第五章 UVA

Uva 12096

自己代码

#include <bits/stdc++.h>using namespace std;typedef set<int> Set;map<Set, int> mark;int cnt = 0;Set s[2010];int ID(Set x){    if(!mark.count(x)) {        mark[x] = cnt++;        s[mark[x]] = x;    }    return mark[x];}int main(){    int t;    cin >> t;    while( t--){        int n;        cin >> n;        cnt = 0;        mark.clear();        stack<int> sta;        for(int i = 0; i < 2000; i++) s[i].clear();        for(int i = 0; i < n; i++){            string quer;            cin >> quer;            if(quer[0] == 'P') sta.push(ID(Set()));            else if(quer[0] == 'D') sta.push(sta.top());            else{                Set x1 = s[sta.top()]; sta.pop();                Set x2 = s[sta.top()]; sta.pop();                Set x;                //printf(" x1 %d  x2 %d\n", ID(x1), ID(x2));                //cout<<quer[0]<<endl;                if(quer[0] == 'U') set_union(x1.begin(), x1.end(), x2.begin(), x2.end(), inserter(x, x.begin()));                if(quer[0] == 'I') set_intersection(x1.begin(), x1.end(), x2.begin(), x2.end(), inserter(x, x.begin()));                if(quer[0] == 'A') x = x2, x.insert(ID(x1));                sta.push(ID(x));            }            printf("%d\n", s[sta.top()].size());        }        printf("***\n");    }    return 0;}

代码库：

#include <iostream>  #include <set>  #include <map>  #include <stack>  #include <vector>  #include <algorithm>  using namespace std;  typedef set<int> Set;  map<Set,int> IDcache;  vector<Set> Setcache;  int ID(Set x)  {      if(IDcache.count(x)) return IDcache[x];      Setcache.push_back(x);      return IDcache[x]=Setcache.size()-1;  }  #define ALL(x) x.begin(),x.end()  #define INS(x) inserter(x,x.begin())  stack<int> s;  int main(void)  {      int n;      cin>>n;      while(n--)      {          int m;          cin>>m;          for(int i=0;i<m;i++)          {              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]=='U') set_union(ALL(x1),ALL(x2),INS(x));                  if(op[0]=='I') set_intersection(ALL(x1),ALL(x2),INS(x));                  if(op[0]=='A') {x=x2;x.insert(ID(x1));}                  s.push(ID(x));              }              cout<<Setcache[s.top()].size()<<endl;          }          cout<<"***"<<endl;      }      return 0;  }  

uva 540

// UVa540 Team Queue// Rujia Liu#include<cstdio>#include<queue>#include<map>using namespace std;const int maxt = 1000 + 10;int main() {  int t, kase = 0;  while(scanf("%d", &t) == 1 && t) {    printf("Scenario #%d\n", ++kase);    // 记录所有人的团队编号    map<int, int> team; // team[x]表示编号为x的人所在的团队编号    for(int i = 0; i < t; i++) {      int n, x;      scanf("%d", &n);      while(n--) { scanf("%d", &x); team[x] = i; }    }    // 模拟    queue<int> q, q2[maxt]; // q是团队的队列，而q2[i]是团队i成员的队列    for(;;) {      int x;      char cmd[10];      scanf("%s", cmd);      if(cmd[0] == 'S') break;      else if(cmd[0] == 'D') {        int t = q.front();        printf("%d\n", q2[t].front()); q2[t].pop();        if(q2[t].empty()) q.pop(); // 团体t全体出队列      }      else if(cmd[0] == 'E') {        scanf("%d", &x);        int t = team[x];        if(q2[t].empty()) q.push(t); // 团队t进入队列        q2[t].push(x);      }    }    printf("\n");  }  return 0;} 

uva 136

// UVa136 Ugly Numbers// Rujia Liu#include<iostream>#include<vector>#include<queue>#include<set>using namespace std;typedef long long LL;const int coeff[3] = {2, 3, 5};int main() {  priority_queue<LL, vector<LL>, greater<LL> > pq;  set<LL> s;  pq.push(1);  s.insert(1);  for(int i = 1; ; i++) {    LL x = pq.top(); pq.pop();    if(i == 1500) {      cout << "The 1500'th ugly number is " << x << ".\n";      break;    }    for(int j = 0; j < 3; j++) {      LL x2 = x * coeff[j];      if(!s.count(x2)) { s.insert(x2); pq.push(x2); }    }  }  return 0;}

uva 400

// UVa400 Unix ls// Rujia Liu#include<iostream>#include<string>#include<algorithm>using namespace std;const int maxcol = 60;const int maxn = 100 + 5;string filenames[maxn];// 输出字符串s，长度不足len时补字符extravoid 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;  while(cin >> n) {    int M = 0;    for(int i = 0; i < n; i++) {      cin >> filenames[i];      M = max(M, (int)filenames[i].length());    }    // 计算列数cols和行数rows    int cols = (maxcol - M) / (M + 2) + 1, rows = (n - 1) / cols + 1;    print("", 60, '-');    cout << "\n";    sort(filenames, filenames+n);    for(int r = 0; r < rows; r++) {      for(int c = 0; c < cols; c++) {        int idx = c * rows + r;        if(idx < n) print(filenames[idx], c == cols-1 ? M : M+2, ' ');      }      cout << "\n";    }  }  return 0;}

