题意:给定一个有向带权图,求两条不相交(无公共点)的路径且路径权值之和最小,路径由1到v
题解:这题的关键就在于每个点只能走一遍,于是我们想到以边换点的思想,用边来代替点,怎么代替呢?
把i拆成i和i',这样经过i就转化为经过i到i'的路径了,从而用最小费用流即可
#include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<queue> #include<vector> #define MAXN 5005 #define ll long long #define INF 0x7f7f7f7f using namespace std; struct Edge{ int from,to,cap,flow,cost; Edge(int u=0,int v=0,int c=0,int f=0,int w=0){ from=u,to=v,cap=c,flow=f,cost=w; } }; struct MCMF{ int n,m,s,t; vector<Edge> edges; vector<int> G[MAXN]; int d[MAXN]; int p[MAXN]; int b[MAXN]; int a[MAXN]; void init(int n,int s,int t){ this->n=n; this->s=s,this->t=t; edges.clear(); for(int i=0;i<=n;i++){ G[i].clear(); } } void AddEdge(int x,int y,int cap,int cost){ edges.push_back(Edge(x,y,cap,0,cost)); edges.push_back(Edge(y,x,0,0,-cost)); m=edges.size(); G[x].push_back(m-2); G[y].push_back(m-1); } int SPFA(int &flow,ll &cost){ memset(d,0x7f,sizeof(d)); memset(b,0,sizeof(b)); queue<int> q; p[s]=0; a[s]=INF; d[s]=0; q.push(s); b[s]=1; while(!q.empty()){ int x=q.front();q.pop(); b[x]=0; for(int i=0;i<G[x].size();i++){ Edge& e=edges[G[x][i]]; if(e.cap>e.flow&&d[e.to]>d[x]+e.cost){ p[e.to]=G[x][i]; a[e.to]=min(a[x],e.cap-e.flow); d[e.to]=d[x]+e.cost; if(!b[e.to]){ b[e.to]=1; q.push(e.to); } } } } if(d[t]==INF){ return 0; } flow+=a[t]; cost+=1LL*d[t]*a[t]; for(int i=t;i!=s;i=edges[p[i]].from){ edges[p[i]].flow+=a[t]; edges[p[i]^1].flow-=a[t]; } return 1; } ll MincostMaxflow(){ int flow=0;ll cost=0; while(SPFA(flow,cost)); return cost; } }D; int n,m; int main() { while(~scanf("%d%d",&n,&m)){ D.init(n<<1,0,n); D.AddEdge(0,1,2,0); for(int i=1;i<=n;i++){ D.AddEdge(i,i+n,1,0); } for(int i=1;i<=m;i++){ int x,y,w; scanf("%d%d%d",&x,&y,&w); if(x==1||x==n) D.AddEdge(x,y,1,1LL*w); else D.AddEdge(x+n,y,1,1LL*w); } printf("%lld ",D.MincostMaxflow()); } return 0; }