紫书第5章 C++STL

例题

例题5-1　大理石在哪儿（Where is the Marble？，Uva 10474）

主要是熟悉一下sort和lower_bound的用法

关于lower_bound：
http://blog.csdn.net/niushuai666/article/details/6734403
此外还有upper_bound
http://blog.csdn.net/niushuai666/article/details/6734650

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <cctype>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long

int a[10005];

int main()
{
int n,q,i,x,cnt=1;
while(~sf("%d%d",&n,&q) && n)
{
pf("CASE# %d:\n",cnt++);
for(i=0;i<n;i++) sf("%d",&a[i]);
sort(a,a+n);
while(q--)
{
sf("%d",&x);
int p = lower_bound(a,a+n,x)-a;
if(a[p]!=x)
{
pf("%d not found\n",x);
continue;
}
else
pf("%d found at %d\n",x,p+1);
}
}
}

例题5-2　木块问题（The Blocks Problem，Uva 101）

主要是熟悉vector的pb和resize，以及字符串结束的处理方法

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <cctype>
#include <vector>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long

vector<int> pile[30];
int n;

void find_block(int x,int &p,int &h)
{
for(p=0;p<n;p++)
{
for(h=0;h<pile[p].size();h++)
{
if(pile[p][h] == x) return;
}
}
}

void clear_block(int p,int h)
{
int i;
for(i = h+1;i<pile[p].size();i++)
{
int b = pile[p][i];
pile[b].pb(b);
}
pile[p].resize(h+1);
}

void onto_block(int p,int h,int p2)
{
int i;
for(i = h;i<pile[p].size();i++)
pile[p2].pb(pile[p][i]);
pile[p].resize(h);
}

void print()
{
int p,h;
for(p=0;p<n;p++)
{
pf("%d:",p);
for(h=0;h<pile[p].size();h++)
{
pf(" %d",pile[p][h]);
}
blank;
}
}

int main()
{
int i,a,b;
sf("%d",&n);
for(i=0;i<n;i++) pile[i].pb(i);
char s[5],s1[5];

while(sf("%s",s) && s[0]!='q')
{

sf("%d%s%d",&a,s1,&b);
int pa,pb,ha,hb;
find_block(a,pa,ha);
find_block(b,pb,hb);
if(pa==pb) continue;
if(!strcmp(s,"move")) clear_block(pa,ha);
if(!strcmp(s1,"onto")) clear_block(pb,hb);
onto_block(pa,ha,pb);
}
print();
}

例题5-3　安迪的第一个字典（Andy's First Dictionary，Uva 10815）

主要是熟悉set的insert还有iterator

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <cctype>
#include <vector>
#include <set>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long

set<string> dict;

int main()
{
string s,buf;
while(cin>>s)
{
int i;
for(i=0;i<s.length();i++)
{
if(isalpha(s[i])) s[i] = tolower(s[i]);
else s[i] = ' ';
}
stringstream ss(s);
while(ss>>buf) dict.insert(buf);
}
for(set<string>::iterator it = dict.begin();it!=dict.end();++it)
cout<<*it<<endl;
}

例题5-4　反片语（Ananagrams，Uva 156）

主要是熟悉map的用法。

给string和vector<string>排序可以用sort(str.begin(),str.end());即字典序

map<string,int> mp可以直接用mp[string] = int;在这道题可以表示某个字符串出现的次数

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <cctype>
#include <vector>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long

map<string,int> cnt;
vector<string> words;

string repr(const string &s)
{
string ans = s;
for(int i=0;i<ans.length();i++)
{
ans[i] = tolower(ans[i]);
}
sort(ans.begin(),ans.end());
return ans;
}

int main()
{
string s;
while(cin>>s)
{
if(s[0]=='#') break;
words.pb(s);
string r = repr(s);
if(!cnt.count(r)) cnt[r] = 0;
cnt[r]++;
}
vector<string> ans;
for(int i = 0;i<words.size();i++)
{
if(cnt[repr(words[i])]==1) ans.pb(words[i]);
}
sort(ans.begin(),ans.end());
for(int i = 0;i<ans.size();i++)
{
cout<<ans[i]<<endl;
}