uva 1592

// UVa1592 Database// Rujia Liu// 本程序只是为了演示STL各种用法，效率较低。实践中一般用C字符串和哈希表来实现。#include<iostream>#include<cstdio>#include<vector>#include<string>#include<map>#include<sstream>using namespace std;typedef pair<int,int> PII;const int maxr = 10000 + 5;const int maxc = 10 + 5;int m, n, db[maxr][maxc], cnt;map<string, int> id;int ID(const string& s) {  if(!id.count(s)) {    id[s] = ++cnt;  }  return id[s];}void find() {  for(int c1 = 0; c1 < m; c1++)    for(int c2 = c1+1; c2 < m; c2++) {      map<PII, int> d;      for(int i = 0; i < n; i++) {        PII p = make_pair(db[i][c1], db[i][c2]);        if(d.count(p)) {          printf("NO\n");          printf("%d %d\n", d[p]+1, i+1);          printf("%d %d\n", c1+1, c2+1);          return;        }        d[p] = i;      }    }  printf("YES\n");}int main() {  string s;  while(getline(cin, s)) {    stringstream ss(s);    if(!(ss >> n >> m)) break;    cnt = 0;    id.clear();    for(int i = 0; i < n; i++) {      getline(cin, s);      int lastpos = -1;      for(int j = 0; j < m; j++) {        int p = s.find(',', lastpos+1);        if(p == string::npos) p = s.length();        db[i][j] = ID(s.substr(lastpos+1, p - lastpos - 1));        lastpos = p;      }    }    find();  }  return 0;}

uva 207

// UVa207 PGA Tour Prize Money// Rujia Liu#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#include<cassert>using namespace std;#define REP(i,n) for(int i = 0; i < (n); i++)const int maxn = 144;const int n_cut = 70;struct Player {  char name[25];  int amateur;  int sc[4];  int sc36, sc72, dq;  int rnds;} player[maxn];int n;double purse, p[n_cut];bool cmp1(const Player& p1, const Player& p2) {  if(p1.sc36 < 0 && p2.sc36 < 0) return false; // equal  if(p1.sc36 < 0) return false; // p2 smaller  if(p2.sc36 < 0) return true; // p1 smaller  return p1.sc36 < p2.sc36;}bool cmp2(const Player& p1, const Player& p2) {  if(p1.dq && p2.dq) {    if(p1.rnds != p2.rnds) return p2.rnds < p1.rnds;    if(p1.sc72 != p2.sc72) return p1.sc72 < p2.sc72;    return strcmp(p1.name, p2.name) < 0;  }  if(p1.dq) return false;  if(p2.dq) return true;  if(p1.sc72 != p2.sc72) return p1.sc72 < p2.sc72;  return strcmp(p1.name, p2.name) < 0;}void print_result() {  printf("Player Name          Place     RD1  RD2");  printf("  RD3  RD4  TOTAL     Money Won\n");  printf("---------------------------------------");  printf("--------------------------------\n");  int i = 0, pos = 0;  while(i < n) {    if(player[i].dq) {      printf("%s           ",player[i].name);      REP(j,player[i].rnds) printf("%-5d", player[i].sc[j]);      REP(j,4-player[i].rnds) printf("     ");      printf("DQ\n");      i++;      continue;    }    int j = i;    int m = 0; // number of tied players    bool have_money = false;    double tot = 0.0; // total pooled money    while(j < n && player[i].sc72 == player[j].sc72) {      if(!player[j].amateur) {        m++;                  if(pos < n_cut) {          have_money = true; // yeah! they still have money          tot += p[pos++];        }      }      j++;    }    // print player [i,j) together because they have the same rank    int rank = i + 1; // rank of all these m players    double amount = purse * tot / m; // if m=0, amount will be nan but we don't use it in that case :)    while(i < j) {      printf("%s ", player[i].name);      char t[5];      sprintf(t, "%d%c", rank, m > 1 && have_money && !player[i].amateur ? 'T' : ' ');      printf("%-10s", t);      REP(e,4) printf("%-5d", player[i].sc[e]);      // with prize      if(!player[i].amateur && have_money) {        printf("%-10d", player[i].sc72);        printf("$%9.2lf\n", amount / 100.0);      } else        printf("%d\n", player[i].sc72);      i++;    }  }}int main() {  int T;   char s[40];  gets(s);  sscanf(s,"%d",&T);  while(T--) {    gets(s); // empty line    // prize    gets(s);    sscanf(s,"%lf", &purse);    REP(i,n_cut) {      gets(s);      sscanf(s, "%lf", &p[i]);    }    // players    gets(s);    sscanf(s, "%d", &n);    assert(n <= 144);    REP(k,n) {      // read a 32-character line      gets(s);      // player name      strncpy(player[k].name, s, 20);            player[k].name[20] = 0;      player[k].amateur = 0;      if(strchr(player[k].name, '*')) {        player[k].amateur = 1;      }      // scores      player[k].sc36 = player[k].sc72 = player[k].dq=0;      memset(player[k].sc, -1, sizeof(player[k].sc));      REP(i,4) {        // raw score        char t[5];        REP(j,3) t[j] = s[20 + i*3 + j]; t[3] = '\0';        // parse        if(!sscanf(t,"%d", &player[k].sc[i])) {          // DQ!          player[k].rnds = i;          player[k].dq = -1;          if(i < 2) player[k].sc36 = -1;          break; // skip other rounds (filled with -1, initially)        } else {                    player[k].sc72 += player[k].sc[i];          if(i < 2)            player[k].sc36 += player[k].sc[i];        }      }    }    // round 1    sort(player, player+n, cmp1);    assert(player[n_cut-1].sc36 >= 0);    for(int i = n_cut-1; i < n; i++)      if(i == n-1 || player[i].sc36 != player[i+1].sc36) { n = i+1; break; }    // round 2    sort(player, player+n, cmp2);    // print result    print_result();    if(T) printf("\n");  }  return 0;}

