POJ2289 Jamie's Contact Groups —— 二分图多重匹配/最大流 + 二分

时间：2017-11-12 17:31:01 阅读：184 评论：0 收藏：0 [点我收藏+]

标签：多重 getchar none ret contact eof sample hose using

题目链接：https://vjudge.net/problem/POJ-2289

Jamie‘s Contact Groups

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 8147		Accepted: 2736

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

Shanghai 2004

题解：

多重匹配：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <string>
 6 #include <vector>
 7 #include <map>
 8 #include <set>
 9 #include <queue>
10 #include <sstream>
11 #include <algorithm>
12 using namespace std;
13 const int INF = 2e9;
14 const int MOD = 1e9+7;
15 const int MAXM = 5e2+10;
16 const int MAXN = 1e3+10;
17 
18 int uN, vN;
19 int num[MAXM], linker[MAXM][MAXN];
20 bool g[MAXN][MAXM], used[MAXM];
21 
22 bool dfs(int u)
23 {
24     for(int v = 0; v<vN; v++)
25     if(g[u][v] && !used[v])
26     {
27         used[v] = true;
28         if(linker[v][0]<num[v])
29         {
30             linker[v][++linker[v][0]] = u;
31             return true;
32         }
33         for(int i = 1; i<=num[v]; i++)
34         if(dfs(linker[v][i]))
35         {
36             linker[v][i] = u;
37             return true;
38         }
39     }
40     return false;
41 }
42 
43 bool hungary(int mid)
44 {
45     for(int i = 0; i<vN; i++)
46     {
47         num[i] = mid;
48         linker[i][0] = 0;
49     }
50     for(int u = 0; u<uN; u++)
51     {
52         memset(used, false, sizeof(used));
53         if(!dfs(u)) return false;
54     }
55     return true;
56 }
57 
58 char tmp[100000];
59 int main()
60 {
61     while(scanf("%d%d", &uN, &vN) && (uN||vN))
62     {
63         memset(g, false, sizeof(g));
64         getchar();
65         for(int i = 0; i<uN; i++)
66         {
67             gets(tmp);
68             int j = 0, len = strlen(tmp);
69             while(tmp[j]!=‘ ‘ && j<len) j++;
70             j++;
71             for(int v = 0; j<=len; j++)
72             {
73                 if(tmp[j]==‘ ‘||j==len)
74                 {
75                     g[i][v] = true;
76                     v = 0;
77                 }
78                 else v = v*10+(tmp[j]-‘0‘);
79             }
80         }
81 
82         int l = 1, r = uN;
83         while(l<=r)
84         {
85             int mid = (l+r)>>1;
86             if(hungary(mid))
87                 r = mid - 1;
88             else
89                 l = mid + 1;
90         }
91         printf("%d\n", l);
92     }
93 }

View Code

最大流：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <string>
  6 #include <vector>
  7 #include <map>
  8 #include <set>
  9 #include <queue>
 10 #include <sstream>
 11 #include <algorithm>
 12 using namespace std;
 13 const int INF = 2e9;
 14 const int MOD = 1e9+7;
 15 const int MAXM = 5e2+10;
 16 const int MAXN = 2e3+10;
 17 
 18 struct Edge
 19 {
 20     int to, next, cap, flow;
 21 }edge[MAXN*MAXN];
 22 int tot, head[MAXN];
 23 
 24 int uN, vN, maze[MAXN][MAXN];
 25 int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
 26 
 27 void add(int u, int v, int w)
 28 {
 29     edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0;
 30     edge[tot].next = head[u]; head[u] = tot++;
 31     edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0;
 32     edge[tot].next = head[v]; head[v] = tot++;
 33 }
 34 
 35 int sap(int start, int end, int nodenum)
 36 {
 37     memset(gap,0,sizeof(gap));
 38     memset(dep,0,sizeof(dep));
 39     memcpy(cur,head,sizeof(head));
 40 
 41     int u = start;
 42     pre[u] = -1;
 43     gap[0] = nodenum;
 44     int maxflow = 0;
 45     while(dep[start]<nodenum)
 46     {
 47         bool flag = false;
 48         for(int i = cur[u]; i!=-1; i=edge[i].next)
 49         {
 50             int v = edge[i].to;
 51             if(edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
 52             {
 53                 flag = true;
 54                 cur[u] = pre[v] = i;
 55                 u = v;
 56                 break;
 57             }
 58         }
 59 
 60         if(flag)
 61         {
 62             if(u==end)
 63             {
 64                 int minn = INF;
 65                 for(int i = pre[u]; i!=-1; i=pre[edge[i^1].to])
 66                     if(minn>edge[i].cap-edge[i].flow)
 67                         minn = edge[i].cap-edge[i].flow;
 68                 for(int i = pre[u]; i!=-1; i=pre[edge[i^1].to])
 69                 {
 70                     edge[i].flow += minn;
 71                     edge[i^1].flow -= minn;
 72                 }
 73                 u = start;
 74                 maxflow += minn;
 75             }
 76         }
 77 
 78         else
 79         {
 80             int minn = nodenum;
 81             for(int i = head[u]; i!=-1; i=edge[i].next)
 82                 if(edge[i].cap-edge[i].flow && dep[edge[i].to]<minn)
 83                 {
 84                     minn = dep[edge[i].to];
 85                     cur[u] = i;
 86                 }
 87             gap[dep[u]]--;
 88             if(gap[dep[u]]==0) break;
 89             dep[u] = minn+1;
 90             gap[dep[u]]++;
 91             if(u!=start) u = edge[pre[u]^1].to;
 92         }
 93     }
 94     return maxflow;
 95 }
 96 
 97 bool test(int mid)
 98 {
 99     tot = 0;
100     memset(head, -1, sizeof(head));
101     for(int i = 0; i<uN; i++)
102     {
103         add(uN+vN, i, 1);
104         for(int j = 0; j<vN; j++)
105             if(maze[i][j])
106                 add(i, uN+j, 1);
107     }
108     for(int i = 0; i<vN; i++)
109         add(uN+i, uN+vN+1, mid);
110 
111     int maxflow = sap(uN+vN, uN+vN+1, uN+vN+2);
112     return maxflow == uN;
113 }
114 
115 char tmp[100000];
116 int main()
117 {
118     while(scanf("%d%d", &uN, &vN) && (uN||vN))
119     {
120         memset(maze, 0, sizeof(maze));
121         getchar();
122         for(int i = 0; i<uN; i++)
123         {
124             gets(tmp);
125             int j = 0, len = strlen(tmp);
126             while(tmp[j]!=‘ ‘ && j<len) j++;
127             j++;
128             for(int v = 0; j<=len; j++)
129             {
130                 if(tmp[j]==‘ ‘||j==len)
131                 {
132                     maze[i][v] = 1;
133                     v = 0;
134                 }
135                 else v = v*10+(tmp[j]-‘0‘);
136             }
137         }
138 
139         int l = 1, r = uN;
140         while(l<=r)
141         {
142             int mid = (l+r)>>1;
143             if(test(mid))
144                 r = mid - 1;
145             else
146                 l = mid + 1;
147         }
148         printf("%d\n", l);
149     }
150 }

View Code

POJ2289 Jamie's Contact Groups —— 二分图多重匹配/最大流 + 二分

标签：多重 getchar none ret contact eof sample hose using

原文地址：http://www.cnblogs.com/DOLFAMINGO/p/7821830.html

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