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题意:给你一副无向图,问从1->n->1这样走一个来回所用的最短路径是多少,每条边只能走一次。
分析:最小费用流问题。把边的长度当成费用,每条边容量为1,由于是无向图,所以每条边要处理两次,即u->v,v->u都要加进去。把图建好后跑一遍流量为2的最小费用流得出最小费用即可。
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m;
const int maxn=2000+100;
const int maxm=20000+100;
const int inf=0x3f3f3f3f;
int head[maxm],cnt;
struct edge
{
int u,v,nxt,flow,cost;
}edge[maxm*5+100];
int dis[maxn],pre[maxm],p[maxn];
int vis[maxn];
void add_edge(int u,int v,int c,int cost)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].flow=c;
edge[cnt].cost=cost;
edge[cnt].nxt=head[u];
head[u]=cnt++;
edge[cnt].u=v;
edge[cnt].v=u;
edge[cnt].flow=0;
edge[cnt].cost=-cost;
edge[cnt].nxt=head[v];
head[v]=cnt++;
}
int spfa(int s,int t)
{
queue<int>q;
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
memset(dis,0x3f3f3f3f,sizeof(dis));
dis[s]=0;
vis[s]=1;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=0;
for(int i=head[u];i!=-1;i=edge[i].nxt)
{
int v=edge[i].v;
if(edge[i].flow&&dis[v]>dis[u]+edge[i].cost)
{
dis[v]=dis[u]+edge[i].cost;
pre[v]=i;
p[v]=u;
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
if(pre[t]==-1)
return 0;
return 1;
}
int min_cost_flow(int s,int t)
{
int ans_flow=0,ans_cost=0;
int u,minn;
while(spfa(s,t))
{
u=t;
minn=inf;
for(int i=pre[u];i!=-1;i=pre[edge[i].u])
{
minn=min(edge[i].flow,minn);
}
u=t;
for(int i=pre[u];i!=-1;i=pre[edge[i].u])
{
edge[i].flow-=minn;
edge[i^1].flow+=minn;
ans_cost+=edge[i].cost;
}
ans_flow+=minn;
if(ans_flow==2)
{
return ans_cost;
}
}
return ans_cost;
}
int main ()
{
scanf("%d%d",&n,&m);
memset(head,-1,sizeof(head));
cnt=0;
add_edge(0,1,2,0);
add_edge(n,n+1,2,0);
while(m--)
{
int xx,yy,zz;
scanf("%d%d%d",&xx,&yy,&zz);
add_edge(xx,yy,1,zz);
add_edge(yy,xx,1,zz);
}
printf("%d\n",min_cost_flow(0,n+1));
}