HDU 1853 Cyclic Tour

Cyclic Tour

Time Limit: 1000ms

Memory Limit: 65535KB

This problem will be judged on HDU. Original ID: 1853
64-bit integer IO format: %I64d Java class name: Main

There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?

Input

There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).

Output

Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.

Sample Input

Sample Output

42
-1

Hint

In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.

Source

HDU 2007 Programming Contest - Final

解题：KM最小权匹配或者费用流

KM算法

 1 #include <bits/stdc++.h>
 2 #define type int
 3 using namespace std;
 4 const int INF = 0x3f3f3f3f;
 5 const int maxn = 310;
 6 type Lx[maxn],Ly[maxn],W[maxn][maxn],slack[maxn];
 7 bool S[maxn],T[maxn];
 8 int n,Link[maxn];
 9 bool match(int u) {
10     S[u] = true;
11     for(int v = 1; v <= n; ++v) {
12         if(T[v]) continue;
13         type d = Lx[u] + Ly[v] - W[u][v];
14         if(!d) { //浮点数时需要用精度判0
15             T[v] = true;
16             if(Link[v] == -1 || match(Link[v])) {
17                 Link[v] = u;
18                 return true;
19             }
20         } else if(slack[v] > d) slack[v] = d;
21     }
22     return false;
23 }
24 void update() {
25     type d = INF;
26     for(int i = 1; i <= n; ++i)
27         if(!T[i] && slack[i] < d)
28             d = slack[i];
29     for(int i = 1; i <= n; ++i) {
30         if(S[i]) Lx[i] -= d;
31         if(T[i]) Ly[i] += d;
32         else slack[i] -= d;
33     }
34 }
35 void KuhnMunkras() {
36     for(int u = 1; u <= n; ++u) {
37         Lx[u] = -INF;
38         Ly[u] = 0;
39         Link[u] = -1;
40         for(int v = 1; v <= n; ++v)
41             Lx[u] = max(Lx[u],W[u][v]);
42     }
43     for(int u = 1; u <= n; ++u) {
44         for(int v = 1; v <= n; ++v) slack[v] = INF;
45         while(true) {
46             memset(S,false,sizeof S);
47             memset(T,false,sizeof T);
48             if(match(u)) break;
49             update();
50         }
51     }
52     type ret = 0;
53     for(int i = 1; i <= n; ++i)
54         if(Link[i] == -1 || W[Link[i]][i] == -INF) {
55             puts("-1");
56             return;
57         } else {
58             ret += W[Link[i]][i];
59         }
60     printf("%d
",-ret);
61 }
62 int m;
63 int main() {
64     int u,v,w;
65     while(~scanf("%d%d",&n,&m)) {
66         for(int i = 1; i <= n; ++i)
67             for(int j = 1; j <= n; ++j)
68                 W[i][j] = -INF;
69         for(int i = 0; i < m; ++i) {
70             scanf("%d%d%d",&u,&v,&w);
71             W[u][v] = max(W[u][v],-w);
72         }
73         KuhnMunkras();
74     }
75     return 0;
76 }

View Code

费用流（最小费用最大流）

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 10010;
 4 const int INF = 0x3f3f3f3f;
 5 struct arc {
 6     int to,flow,cost,next;
 7     arc(int x = 0,int y = 0,int z = 0,int nxt = -1) {
 8         to = x;
 9         flow = y;
10         cost = z;
11         next = nxt;
12     }
13 } e[1000010];
14 int head[maxn],d[maxn],p[maxn],tot,S,T;
15 int n,m;
16 void add(int u,int v,int flow,int cost) {
17     e[tot] = arc(v,flow,cost,head[u]);
18     head[u] = tot++;
19     e[tot] = arc(u,0,-cost,head[v]);
20     head[v] = tot++;
21 }
22 bool inq[maxn];
23 bool spfa() {
24     memset(p,-1,sizeof p);
25     memset(d,0x3f,sizeof d);
26     queue<int>q;
27     memset(inq,false,sizeof inq);
28     d[S] = 0;
29     q.push(S);
30     while(!q.empty()) {
31         int u = q.front();
32         q.pop();
33         inq[u] = false;
34         for(int i = head[u]; ~i; i = e[i].next) {
35             if(e[i].flow && d[e[i].to] > d[u] + e[i].cost) {
36                 d[e[i].to] = d[u] + e[i].cost;
37                 p[e[i].to] = i;
38                 if(!inq[e[i].to]) {
39                     inq[e[i].to] = true;
40                     q.push(e[i].to);
41                 }
42             }
43         }
44     }
45     return p[T] > -1;
46 }
47 void solve() {
48     int ret = 0,flow = 0;
49     while(spfa()) {
50         int minF = INF;
51         for(int i = p[T]; ~i; i = p[e[i^1].to])
52             minF = min(minF,e[i].flow);
53         for(int i = p[T]; ~i; i = p[e[i^1].to]) {
54             e[i].flow -= minF;
55             e[i^1].flow += minF;
56         }
57         ret += d[T]*minF;
58         flow += minF;
59     }
60     if(flow != n) puts("-1");
61     else printf("%d
",ret);
62 }
63 
64 int main() {
65     int u,v,w;
66     while(~scanf("%d%d",&n,&m)) {
67         memset(head,-1,sizeof head);
68         S = n*2 + 2;
69         T = S + 1;
70         for(int i = tot = 0; i < m; ++i) {
71             scanf("%d%d%d",&u,&v,&w);
72             add(u,v+n,1,w);
73         }
74         for(int i = 1; i <= n; ++i) {
75             add(S,i,1,0);
76             add(i+n,T,1,0);
77         }
78         solve();
79     }
80     return 0;
81 }

View Code