only( WORD &m )
{
for ( WORD::iterator i=m.begin() ; i!=m.end(); i++ )
{
vector<int>::iterator new_end=unique(i->second.begin(),i->second.end());
i->second.erase(new_end,i->second.end());
}
}
void sort_list ( WORD m , map<int,string> &mm )
{
for ( WORD::iterator iter_m = m.begin() ; iter_m != m.end() ; iter_m++ )
mm[iter_m->second .size ()]=iter_m->first;
}
void output ( WORD m , map<int,string> mm )
{
string keyword;
for ( map<int,string>::reverse_iterator iter=mm.rbegin() ; iter!=mm.rend() ; iter++ )
{
cout<<setw(10)<<iter->second<<":"
<<"(";
for ( vector<int>::iterator iter_vector=m[iter->second].begin() ; iter_vector!=m[iter->second].end(); iter_vector++ )
{
if ( iter_vector!=m[iter->second].begin() )
cout<<",";
cout<<*iter_vector;
}
cout<<")"<<endl;
}
}
int main ()
{
WORD m;
map<int , string> mm;
init(m);
count(m);
sort_list(m,mm);
only(m);
output(m,mm);
}
2008年3月31日
我就是不理解这个投票规则，对这题整整困惑了12个小时，从下午5点到凌晨5点。。。。。。。
Long March Voting
Description
Instant run-off voting is a system for selecting the most preferred candidate in an election. At the beginning of the process, each voter ranks the candidates from most preferred to least preferred. A series of automated voting rounds are then held to determine the overall winner.
In each round, each voter casts a single vote for his most preferred remaining candidate. If a candidate receives strictly more than 50% of the votes cast in that round, that candidate is declared the winner of the election. Otherwise, the candidate with the fewest votes in that round is eliminated, and another round is held. If multiple candidates are tied for the least number of votes, they are all eliminated. If all the candidates are eliminated, the election ends without a winner.
You are given the preferences of the voters in an election, and you must determine the outcome. There are M candidates numbered 0 to M-1, inclusive. The preferences are given in N lines, where each element describes the preferences of a single voter. This is a permutation of the digits 0 to M-1 in decreasing order of preference. In other words, the first digit is the voter's most preferred candidate, the second digit is his second most preferred candidate, and so on.
Input
There are several test cases,each test case begins with a integer N(1<=N<=50),means there are N voters.The next N lines,each contains a string with the same lenth M(1<=M<=10).Each element of a voter will be a permutation of the digits between 0 and M-1. There is a blank line between each test case.
Output
For each test case,output the number of the candidate who wins the election, or -1 if the election ends without a winner.
Sample Input
5
120
102
210
021
012
8
3120
3012
1032
3120
2031
2103
1230
1230
Sample Output
1
-1
Hint:
Case 1:
Nobody gets an absolute majority in the first round and candidate 2 is eliminated. Candidate 1 then receives 3 votes in the next round, giving an absolute majority.
Case 2:
Candidate 0 is eliminated in the first round of voting. Candidate 2 is eliminated in the second round. In the third round, candidates 1 and 3 get 4 votes each. Neither candidate receives an absolute majority, and they are both eliminated for having the least number of votes, so the election ends without a winner.
