题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2426
For any school, it is hard to find a feasible accommodation plan with every student assigned to a suitable apartment while keeping everyone happy, let alone an optimal one. Recently the president of University ABC, Peterson, is facing a similar problem. While Peterson does not like the idea of delegating the task directly to the class advisors as so many other schools are doing, he still wants to design a creative plan such that no student is assigned to a room he/she dislikes, and the overall quality of the plan should be maximized. Nevertheless, Peterson does not know how this task could be accomplished, so he asks you to solve this so-called "interesting" problem for him.
Suppose that there are N students and M rooms. Each student
is asked to rate some rooms (not necessarily all M rooms) by stating how he/she
likes the room. The rating can be represented as an integer, positive value
meaning that the student consider the room to be of good quality, zero
indicating neutral, or negative implying that the student does not like living
in the room. Note that you can never assign a student to a room which he/she has
not rated, as the absence of rating indicates that the student cannot live in
the room for other reasons.
With limited information available, you‘ve
decided to simply find an assignment such that every student is assigned to a
room he/she has rated, no two students are assigned to the same room, and the
sum of rating is maximized while satisfying Peterson‘s requirement. The question
is … what exactly is the answer?
题意描述：学校里有n个学生和m个公寓房间，每个学生对一些房间有一些打分，如果分数为正，说明学生喜欢这个房间，若为0，对这个房间保持中立，若为负，则不喜欢这个房间。学生不会住进不喜欢的房间和没有打分的房间。问安排这n个学生来求最大的分数。
算法分析：KM算法。n个学生作为X集，m个房间作为Y集，然后调用KM算法就可以了。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #define inf 0x7fffffff
8 using namespace std;
9 const int maxn=500+10;
10
11 int n,m;
12 int lx[maxn],ly[maxn],visx[maxn],visy[maxn];
13 int link[maxn],slack[maxn],w[maxn][maxn];
14
15 int dfs(int x)
16 {
17 visx[x]=1;
18 for (int y=1 ;y<=m ;y++) if (w[x][y]!=-1)
19 {
20 if (visy[y]) continue;
21 int t=lx[x]+ly[y]-w[x][y];
22 if (t==0)
23 {
24 visy[y]=1;
25 if (link[y]==-1 || dfs(link[y]))
26 {
27 link[y]=x;
28 return 1;
29 }
30 }
31 else if (slack[y]>t) slack[y]=t;
32 }
33 return 0;
34 }
35
36 int KM()
37 {
38 memset(link,-1,sizeof(link));
39 memset(ly,0,sizeof(ly));
40 for (int x=1 ;x<=n ;x++)
41 {
42 lx[x]=-inf;
43 for (int y=1 ;y<=m ;y++)
44 lx[x]=max(lx[x],w[x][y]);
45 }
46 for (int x=1 ;x<=n ;x++)
47 {
48 for (int i=1 ;i<=m ;i++) slack[i]=inf;
49 int flag=0;
50 for (int i=1 ;i<=m ;i++) if (w[x][i]!=-1) flag=1;
51 while (flag)
52 {
53 memset(visx,0,sizeof(visx));
54 memset(visy,0,sizeof(visy));
55 if (dfs(x)) break;
56 int d=inf;
57 for (int i=1 ;i<=m ;i++)
58 if (!visy[i] && d>slack[i]) d=slack[i];
59 for (int i=1 ;i<=n ;i++)
60 if (visx[i]) lx[i] -= d;
61 for (int i=1 ;i<=m ;i++)
62 {
63 if (visy[i]) ly[i] += d;
64 else slack[i] -= d;
65 }
66 }
67 }
68 int ans=0;
69 int vis[maxn];
70 memset(vis,0,sizeof(vis));
71 for (int i=1 ;i<=m ;i++)
72 {
73 if (link[i]!=-1)
74 {
75 ans += w[link[i] ][i];
76 vis[link[i] ]=1;
77 }
78 }
79 int i=1;
80 for (i=1 ;i<=n ;i++)
81 if (vis[i]==0) return -1;
82 return ans;
83 }
84
85 int main()
86 {
87 int e;
88 int ncase=1;
89 while (scanf("%d%d%d",&n,&m,&e)!=EOF)
90 {
91 memset(w,-1,sizeof(w));
92 int a,b,c;
93 for (int i=0 ;i<e ;i++)
94 {
95 scanf("%d%d%d",&a,&b,&c);
96 a++ ;b++ ;
97 if (c>=0) w[a][b]=c;
98 }
99 printf("Case %d: %d\n",ncase++,KM());
100 }
101 return 0;
102 }