lrj紫书第五章

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UVA-1592

技术分享
 1 // UVa1592 Database
 2 // Rujia Liu
 3 // 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。
 4 
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<vector>
 8 #include<string>
 9 #include<map>
10 #include<sstream>
11 using namespace std;
12 
13 typedef pair<int,int> PII;
14 
15 const int maxr = 10000 + 5;
16 const int maxc = 10 + 5;
17 
18 int m, n, db[maxr][maxc], cnt;
19 
20 map<string, int> id;
21 int ID(const string& s) {
22   if(!id.count(s)) {
23     id[s] = ++cnt;
24   }
25   return id[s];
26 }
27 
28 void find() {
29   for(int c1 = 0; c1 < m; c1++)
30     for(int c2 = c1+1; c2 < m; c2++) {
31       map<PII, int> d;
32       for(int i = 0; i < n; i++) {
33         PII p = make_pair(db[i][c1], db[i][c2]);
34         if(d.count(p)) {
35           printf("NO\n");
36           printf("%d %d\n", d[p]+1, i+1);
37           printf("%d %d\n", c1+1, c2+1);
38           return;
39         }
40         d[p] = i;
41       }
42     }
43   printf("YES\n");
44 }
45 
46 
47 int main() {
48   string s;
49   while(getline(cin, s)) {
50     stringstream ss(s);
51     if(!(ss >> n >> m)) break;
52     cnt = 0;
53     id.clear();
54     for(int i = 0; i < n; i++) {
55       getline(cin, s);
56       int lastpos = -1;
57       for(int j = 0; j < m; j++) {
58         int p = s.find(,, lastpos+1);
59         if(p == string::npos) p = s.length();
60         db[i][j] = ID(s.substr(lastpos+1, p - lastpos - 1));
61         lastpos = p;
62       }
63     }
64     find();
65   }
66   return 0;
67 }
View Code

UVA-207

技术分享
  1 // UVa207 PGA Tour Prize Money
  2 // Rujia Liu
  3 #include<cstdio>
  4 #include<cstdlib>
  5 #include<cstring>
  6 #include<cmath>
  7 #include<algorithm>
  8 #include<cassert>
  9 using namespace std;
 10 
 11 #define REP(i,n) for(int i = 0; i < (n); i++)
 12 
 13 const int maxn = 144;
 14 const int n_cut = 70;
 15 
 16 struct Player {
 17   char name[25];
 18   int amateur;
 19   int sc[4];
 20   int sc36, sc72, dq;
 21   int rnds;
 22 } player[maxn];
 23 
 24 int n;
 25 double purse, p[n_cut];
 26 
 27 bool cmp1(const Player& p1, const Player& p2) {
 28   if(p1.sc36 < 0 && p2.sc36 < 0) return false; // equal
 29   if(p1.sc36 < 0) return false; // p2 smaller
 30   if(p2.sc36 < 0) return true; // p1 smaller
 31   return p1.sc36 < p2.sc36;
 32 }
 33 
 34 bool cmp2(const Player& p1, const Player& p2) {
 35   if(p1.dq && p2.dq) {
 36     if(p1.rnds != p2.rnds) return p2.rnds < p1.rnds;
 37     if(p1.sc72 != p2.sc72) return p1.sc72 < p2.sc72;
 38     return strcmp(p1.name, p2.name) < 0;
 39   }
 40   if(p1.dq) return false;
 41   if(p2.dq) return true;
 42   if(p1.sc72 != p2.sc72) return p1.sc72 < p2.sc72;
 43   return strcmp(p1.name, p2.name) < 0;
 44 }
 45 
 46 void print_result() {
 47   printf("Player Name          Place     RD1  RD2");
 48   printf("  RD3  RD4  TOTAL     Money Won\n");
 49   printf("---------------------------------------");
 50   printf("--------------------------------\n");
 51 
 52   int i = 0, pos = 0;
 53   while(i < n) {
 54     if(player[i].dq) {
 55       printf("%s           ",player[i].name);
 56       REP(j,player[i].rnds) printf("%-5d", player[i].sc[j]);
 57       REP(j,4-player[i].rnds) printf("     ");
 58       printf("DQ\n");
 59       i++;
 60       continue;
 61     }
 62 
 63     int j = i;
 64     int m = 0; // number of tied players
 65     bool have_money = false;
 66     double tot = 0.0; // total pooled money
 67     while(j < n && player[i].sc72 == player[j].sc72) {
 68       if(!player[j].amateur) {
 69         m++;          
 70         if(pos < n_cut) {
 71           have_money = true; // yeah! they still have money
 72           tot += p[pos++];
 73         }
 74       }
 75       j++;
 76     }
 77 
 78     // print player [i,j) together because they have the same rank
 79     int rank = i + 1; // rank of all these m players
 80     double amount = purse * tot / m; // if m=0, amount will be nan but we don‘t use it in that case :)
 81     while(i < j) {
 82       printf("%s ", player[i].name);
 83       char t[5];
 84       sprintf(t, "%d%c", rank, m > 1 && have_money && !player[i].amateur ? T :  );
 85       printf("%-10s", t);
 86       REP(e,4) printf("%-5d", player[i].sc[e]);
 87 
 88       // with prize
 89       if(!player[i].amateur && have_money) {
 90         printf("%-10d", player[i].sc72);
 91         printf("$%9.2lf\n", amount / 100.0);
 92       } else
 93         printf("%d\n", player[i].sc72);
 94       i++;
 95     }
 96   }
 97 }
 98 
 99 int main() {
100   int T; 
101   char s[40];
102 
103   gets(s);
104   sscanf(s,"%d",&T);
105   while(T--) {
106     gets(s); // empty line
107 
108     // prize
109     gets(s);
110     sscanf(s,"%lf", &purse);
111     REP(i,n_cut) {
112       gets(s);
113       sscanf(s, "%lf", &p[i]);
114     }
115 
116     // players
117     gets(s);
118     sscanf(s, "%d", &n);
119     assert(n <= 144);
120     REP(k,n) {
121       // read a 32-character line
122       gets(s);
123 
124       // player name
125       strncpy(player[k].name, s, 20);      
126       player[k].name[20] = 0;
127       player[k].amateur = 0;
128       if(strchr(player[k].name, *)) {
129         player[k].amateur = 1;
130       }
131 
132       // scores
133       player[k].sc36 = player[k].sc72 = player[k].dq=0;
134       memset(player[k].sc, -1, sizeof(player[k].sc));
135       REP(i,4) {
136         // raw score
137         char t[5];
138         REP(j,3) t[j] = s[20 + i*3 + j]; t[3] = \0;
139 
140         // parse
141         if(!sscanf(t,"%d", &player[k].sc[i])) {
142           // DQ!
143           player[k].rnds = i;
144           player[k].dq = -1;
145           if(i < 2) player[k].sc36 = -1;
146           break; // skip other rounds (filled with -1, initially)
147         } else {          
148           player[k].sc72 += player[k].sc[i];
149           if(i < 2)
150             player[k].sc36 += player[k].sc[i];
151         }
152       }
153     }
154 
155     // round 1
156     sort(player, player+n, cmp1);
157     assert(player[n_cut-1].sc36 >= 0);
158     for(int i = n_cut-1; i < n; i++)
159       if(i == n-1 || player[i].sc36 != player[i+1].sc36) { n = i+1; break; }
160 
161     // round 2
162     sort(player, player+n, cmp2);
163     
164     // print result
165     print_result();
166  
167     if(T) printf("\n");
168   }
169   
170   return 0;
171 }
View Code

 

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