uva 814

// UVa814 The Letter Carrier's Rounds// Rujia Liu#include<iostream>#include<string>#include<vector>#include<set>#include<map>using namespace std;void parse_address(const string& s, string& user, string& mta) {  int k = s.find('@');  user = s.substr(0, k);  mta = s.substr(k+1);}int main() {  int k;  string s, t, user1, mta1, user2, mta2;  set<string> addr;  // 输入所有MTA，转化为地址列表  while(cin >> s && s != "*") {    cin >> s >> k;    while(k--) { cin >> t; addr.insert(t + "@" + s); }  }  while(cin >> s && s != "*") {    parse_address(s, user1, mta1); // 处理发件人地址    vector<string> mta; // 所有需要连接的mta，按照输入顺序    map<string, vector<string> > dest; // 每个mta需要发送的用户    set<string> vis;    while(cin >> t && t != "*") {      parse_address(t, user2, mta2); // 处理收件人地址      if(vis.count(t)) continue;     // 重复的收件人      vis.insert(t);      if(!dest.count(mta2)) { mta.push_back(mta2); dest[mta2] = vector<string>(); }      dest[mta2].push_back(t);    }    getline(cin, t); // 把"*"这一行的回车吃掉    // 输入邮件正文    string data;    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 " << mta1 << " and " << mta2 <<endl;      cout << "     HELO " << mta1 << "\n";      cout << "     250\n";      cout << "     MAIL FROM:<" << s << ">\n";      cout << "     250\n";      bool ok = false;      for(int i = 0; i < users.size(); i++) {        cout << "     RCPT TO:<" << users[i] << ">\n";        if(addr.count(users[i])) { ok = true; cout << "     250\n"; }        else cout << "     550\n";      }      if(ok) {        cout << "     DATA\n";        cout << "     354\n";        cout << data;        cout << "     .\n";        cout << "     250\n";      }      cout << "     QUIT\n";      cout << "     221\n";    }  }  return 0;}

uva 221

// UVa221 Urban Elevations// Rujia Liu#include<cstdio>#include<algorithm>using namespace std;const int maxn = 100 + 5;struct Building {  int id;  double x, y, w, d, h;  bool operator < (const Building& rhs) const {    return x < rhs.x || (x == rhs.x && y < rhs.y);  }} b[maxn];int n;double x[maxn*2];bool cover(int i, double mx) {  return b[i].x <= mx && b[i].x+b[i].w >= mx;}// 判断建筑物i在x=mx处否可见bool visible(int i, double mx) {  if(!cover(i, mx)) return false;  for(int k = 0; k < n; k++)    if(b[k].y < b[i].y && b[k].h >= b[i].h && cover(k, mx)) return false;  return true;}int main() {  int kase = 0;  while(scanf("%d", &n) == 1 && n) {    for(int i = 0; i < n; i++) {      scanf("%lf%lf%lf%lf%lf", &b[i].x, &b[i].y, &b[i].w, &b[i].d, &b[i].h);      x[i*2] = b[i].x; x[i*2+1] = b[i].x + b[i].w;      b[i].id = i+1;    }    sort(b, b+n);    sort(x, x+n*2);    int m = unique(x, x+n*2) - x; // x坐标排序后去重，得到m个坐标    if(kase++) printf("\n");    printf("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, (x[j] + x[j+1]) / 2)) { vis = true; break; }      if(vis) printf(" %d", b[i].id);    }    printf("\n");  }  return 0;}