[CF151B]Phone Numbers(暴力,模拟)

题目链接:http://codeforces.com/contest/151/problem/B

题意:三种电话号码:假如数字都相同,则说明是taxi;严格下降的序列则说明是pizza;其他是girl。输出拥有某种号码最多的那个人的姓名,假如有同样多的,要按照输入顺序输出。

暴力搞,统计每个人各种电话号码有多少,排序三遍。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye!
  3 ┓┏┓┏┓┃キリキリ♂ mind!
  4 ┛┗┛┗┛┃\○/
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%I64d", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onecnt(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef pair<LL, LL> pll;
 65 typedef map<string, int> msi;
 66 typedef vector<int> vi;
 67 typedef vector<LL> vl;
 68 typedef vector<vl> vvl;
 69 typedef vector<bool> vb;
 70 
 71 const int maxn = 110;
 72 typedef struct Node {
 73     string name;
 74     int n, idx;
 75     int t, p, g;
 76 }Node;
 77 int n;
 78 Node p[maxn];
 79 
 80 int judge(string tmp) {
 81     if(tmp[0]==tmp[1]&&tmp[1]==tmp[3]&&tmp[3]==tmp[4]&&tmp[4]==tmp[6]&&tmp[6]==tmp[7]) return 1;
 82     if(tmp[0]>tmp[1]&&tmp[1]>tmp[3]&&tmp[3]>tmp[4]&&tmp[4]>tmp[6]&&tmp[6]>tmp[7]) return 2;
 83     return 3;
 84 }
 85 
 86 bool cmp1(Node a, Node b) {
 87     if(a.t == b.t) return a.idx < b.idx;
 88     return a.t > b.t;
 89 }
 90 
 91 bool cmp2(Node a, Node b) {
 92     if(a.p == b.p) return a.idx < b.idx;
 93     return a.p > b.p;
 94 }
 95 
 96 bool cmp3(Node a, Node b) {
 97     if(a.g == b.g) return a.idx < b.idx;
 98     return a.g > b.g;
 99 }
100 
101 int main() {
102     // FRead();
103     string tmp;
104     while(~Rint(n)) {
105         int cur;
106         For(i, 1, n+1) {
107             cin >> p[i].n >> p[i].name;
108             p[i].t = p[i].p = p[i].g = 0;
109             p[i].idx = i;
110             Rep(j, p[i].n) {
111                 cin >> tmp;
112                 int q = judge(tmp);
113                 if(q == 1) p[i].t++;
114                 else if(q == 2) p[i].p++;
115                 else if(q == 3) p[i].g++;
116             }
117         }
118 
119         sort(p+1,p+n+1,cmp1);
120         printf("If you want to call a taxi, you should call: ");
121         cout << p[1].name;
122         cur = p[1].t;
123         For(i, 2, n+1) {
124             if(p[i].t == cur) cout << ", " << p[i].name;
125             else break;
126         }
127         cout << "." << endl;
128 
129         sort(p+1,p+n+1,cmp2);
130         printf("If you want to order a pizza, you should call: ");
131         cout << p[1].name;
132         cur = p[1].p;
133         For(i, 2, n+1) {
134             if(p[i].p == cur) cout << ", " << p[i].name;
135             else break;
136         }
137         cout << "." << endl;
138 
139         sort(p+1,p+n+1,cmp3);
140         printf("If you want to go to a cafe with a wonderful girl, you should call: ");
141         cout << p[1].name;
142         cur = p[1].g;
143         For(i, 2, n+1) {
144             if(p[i].g == cur) cout << ", " << p[i].name;
145             else break;
146         }
147         cout << "." << endl;
148     }
149     RT 0;
150 }

 

posted @ 2016-08-22 20:08  Kirai  阅读(284)  评论(0编辑  收藏  举报