题目链接：https://vjudge.net/problem/HDU-3488
Tour
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3720 Accepted Submission(s): 1777
Every city should be just in one route.
A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)
The total distance the N roads you have chosen should be minimized.
Input An integer T in the first line indicates the number of the test cases.
In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W.
It is guaranteed that at least one valid arrangement of the tour is existed.
A blank line is followed after each test case.
Output For each test case, output a line with exactly one integer, which is the minimum total distance.
Sample Input 16 91 2 52 3 53 1 103 4 124 1 84 6 115 4 75 6 96 5 4
Sample Output 42
Source 2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT
Recommend zhouzeyong
题解：
题意：有n座城市，m条带权有向边。有人想要游历所有城市，于是制定了计划：游历的路径是一个或者多个环,且所有城市都必须仅存在于一个环中。问怎样设计路线使得总路程最短？
我们可以用最大权匹配去求：
1.对于每个点u，我们拆成u和u'。u代表着出点，在二分图的左侧；u'代表着入点，在二分图的右侧。
2.如果有一条有向边u-->v，那么在二分图中，我们加一条边u-->v'，并且权值取反。
3.利用KM()算法，求出最大权匹配，再将结果取反，即为答案。
问：为何KM()算法求出来的就一定是一个或多个环呢？
1.可知，题目已经说明了必定有解。那么对应的二分图必定存在完全匹配，完全匹配也必定是最大匹配。
2.我们用KM()算法求出了最大权匹配。根据定理：最大权匹配必定为最大匹配。所以，如果最大匹配是完全匹配，那么最大权匹配也是完全匹配。
3.因为最大权匹配是完全匹配。所以所求出的解必定是一个或多个环。
代码如下：
1 #include <bits/stdc++.h>View Code
2 using namespace std;
3 typedef long long LL;
4 const int INF = 0x3f3f3f3f;
5 const LL LNF = 9e18;
6 const int mod = 1e9+7;
7 const int MAXN = 2e2+10;
8
9 int nx, ny;
10 int g[MAXN][MAXN];
11 int linker[MAXN], lx[MAXN], ly[MAXN];
12 int slack[MAXN];
13 bool visx[MAXN], visy[MAXN];
14
15 bool DFS(int x)
16 {
17 visx[x] = true;
18 for(int y = 1; y<=ny; y++)
19 {
20 if(visy[y]) continue;
21 int tmp = lx[x] + ly[y] - g[x][y];
22 if(tmp==0)
23 {
24 visy[y] = true;
25 if(linker[y]==-1 || DFS(linker[y]))
26 {
27 linker[y] = x;
28 return true;
29 }
30 }
31 else
32 slack[y] = min(slack[y], tmp);
33 }
34 return false;
35 }
36
37 int KM()
38 {
39 memset(linker, -1, sizeof(linker));
40 memset(ly, 0, sizeof(ly));
41 for(int i = 1; i<=nx; i++)
42 {
43 lx[i] = -INF;
44 for(int j = 1; j<=ny; j++)
45 lx[i] = max(lx[i], g[i][j]);
46 }
47
48 for(int x = 1; x<=nx; x++)
49 {
50 for(int i = 1; i<=ny; i++)
51 slack[i] = INF;
52 while(true)
53 {
54 memset(visx, 0, sizeof(visx));
55 memset(visy, 0, sizeof(visy));
56
57 if(DFS(x)) break;
58 int d = INF;
59 for(int i = 1; i<=ny; i++)
60 if(!visy[i])
61 d = min(d, slack[i]);
62
63 for(int i = 1; i<=nx; i++)
64 if(visx[i])
65 lx[i] -= d;
66 for(int i = 1; i<=ny; i++)
67 {
68 if(visy[i]) ly[i] += d;
69 else slack[i] -= d;
70 }
71 }
72 }
73
74 int res = 0;
75 for(int i = 1; i<=ny; i++)
76 if(linker[i]!=-1)
77 res += g[linker[i]][i];
78 return res;
79 }
80
81 int main()
82 {
83 int T, n, m;
84 scanf("%d", &T);
85 while(T--)
86 {
87 scanf("%d%d", &n,&m);
88 nx = ny = n;
89 memset(g, 0, sizeof(g));
90 for(int i = 1; i<=nx; i++)
91 for(int j = 1; j<=ny; j++)
92 g[i][j] = -INF;
93 for(int i = 1; i<=m; i++)
94 {
95 int u, v, w;
96 scanf("%d%d%d", &u, &v, &w);
97 g[u][v] = max(g[u][v], -w);
98 }
99
100 printf("%d\n", -KM());
101 }
102 }