POJ3189 Steady Cow Assignment —— 二分图多重匹配/最大流 + 二分

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标签：开始 add 链接 src isp point search open eof

题目链接：https://vjudge.net/problem/POJ-3189

Steady Cow Assignment

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6979		Accepted: 2418

Description

Farmer John‘s N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. Some cows really like their current barn, and some are not so happy.

FJ would like to rearrange the cows such that the cows are as equally happy as possible, even if that means all the cows hate their assigned barn.

Each cow gives FJ the order in which she prefers the barns. A cow‘s happiness with a particular assignment is her ranking of her barn. Your job is to find an assignment of cows to barns such that no barn‘s capacity is exceeded and the size of the range (i.e., one more than the positive difference between the the highest-ranked barn chosen and that lowest-ranked barn chosen) of barn rankings the cows give their assigned barns is as small as possible.

Input

Line 1: Two space-separated integers, N and B

Lines 2..N+1: Each line contains B space-separated integers which are exactly 1..B sorted into some order. The first integer on line i+1 is the number of the cow i‘s top-choice barn, the second integer on that line is the number of the i‘th cow‘s second-choice barn, and so on.

Line N+2: B space-separated integers, respectively the capacity of the first barn, then the capacity of the second, and so on. The sum of these numbers is guaranteed to be at least N.

Output

Line 1: One integer, the size of the minumum range of barn rankings the cows give their assigned barns, including the endpoints.

Sample Input

Sample Output

Hint

Explanation of the sample:

Each cow can be assigned to her first or second choice: barn 1 gets cows 1 and 5, barn 2 gets cow 2, barn 3 gets cow 4, and barn 4 gets cows 3 and 6.

Source

USACO 2006 February Gold

题解：

题意：有n头牛，安排到m个牲棚里住。每头牛对每个牲棚都有一个好感度排名。主人为了使得这些牛尽可能满意，规定了获得最低好感度的牛和获得最高好感度的牛的好感度差值最小（即好感度跨度最小）。

1.二分跨度。然后对于每个跨度，枚举最低好感度（最高好感度也就可以求出），然后开始建图：如果某头牛对某个牲棚的好感度在这个范围内，则连上边；否则不连。

2.用二分图多重匹配或者最大流，求出是否每头牛都可以被安排到某个牲棚中。如果可以，则缩小跨度，否则增大跨度。

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 = 20+10;
16 const int MAXN = 1e3+10;
17 
18 int uN, vN, Rank[MAXN][MAXM];
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 = 1; 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()
44 {
45     for(int i = 1; i<=vN; i++)
46         linker[i][0] = 0;
47     for(int u = 1; u<=uN; u++)
48     {
49         memset(used, false, sizeof(used));
50         if(!dfs(u)) return false;
51     }
52     return true;
53 }
54 
55 bool test(int mid)
56 {
57     for(int down = 1; down<=vN-mid+1; down++)
58     {
59         int up = down+mid-1;
60         memset(g, false, sizeof(g));
61         for(int i = 1; i<=uN; i++)
62         for(int j = down; j<=up; j++)
63                 g[i][Rank[i][j]] = true;
64 
65         if(hungary()) return true;
66     }
67     return false;
68 }
69 
70 int main()
71 {
72     while(scanf("%d%d", &uN, &vN)!=EOF)
73     {
74         for(int i = 1; i<=uN; i++)
75         for(int j = 1; j<=vN; j++)
76             scanf("%d", &Rank[i][j]);
77 
78         for(int i = 1; i<=vN; i++)
79             scanf("%d", &num[i]);
80 
81         int l = 1, r = vN;
82         while(l<=r)
83         {
84             int mid = (l+r)>>1;
85             if(test(mid))
86                 r = mid - 1;
87             else
88                 l = mid + 1;
89         }
90         printf("%d\n", l);
91     }
92 }

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 = 20+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, Rank[MAXN][MAXM], num[MAXM];
 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     for(int down = 1; down<=vN-mid+1; down++)
100     {
101         tot = 0;
102         memset(head, -1, sizeof(head));
103         for(int i = 1; i<=uN; i++)
104         {
105             add(0, i, 1);
106             int up = down+mid-1;
107             for(int j = down; j<=up; j++)
108                 add(i, uN+Rank[i][j], 1);
109         }
110         for(int i = 1; i<=vN; i++)
111             add(uN+i, uN+vN+1, num[i]);
112 
113         int maxflow = sap(0, uN+vN+1, uN+vN+2);
114         if(maxflow==uN) return true;
115     }
116     return false;
117 }
118 
119 int main()
120 {
121     while(scanf("%d%d", &uN, &vN)!=EOF)
122     {
123         for(int i = 1; i<=uN; i++)
124         for(int j = 1; j<=vN; j++)
125             scanf("%d", &Rank[i][j]);
126 
127         for(int i = 1; i<=vN; i++)
128             scanf("%d", &num[i]);
129 
130         int l = 1, r = vN;
131         while(l<=r)
132         {
133             int mid = (l+r)>>1;
134             if(test(mid))
135                 r = mid - 1;
136             else
137                 l = mid + 1;
138         }
139         printf("%d\n", l);
140     }
141 }

View Code

POJ3189 Steady Cow Assignment —— 二分图多重匹配/最大流 + 二分

标签：开始 add 链接 src isp point search open eof

原文地址：http://www.cnblogs.com/DOLFAMINGO/p/7825303.html

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