EOJ 2239 Friends

http://www.acm.cs.ecnu.edu.cn/problem.php?problemid=2239

题目的大概意思就是一些人要投票去三个地方，要注意的是同一个人如果同时有两或以上个想去的地方，那他就会弃权，其他的只要按照题目给的逻辑就可以了

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <stack>
 8 #include <vector>
 9 #include <string>
10 #include <set>
11 #include <cctype>
12 #include <cstdlib>
13 #include <map>
14 #define maxn 100005
15 #define inf 0x3f3f3f
16 #define LL long long
17 using namespace std;
18  
19 string name[8] = {"Anne", "Bob", "Karin", "Dave", "Charly", "Edward", "Frank"};
20 int mark[9]; //mark用于标记是否到场
21 string stay[4] = {"cinema", "cocktail bar", "disco", "stay at the Hacienda"};
22 char s[1005];
23 int note[4]; //投票的
24  
25 void p(string st){
26     for (int i = 0; i < 7; i++)
27         if (st == name[i]) mark[i]++;
28 }
29  
30 void init(){
31     memset(mark, 0, sizeof(mark));
32     memset(note, 0, sizeof(note));
33     string st;
34     for (int i = 0; s[i]; i++){
35         if (s[i] != ' ') st += s[i];
36         else{
37             p(st);
38             st = "";
39         }
40     }
41     p(st);
42 }
43  
44 void solve(){
45     if (mark[0]) note[0]++; //Anne
46     if (mark[1]){ // Bob
47         int f1 = 0, f2 = 0;
48         if (mark[2]) f1 = 1;
49         if (mark[3] || mark[5] || !mark[0]) f2 = 1;
50         if (f1 == 1 && f2 == 0) note[2]++;
51         else if (f1 == 0 && f2 == 1) note[1]++;
52     } 
53     if (mark[2]){//Karin
54         if (mark[4]) note[2]++;
55         else if (mark[0]) note[0]++;
56         else note[1]++;
57     } 
58     //if (mark[3]) //Dave
59     if (mark[4]){ //Charly
60         if (mark[0]) note[0]++;
61     }
62     if (mark[5]){ //Edward
63         if (mark[0] && !mark[4]) note[1]++;
64         else note[0]++;
65     }
66     if (mark[6]){ //Frank
67         if (!mark[0] && !mark[1]) note[0]++;
68         if (mark[0]) note[2]++;
69     }    
70     if (note[0] > note[1] && note[0] > note[2]) cout << stay[0] << endl;
71     else if (note[1] > note[0] && note[1] > note[2]) cout << stay[1] << endl;
72     else if (note[2] > note[0] && note[2] > note[1]) cout << stay[2] << endl;
73     else cout << stay[3] << endl; 
74 }
75  
76 int main(){
77     int T;
78     cin >> T;
79     getchar();
80     for (int cnt = 1; cnt <= T; cnt++){
81         if (cnt != 1) printf("
");
82         gets(s);
83         init();
84         printf("Scenario #%d:
", cnt);
85         solve();
86     }
87     return 0;
88 }

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