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  • hdu 4194(模拟)

    符合三者之一的则不满足规定,求不满足规定的个数。直接模拟。

    1.被同一个人审查多次

    2.被和自己同一组织的审查

    3.被审查次数不等于k

    代码如下:

     1 /**************************************************
     2  * Author     : xiaohao Z
     3  * Blog     : http://www.cnblogs.com/shu-xiaohao/
     4  * Last modified : 2014-06-28 17:36
     5  * Filename     : hdu_4194.cpp
     6  * Description     : 
     7  * ************************************************/
     8 
     9 #include <iostream>
    10 #include <cstdio>
    11 #include <cstring>
    12 #include <cstdlib>
    13 #include <cmath>
    14 #include <algorithm>
    15 #include <queue>
    16 #include <stack>
    17 #include <vector>
    18 #include <set>
    19 #include <map>
    20 #define MP(a, b) make_pair(a, b)
    21 #define PB(a) push_back(a)
    22 
    23 using namespace std;
    24 typedef long long ll;
    25 typedef pair<int, int> pii;
    26 typedef pair<unsigned int,unsigned int> puu;
    27 typedef pair<int, double> pid;
    28 typedef pair<ll, int> pli;
    29 typedef pair<int, ll> pil;
    30 
    31 const int INF = 0x3f3f3f3f;
    32 const double eps = 1E-6;
    33 const int LEN = 1001;
    34 map<string, int> mp;
    35 int n, k, top;
    36 int num[LEN];
    37 vector<int> tab[LEN];
    38 
    39 int ch(string s){
    40     if(!mp.count(s)) mp[s] = top++;
    41     return mp[s];
    42 }
    43 
    44 int main()
    45 {
    46 //    freopen("in.txt", "r", stdin);
    47 
    48     char str[1010];
    49     int tmp;
    50     while(scanf("%d%d", &k, &n) != EOF){
    51         if(!k && !n) break;
    52         for(int i=0; i<LEN; i++) tab[i].clear();
    53         top = 0; mp.clear();
    54         for(int i=0; i<n; i++){
    55             scanf("%s", &str);
    56             num[i] = ch(str);
    57             for(int j=0; j<k; j++){
    58                 scanf("%d", &tmp);
    59                 tmp --;
    60                 tab[tmp].PB(i);
    61             }
    62         }
    63         int ans = 0;
    64         for(int i=0; i<n; i++){
    65             if(tab[i].size() != k) ans ++;
    66             else{
    67                 map<int, int> st;
    68                 int cc = 0;
    69                 for(int j=0; j<tab[i].size(); j++){
    70                     int x = tab[i][j], y = num[i];
    71                     if(num[x] == y || st.count(x)){
    72                         ans ++;break;
    73                     }else st[x] = 1;
    74                 }
    75             }
    76         }
    77         if(ans == 0) puts("NO PROBLEMS FOUND");
    78         else if(ans == 1) puts("1 PROBLEM FOUND");
    79         else printf("%d PROBLEMS FOUND
    ", ans);
    80     }
    81     return 0;
    82 }
    View Code
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  • 原文地址:https://www.cnblogs.com/shu-xiaohao/p/3813972.html
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