题意:给定v个点和e条边的有向加权图,求1~v的两条不相交(除了终点和起点没有公共点)的路径,使得路径最小。
题解:把2到v-1的每个结点拆成i和i‘两个结点,中间连一条容量为1,费用为0的边,然后求1到v的流量为2的最小费用流即可。
附上代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn=2e3+50;
const int inf=0x3f3f3f3f;
struct Edge{
int from,to,cap,flow,cost;
Edge(int _from,int _to,int _cap,int _flow,int _cost):from(_from),to(_to),cap(_cap),flow(_flow),cost(_cost){}
};
struct MCMF{
int n,m;
vector<Edge>edges;
vector<int>G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
void init(int n){
this->n=n;
for(int i=0;i<n;i++){
G[i].clear();
}
edges.clear();
}
void addedge(int from,int to,int cap,int cost){
edges.push_back(Edge(from,to,cap,0,cost));
edges.push_back(Edge(to,from,0,0,-cost));
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool bellmanford(int s,int t,int flow_limit,int &flow,int &cost){
for(int i=0;i<n;i++){
d[i]=inf;
}
memset(inq,0,sizeof(inq));
d[s]=0;inq[s]=1;
p[s]=0;a[s]=inf;
queue<int>Q;
Q.push(s);
while(!Q.empty()){
int u=Q.front();Q.pop();
inq[u]=0;
for(int i=0;i<G[u].size();i++){
Edge&e=edges[G[u][i]];
if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
d[e.to]=d[u]+e.cost;
p[e.to]=G[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if(!inq[e.to]){
Q.push(e.to);
inq[e.to]=1;
}
}
}
}
if(d[t]==inf){
return false;
}
if(flow+a[t]>flow_limit){
a[t]=flow_limit-flow;
}
flow+=a[t];
cost+=d[t]*a[t];
for(int u=t;u!=s;u=edges[p[u]].from){
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
}
return true;
}
int mincostflow(int s,int t,int flow_limit,int &cost){
int flow=0;
cost=0;
while(flow<flow_limit&&bellmanford(s,t,flow_limit,flow,cost));
return flow;
}
};
MCMF g;
int main()
{
int n,m,a,b,c;
while(scanf("%d%d",&n,&m)==2&&n){
g.init(n*2-2);
for(int i=2;i<=n-1;i++){
g.addedge(i-1,i+n-2,1,0);
}
while(m--){
scanf("%d%d%d",&a,&b,&c);
if(a!=1&&a!=n){
a+=n-2;
}else{
a--;
}
b--;
g.addedge(a,b,1,c);
}
int cost;
g.mincostflow(0,n-1,2,cost);
printf("%d\n",cost);
}
return 0;
}