Jamie's Contact Groups
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 9227 | Accepted: 3180 |
Description
Jamie is a very popular girl and has quite a lot of friends, so she always keeps a very long contact list in her cell phone. The contact list has become so long that it often takes a long time for her to browse through the whole list to find a friend's number. As Jamie's best friend and a programming genius, you suggest that she group the contact list and minimize the size of the largest group, so that it will be easier for her to search for a friend's number among the groups. Jamie takes your advice and gives you her entire contact list containing her friends' names, the number of groups she wishes to have and what groups every friend could belong to. Your task is to write a program that takes the list and organizes it into groups such that each friend appears in only one of those groups and the size of the largest group is minimized.
Input
There will be at most 20 test cases. Ease case starts with a line containing two integers N and M. where N is the length of the contact list and M is the number of groups. N lines then follow. Each line contains a friend's name and the groups the friend could belong to. You can assume N is no more than 1000 and M is no more than 500. The names will contain alphabet letters only and will be no longer than 15 characters. No two friends have the same name. The group label is an integer between 0 and M - 1. After the last test case, there is a single line `0 0' that terminates the input.
Output
For each test case, output a line containing a single integer, the size of the largest contact group.
Sample Input
3 2 John 0 1 Rose 1 Mary 1 5 4 ACM 1 2 3 ICPC 0 1 Asian 0 2 3 Regional 1 2 ShangHai 0 2 0 0
Sample Output
2 2
Source
This problem has to do for a while, he is to match the minimum packet after the maximum demand for all matches, although it is obviously a half, but ...... I feel old to change the code to match over ... and the most deadly is my own test answers all had trained, trained, trained, my first thought: As the subject of the request so that the upper bound of the smallest group, then we may be constrained by judging the conditions, how to rein it If we have not found a matching set of people, that direct match because one must be a minimum, if not the first match, then let him go match to match others, matching rule is still the case, but If there is a node does not satisfy the above-mentioned circumstances, meaning that all groups are attached to this node have been matched by someone else, then its first direct elections to match it, but the problem here, because we can not the first match to determine who is the minimum after the match, because this will certainly be added to the minimum value will not affect the answer .so below half the answer.
Dichotomous answer:
Binary answer, each with matching Check, if the upper bound can be changed so that a complete match, then the change r, L or changed, then ok. Output mid.
1 /************************************************************************* 2 > File Name: poj-2289.jamies_contact_groups.cpp 3 > Author: CruelKing 4 > Mail: [email protected] 5 > Created Time: 2019年09月03日 星期二 21时09分58秒 6 ************************************************************************/ 7 /* 8 #include <iostream> 9 #include <sstream> 10 #include <string> 11 #include <cstring> 12 using namespace std; 13 14 const int maxn = 1000 + 5, maxm = 500 + 5, inf = 0x3f3f3f3f; 15 16 int n, m; 17 string str; 18 char name[20]; 19 int linker[maxm][maxn]; 20 bool used[maxm], g[maxn][maxm]; 21 22 bool dfs(int u) { 23 for(int v = 0; v < m; v ++) { 24 if(g[u][v] && !used[v]) { 25 used[v] = true; 26 if(linker[v][0] == 0) { 27 linker[v][++ linker[v][0]] = u; 28 return true; 29 } 30 for(int i = 1; i <= linker[v][0]; i ++) { 31 if(dfs(linker[v][i])) { 32 linker[v][i] = u; 33 return true; 34 } 35 } 36 linker[v][++ linker[v][0]] = u; 37 return true; 38 } 39 } 40 return false; 41 } 42 43 int main() { 44 while(cin >> n >> m && (n || m)) { 45 memset(g, false, sizeof g); 46 for(int i = 0; i < n; i ++) { 47 cin >> name; 48 getline(cin, str); 49 stringstream ss; 50 ss << str; 51 int x; 52 while(ss >> x) 53 g[i][x] = true; 54 } 55 int res = 0; 56 for(int i = 0; i < m; i ++) linker[i][0] = 0; 57 for(int i = 0; i < n; i ++) { 58 memset(used, false, sizeof used); 59 if(dfs(i)) res ++; 60 } 61 int M = 0; 62 for(int i = 0; i < m; i ++) { 63 if(M < linker[i][0]) M = linker[i][0]; 64 } 65 cout << M << endl; 66 } 67 return 0; 68 } 69 */ 70 #include <iostream> 71 #include <sstream> 72 #include <string> 73 #include <cstring> 74 #define mid ((l + r) >> 1) 75 using namespace std; 76 77 const int maxn = 1000 + 5, maxm = 500 + 5, inf = 0x3f3f3f3f; 78 79 int n, m, l, r; 80 string str; 81 char name[20]; 82 int linker[maxm][maxn]; 83 bool used[maxm], g[maxn][maxm]; 84 85 bool dfs(int u) { 86 for(int v = 0; v < m; v ++) { 87 if(g[u][v] && !used[v]) { 88 used[v] = true; 89 if(linker[v][0] < mid) { 90 linker[v][++ linker[v][0]] = u; 91 return true; 92 } 93 for(int i = 1; i <= mid; i ++) { 94 if(dfs(linker[v][i])) { 95 linker[v][i] = u; 96 return true; 97 } 98 } 99 } 100 } 101 return false; 102 } 103 104 int main() { 105 while(cin >> n >> m && (n || m)) { 106 memset(g, false, sizeof g); 107 for(int i = 0; i < n; i ++) { 108 cin >> name; 109 getline(cin, str); 110 stringstream ss; 111 ss << str; 112 int x; 113 while(ss >> x) 114 g[i][x] = true; 115 } 116 /* 117 int res = 0; 118 l = 0, r = maxn << 1; 119 for(int i = 0; i < m; i ++) linker[i][0] = 0; 120 for(int i = 0; i < n; i ++) { 121 memset(used, false, sizeof used); 122 if(dfs(i)) res ++; 123 } 124 cout << res << endl; 125 */ 126 l= 0, r = n; 127 int res; 128 while(l <= r) { 129 res = 0; 130 for(int i = 0; i < m; i ++) linker[i][0] = 0; 131 for(int i = 0; i < n; i ++) { 132 memset(used, false, sizeof used); 133 if(dfs(i)) res ++; 134 } 135 if(n == res) r = mid - 1; 136 else l = mid + 1; 137 } 138 cout << mid + 1 << endl; 139 } 140 return 0; 141 }