}

例题5-5　集合栈计算机（The Set Stack Computer，ACM/ICPC NWERC2006，UVa12096）

map映射+vector，inserter插入迭代器的使用，set_union和set_intersection

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())

typedef set<int> Set;
map<Set,int> IDcache;
vector<Set> Setcache;

int ID(Set x)
{
if(IDcache.count(x)) return IDcache[x];
Setcache.pb(x);
return IDcache[x] = Setcache.size() - 1;
}

int main()
{
stack<int> s;
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
while(n--)
{
string op;
cin>>op;
if(op[0]=='P') s.push(ID(Set()));
else if(op[0]=='D') s.push(s.top());
else
{
Set x1 = Setcache[s.top()];s.pop();
Set x2 = Setcache[s.top()];s.pop();
Set x;
if(op[0]=='A')
{
x = x2;
x.insert(ID(x1));
}
if(op[0]=='U') set_union(ALL(x1),ALL(x2),INS(x));
if(op[0]=='I') set_intersection(ALL(x1),ALL(x2),INS(x));
s.push(ID(x));
}
cout<<Setcache[s.top()].size()<<endl;
}
cout<<"***"<<endl;
}

}

例题5-6　团体队列（Team Queue，UVa540）

队列的使用

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())

int main()
{
int t,x,i,kase=1;
while(sf("%d",&t),t)
{
pf("Scenario #%d\n",kase++);
map<int,int> team;
for(i=0;i<t;i++)
{
int n;
sf("%d",&n);
while(n--)
{
sf("%d",&x);
team[x] = i;
}
}
queue<int> q,q2[1010];
string op;
while(cin>>op && op[0]!='S')
{
if(op[0] == 'E')
{
int r;
sf("%d",&r);
int y = team[r];
if(q2[y].empty()) q.push(y);
q2[y].push(r);
}
if(op[0] == 'D')
{
int y = q.front();
pf("%d\n",q2[y].front());
q2[y].pop();
if(q2[y].empty())
q.pop();
}
}
blank;
}
}

例题5-7　丑数（Ugly Numbers，Uva 136）

优先队列的应用

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

int con[]={2,3,5};

int main()
{
int i,j;
set<LL> s;
pqueue<LL,vector<LL>,greater<LL> >pq;
s.insert(1);
pq.push(1);
for(i=1;;i++)
{
LL t = pq.top();pq.pop();
if(i==1500)
{
pf("The 1500'th ugly number is %I64d.\n",t);
break;
}
for(j=0;j<3;j++)
{
LL x = t*con[j];
if(!s.count(x))
{
s.insert(x);
pq.push(x);
}
}
}
}

例题5-8　Unixls命令（Unix ls，UVa400）

对于这样控制格式的要求（比如对齐）可以用规定长度然后填充的方式输出

col求出来后，row应该是(n-1)/col+1而不是n/col

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

const int maxcol = 60;
string filenames[105];

void print(const string& s,int len,char extra)
{
cout<<s;
for(int i = 0;i<len-s.length();i++)
cout<<extra;
}

int main()
{
int n,i,j;
while(~sf("%d",&n))
{
int ma = 0;
for(i=0;i<n;i++)
{
cin>>filenames[i];
ma = max(ma,(int)filenames[i].length());
}
int col = (maxcol-ma)/(ma+2) + 1;
int row = (n-1)/col +1;
print("",60,'-');
blank;
sort(filenames,filenames+n);
for(i = 0;i<row;i++)
{
for(j=0;j<col;j++)
{
int idx = j*row + i;
if(idx<n) print(filenames[idx],j==col-1?ma:ma+2,' ');
}
blank;
}
}
}

例题5-9　数据库（Database，ACM/ICPC NEERC 2009，UVa1592）

这题的思路是：

因为字符串比较的话比较慢，所以可以先做一个预处理，将所有字符串用一个ID表示

读取字符是这样的：因为字符用,和回车分隔，所以可以用getchar每个字符读取，遇到,或\n就处理

处理字符可以用map映射，如果count==0就pb到ID集vector里

然后就是遍历。先用一个结构保存两个ID。三重循环遍历，如果相同的话则输出

struct里小于号的定义很不错，return x<r.x || x==r.x && y<r.y;先看第一个的大小，相等再看第二个

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

const int ROW = 10000+10;
const int COL = 10+5;

int m,n;

struct node
{
int x,y;
node(int xx,int yy):x(xx),y(yy){}
bool operator <(const node& r) const { return x<r.x || x==r.x && y<r.y;}
};

map<string,int> IDcache;
vector<string> stringSet;
vector<int> Text[ROW];
map<node,int> data;

int ID(string str)
{
//cout<<"str: "<<str<<endl;
if(IDcache.count(str)) return IDcache[str];
stringSet.pb(str);
//pf("id: %d\n",stringSet.size()-1);
return IDcache[str] = stringSet.size()-1;
}

void read()
{
string str;
int i,j;
char ch = getchar();
for(i=0;i<n;i++)
{
for(;;)
{
ch = getchar();
if(ch=='\r' || ch == '\n')
{
if(!str.empty()) Text[i].pb(ID(str));
str.clear();
break;
}
if(ch!=',') str+=ch;
else
{
Text[i].pb(ID(str));str.clear();
}
}
//          cout<<"i "<<i<<endl;
//          for(j=0;j<Text[i].size();j++)
//               cout<<"j "<<j<<" Text: "<<Text[i][j]<<endl;
}

