POJ2112 Optimal Milking —— 二分图多重匹配/最大流 + 二分

时间：2017-11-12 17:28:14 阅读：149 评论：0 收藏：0 [点我收藏+]

标签：sed row closed positive ber travel struct cep his

题目链接：https://vjudge.net/problem/POJ-2112

Optimal Milking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 18555		Accepted: 6626
Case Time Limit: 1000MS

Description

FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow locations are named by ID numbers K+1..K+C.

Each milking point can "process" at most M (1 <= M <= 15) cows each day.

Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse several paths on the way to their milking machine.

Input

* Line 1: A single line with three space-separated integers: K, C, and M.

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line.

Output

A single line with a single integer that is the minimum possible total distance for the furthest walking cow.

Sample Input

Sample Output

Source

USACO 2003 U S Open

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 = 4e2+10;
17 
18 int uN, vN, m, N, maze[MAXN][MAXN];
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     memset(g, false, sizeof(g));
46     for(int i = vN; i<N; i++)
47         for(int j = 0; j<vN; j++)
48             if(maze[i][j]<=mid)
49                 g[i][j] = true;
50 
51     for(int i = 0; i<vN; i++)
52     {
53         num[i] = m;
54         linker[i][0] = 0;
55     }
56     for(int u = vN; u<N; u++)
57     {
58         memset(used, false, sizeof(used));
59         if(!dfs(u)) return false;
60     }
61     return true;
62 }
63 
64 void Flyod()
65 {
66     for(int k = 0; k<N; k++)
67         for(int i = 0; i<N; i++)
68             for(int j = 0; j<N; j++)
69                 maze[i][j] = min(maze[i][j], maze[i][k]+maze[k][j]);
70 }
71 
72 int main()
73 {
74     while(scanf("%d%d%d", &vN, &uN, &m)!=EOF)
75     {
76         N = uN + vN;
77         for(int i = 0; i<N; i++)
78         for(int j = 0; j<N; j++)
79         {
80             scanf("%d", &maze[i][j]);
81             if(maze[i][j]==0) maze[i][j] = INF/2;
82         }
83 
84         Flyod();
85         int l = 1, r = 200*400;
86         while(l<=r)
87         {
88             int mid = (l+r)>>1;
89             if(hungary(mid))
90                 r = mid - 1;
91             else
92                 l = mid + 1;
93         }
94         printf("%d\n", l);
95     }
96 }

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 = 4e2+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, m, N, maze[MAXN][MAXN];
 25 int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
 26 void add(int u, int v, int w)
 27 {
 28     edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0;
 29     edge[tot].next = head[u]; head[u] = tot++;
 30     edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0;
 31     edge[tot].next = head[v]; head[v] = tot++;
 32 }
 33 
 34 int sap(int start, int end, int nodenum)
 35 {
 36     memset(gap,0,sizeof(gap));
 37     memset(dep,0,sizeof(dep));
 38     memcpy(cur,head,sizeof(head));
 39 
 40     int u = start;
 41     pre[u] = -1;
 42     gap[0] = nodenum;
 43     int maxflow = 0;
 44     while(dep[start]<nodenum)
 45     {
 46         bool flag = false;
 47         for(int i = cur[u]; i!=-1; i=edge[i].next)
 48         {
 49             int v = edge[i].to;
 50             if(edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
 51             {
 52                 flag = true;
 53                 cur[u] = pre[v] = i;
 54                 u = v;
 55                 break;
 56             }
 57         }
 58 
 59         if(flag)
 60         {
 61             if(u==end)
 62             {
 63                 int minn = INF;
 64                 for(int i = pre[u]; i!=-1; i=pre[edge[i^1].to])
 65                     if(minn>edge[i].cap-edge[i].flow)
 66                         minn = edge[i].cap-edge[i].flow;
 67                 for(int i = pre[u]; i!=-1; i=pre[edge[i^1].to])
 68                 {
 69                     edge[i].flow += minn;
 70                     edge[i^1].flow -= minn;
 71                 }
 72                 u = start;
 73                 maxflow += minn;
 74             }
 75         }
 76 
 77         else
 78         {
 79             int minn = nodenum;
 80             for(int i = head[u]; i!=-1; i=edge[i].next)
 81                 if(edge[i].cap-edge[i].flow && dep[edge[i].to]<minn)
 82                 {
 83                     minn = dep[edge[i].to];
 84                     cur[u] = i;
 85                 }
 86             gap[dep[u]]--;
 87             if(gap[dep[u]]==0) break;
 88             dep[u] = minn+1;
 89             gap[dep[u]]++;
 90             if(u!=start) u = edge[pre[u]^1].to;
 91         }
 92     }
 93     return maxflow;
 94 }
 95 
 96 bool test(int mid)
 97 {
 98     tot = 0;
 99     memset(head, -1, sizeof(head));
100     for(int i = vN; i<N; i++)
101     {
102         add(N, i, 1);
103         for(int j = 0; j<vN; j++)
104             if(maze[i][j]<=mid)
105                 add(i, j, 1);
106     }
107     for(int i = 0; i<vN; i++)
108         add(i, N+1, m);
109 
110     int maxflow = sap(N, N+1, N+2);
111     return maxflow == uN;
112 }
113 
114 void Flyod()
115 {
116     for(int k = 0; k<N; k++)
117         for(int i = 0; i<N; i++)
118             for(int j = 0; j<N; j++)
119                 maze[i][j] = min(maze[i][j], maze[i][k]+maze[k][j]);
120 }
121 
122 int main()
123 {
124     while(scanf("%d%d%d", &vN, &uN, &m)!=EOF)
125     {
126         N = uN + vN;
127         for(int i = 0; i<N; i++)
128         for(int j = 0; j<N; j++)
129         {
130             scanf("%d", &maze[i][j]);
131             if(maze[i][j]==0) maze[i][j] = INF/2;
132         }
133 
134         Flyod();
135         int l = 1, r = 200*400;
136         while(l<=r)
137         {
138             int mid = (l+r)>>1;
139             if(test(mid))
140                 r = mid - 1;
141             else
142                 l = mid + 1;
143         }
144         printf("%d\n", l);
145     }
146 }

View Code

POJ2112 Optimal Milking —— 二分图多重匹配/最大流 + 二分

标签：sed row closed positive ber travel struct cep his

原文地址：http://www.cnblogs.com/DOLFAMINGO/p/7821843.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行