这题的投票，每轮都是从第一个数开始找的，找到第一个没有被淘汰的人。
我一开始以为之前几轮选的人，在后面不能被选了。。。。
2008年3月27日
#include<iostream>
#include<string>
#include<iomanip>
#include<vector>
#include<algorithm>
const int MAXN=100;
using namespace std;
int cmp ( const void* p1 , const void *p2 )
{
return *( double* )p1>*(double*)p2?1:-1;
}
typedef struct COLOR
{
string col;
double mean,median,sum,value[MAXN];
int num,p;
}COL;
int find ( vector<COL> &str , COL s )
{
for ( vector<COL>::size_type i = 0; i != str.size(); i++ )
if ( str[i].col==s.col )
{
str[i].sum+=s.sum;
str[i].num++;
str[i].value[str[i].p++]=s.sum;
return 1;
}
return 0;
}
void add ( vector<COL>& str , COL s )
{
if ( !find ( str , s ) )
{
s.p=1;
s.num=1;
s.value[0]=s.sum;
str.push_back(s);
}
}
void output ( vector<COL> str )
{
double s=0,v[MAXN],median;
int n=0,q,j=0;
for (vector<COL>::size_type i = 0; i != str.size(); ++i )
{
for (q=0 ; q<str[i].p ; q++)
v[j++]=str[i].value[q];
s+=str[i].sum;
n+=str[i].p;
qsort(str[i].value,str[i].num,sizeof(str[i].value[0]),cmp);
if (str[i].num%2)
str[i].median=str[i].value[(str[i].num-1)/2];
else
str[i].median=(str[i].value[str[i].num/2]+str[i].value[str[i].num/2-1])/2;
cout<<str[i].col<<"\t"<<": "<<"sum="<<"\t"
<<setw(10)<<str[i].sum<<"\t"<<"mean="<<"\t"
<<setw(10)<<str[i].sum/str[i].num<<"\t"<<"median="<<"\t"
<<setw(10)<<str[i].median<<endl;
/*
for ( int j=0 ; j<str[i].p ; j++ )
cout<<str[i].value[j]<<endl;
*/
}
qsort(v,j,sizeof(v[0]),cmp);
if (n%2!=0)
median=v[(n-1)/2];
else
median=(v[n/2]+v[n/2-1])/2;
cout<<"============================================================================"<<endl;
cout<<"ALL"<<"\t"<<": "<<"sum="<<"\t"
<<setw(10)<<s<<"\t"<<"mean="<<"\t"
<<setw(10)<<s/n<<"\t"<<"median="<<"\t"
<<setw(10)<<median<<endl;
/*
for ( j=0 ; j<n ;j++ )
cout<<v[j]<<endl;
*/
}
int main ( )
{
vector<COL> str;
COL s;
while ( cin>>s.col>>s.sum )
{
add(str,s);
}
output( str );
}
摘要: 0 前言: STL，为什么你必须掌握
对于程序员来说，数据结构是必修的一门课。从查找到排序，从链表到二叉树，几乎所有的算法和原理都需要理解，理解不了也要死记硬背下来。幸运的是这些理论都已经比较成熟，算法也基本固定下来，不需要你再去花费心思去考虑其算法原理，也不用再去验证其准确性。不过，等你开始应用计算机语言来工作的时候，你会发现，面对不同的需求你需要一次又一次去用代码重复实现这些已经成熟的... 阅读全文
助教给我的邮件中这样说：
Jerry Huang 致 我
显示详细信息 3月21日 (6天前)
Hi,
我编译了你们的代码，进行了测试，好像和希望的结果差距比较大，请你们再检查确认一下。
如果是提交错了，请重新提交。
Thanks
huang
结果我只打了70分。。。。。
1#include<iostream>
2using namespace std;
3/* 判断字符串是否在引号里面 */
4int qutation ( char c , bool &f ,bool f1 )
5{
6 char temp;
7 if ( c=='"' && f1==false)
8 {
9 f=true;
10 cout<<c;
11 while ( f==true )
12 {
13 cin.get(temp);
14 if ( temp=='"')
15 f=false;
16 cout<<temp;
17 }
18 return 1; //发现引号
19 }
20 return 0; //没有发现引号
21}
22void cut_add ( )
23{
24 char c,temp;
25 bool f1,f2,f3;
26 f1=f2=f3=false; // f1标记block注释,f2标记引号,f3标记line注释
27 while ( cin.get(c) )
28 {
29 if ( !qutation(c,f2,f1) ) // 没有出现引号
30 {
31 /*判断注释开头*/
32 if ( c=='/' )
33 {
34 cin.get(temp);
35 if ( temp=='*' )
36 f1=true;//找到了block注释的开头
37 else
38 if ( temp=='/' )
39 f3=true;//找到了line注释的开头
40 /*当没有找到注释的开头时，执行else部分*/
41 else
42 {
43 cout<<c;
44 cin.putback(temp);
45 }
46 }
47 else
48 if ( f1==false && f3==false )
49 cout<<c;
50 /*判断注释结尾*/
51 if ( c=='*' )
52 {
53 cin.get(temp);
54 if ( temp=='/' )
55 f1=false; //关闭block注释
56 else
57 {
58 cout<<c;
59 cin.putback(temp);
60 }
61 }
62 else
63 if ( c=='\n' )
64 {
65 f3=false; //关闭line注释
66 cout<<c;
67 }
68 }
69 }
70}
71int main ( )
72{
73 cut_add();
74}
我不知道什么原因，手动输入可以的，但是用文件输入输出的话，输出就停不了了。。。。
题目：
就是给你个.cpp文件，这是加注释的，然后让你生成一个.txt文件，除去代码的注释。
用命令行输入：
erasecomment < DataIn.cpp > result.txt