}

void sol()
{
int c1,c2,r,i,j,k;
for(c1=0;c1<m;c1++)
{
for(c2=c1+1;c2<m;c2++)
{
data.clear();
for(r=0;r<n;r++)
{
int x = Text[r][c1];
int y = Text[r][c2];
node p(x,y);
if(data.count(p))
{
pf("NO\n");
pf("%d %d\n%d %d\n",data[p]+1,r+1,c1+1,c2+1);
return;
}
else
data[p]= r;
}
}
}
pf("YES\n");
}

int main()
{
int i,j;
while(~sf("%d%d",&n,&m))
{
read();
sol();
for(i=0;i<n;i++) Text[i].clear();
IDcache.clear();
stringSet.clear();
}
}

例题5-10　PGA巡回赛的奖金（PGA Tour Prize Money，ACM/ICPC World Finals1990，UVa207）

例题5-11　邮件传输代理的交互（The Letter Carrier's Rounds, ACM/ICPC World Finals 1999, UVa814）

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

set<string> addr;

void getn(string &s,string &user,string &mta)
{
int k = s.find('@');
user = s.substr(0,k);
mta = s.substr(k+1);
}

int main()
{
string s,mta,user,ma,na,user1,mta1,t;
int n;
while(cin>>s && s[0]!='*')
{
cin>>ma>>n;
while(n--)
{
cin>>na;
addr.insert(na+"@"+ma);
}
}

while(cin>>s && s!="*")
{
getn(s,user,mta);

vector<string> MTA;
map<string,vector<string> > dest;//用户
set<string> vis;
while(cin>>t && t!="*")
{
getn(t,user1,mta1);
if(vis.count(t)) continue;
vis.insert(t);
if(!dest.count(mta1))
{
MTA.pb(mta1);
dest[mta1] = vector<string>();
}
dest[mta1].pb(t);
}
string data;
getline(cin,t);

while(getline(cin,t) && t[0]!='*')
{
data+="     " + t +"\n";
}

for(int i = 0;i<MTA.size();i++)
{
string mta2 = MTA[i];
vector<string> users = dest[mta2];
cout<<"Connection between "<<mta<<" and "<<mta2<<endl;
cout<<"     HELO "<<mta<<endl;
pf("     250\n");
cout<<"     MAIL FROM:<"<<s<<">\n";
pf("     250\n");
bool ok = false;
for(int j = 0;j<users.size();j++)
{
cout<<"     RCPT TO:<"<<users[j]<<">"<<endl;
if(addr.count(users[j]))
{
ok = true;
pf("     250\n");
}
else
pf("     550\n");
}
if(ok)
{
cout<<"     DATA\n     354\n"<<data;
pf("     .\n     250\n");
}
pf("     QUIT\n     221\n");
}
}
}

例题5-12　城市正视图（Urban Elevations, ACM/ICPC World Finals 1992, UVa221）

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

int n;
double s[105*2];

struct building
{
int id;
double x,y,w,h,d;
bool operator <(const building& a) const
{
return x<a.x || (x==a.x && y<a.y);
}
//一定要是小于号，不然会报错
}b[105];

bool cover(int i,double mx)
{
return b[i].x<=mx && b[i].x+b[i].w>=mx;
}

bool visible(int i,double mx)
{
if(!cover(i,mx)) return false;
for(int k = 0;k<n;k++)
{
if(cover(k,mx) && b[k].y<b[i].y && b[k].h>=b[i].h)
return false;
}
return true;
}

int main()
{
int kase = 0;
while(sf("%d",&n)==1 && n)
{
for(int i = 0;i<n;i++)
{
sf("%lf%lf%lf%lf%lf",&b[i].x,&b[i].y,&b[i].w,&b[i].d,&b[i].h);
s[i*2]=b[i].x;
s[i*2+1]=b[i].x+b[i].w;
b[i].id=i+1;
}
sort(b,b+n);
sort(s,s+n*2);
int m = unique(s,s+n*2)-s;

if(kase++) blank;
pf("For map #%d, the visible buildings are numbered as follows:\n%d",kase,b[0].id);

for(int i=1;i<n;i++)
{
bool vis = false;
for(int j = 0;j<m-1;j++)
if(visible(i,(s[j]+s[j+1])/2)){vis=true;break;}
if(vis){pf(" %d",b[i].id);};
}
blank;

}
}