題目鏈接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=569&problem=4277&mosmsg=Submission+received+with+ID+2182664

題目大意:給定一個帶權有向圖(v,e)，求兩條互不相交的從1到v的路徑(不經過同一個點)，使兩條路徑的權和最小.

分析:即求從1到v的最小費用流,到v的流量為2,且每個點只能經過一次.加一個從v出發的點v‘,費用為0,容量為2,求1到v‘的最小費用流即滿足前一條件;將2到v-1每個點拆分成兩個點,費用為0,容量為1,再求從1到v‘的最小費用流即可.

　　一定要註意求mcmf時設置的INF要恰當!!!不然就溢出然後瘋狂WA了!!!

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<vector>
 5 #include<queue>
 6 using namespace std;
 7 const int maxn=2005,maxm=20005,INF=1000000;
 8 struct Edge{
 9     int from,to,cap,flow,cost;
10     Edge(int u,int
 
 v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}
11 };
12 
13 struct MCMF{
14     int n,m;
15     vector<Edge> edges;
16     vector<int> G[maxn];
17     int inq[maxn],d[maxn],p[maxn],a[maxn];
18 
19     void init(int n){
20         this->n=n;
21         this->m=0;
22         for
 
(int i=0;i<n;i++)G[i].clear();
23         edges.clear();
24     }
25 
26     void AddEdge(int from,int to,int cap,int cost){
27         edges.push_back(Edge(from,to,cap,0,cost));
28         edges.push_back(Edge(to,from,0,0,-cost));
29         G[from].push_back(this->m++);
30         G[to].push_back(this->m++);
31     }
32 
33     bool BellmanFord(int s,int t,int &flow,long long &cost){
34         for(int i=0;i<n;i++)d[i]=INF;
35         memset(inq,0,sizeof(inq));
36         d[s]=0;inq[s]=1;p[s]=0;a[s]=INF;
37         queue<int> Q;
38         Q.push(s);
39         while(!Q.empty()){
40             int u=Q.front();Q.pop();
41             inq[u]=0;
42 
43             for(int i=0;i<G[u].size();i++){
44                 Edge &e=edges[G[u][i]];
45                 if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
46                     d[e.to]=d[u]+e.cost;
47                     p[e.to]=G[u][i];
48                     a[e.to]=min(a[u],e.cap-e.flow);
49                     if(!inq[e.to]){Q.push(e.to);inq[e.to]=1;}
50                 }
51             }
52         }
53         if(d[t]==INF)return false;
54         flow+=a[t];
55         cost+=(long long)d[t]*(long long)a[t];
56         for(int u=t;u!=s;u=edges[p[u]].from){
57             edges[p[u]].flow+=a[t];
58             edges[p[u]^1].flow-=a[t];
59         }
60         return true;
61     }
62 
63     int Mincost(int s,int t,long long &cost){
64         int flow=0;cost=0;
65         while(BellmanFord(s,t,flow,cost));
66         return flow;
67     }
68 };
69 
70 int main(){
71     //freopen("e:\\in.txt","r",stdin);
72     int n,m,a,b,c;
73     MCMF mcmf;
74     mcmf.init(2*n);
75     for(int i=1;i<n-1;i++){
76         mcmf.AddEdge(i,i+n,1,0);
77     }
78     for(int i=0;i<m;i++){
79         cin>>a>>b>>c;
80         a--;b--;
81         if(a==0||a==n-1){
82             mcmf.AddEdge(a,b,1,c);
83         }
84         else{
85             mcmf.AddEdge(a+n,b,1,c);
86         }
87     }
88     mcmf.AddEdge(n-1,2*n-1,2,0);
89     long long cost=0;
90     mcmf.Mincost(0,2*n-1,cost);
91     cout<<cost<<endl;
92     return 0;
93 }

UVa 1658,Admiral (拆點+限制最小費用流)