题意就是让你求两次1到n的最短路。这题应该可以用最短路来求解吧,只需要将第一次用到的边删去即可。我这里是按照算法竞赛入门经典里面提到拆点+最小费最大流。
#include<bits/stdc++.h>
using namespace std;
const int N=1024*4;
const int inf=1<<24;
struct Edge
{
int from,to,cap,flow,cost;
};
vector<Edge>edges;
vector<int>G[N];
int n,m;
int inq[N],p[N],d[N],a[N];
void AddEdge(int from, int to,int cap, int cost)
{
Edge tp;
tp.from=from,tp.to=to,tp.cap=cap,tp.flow=0,tp.cost=cost;
edges.push_back(tp);
tp.from=to,tp.to=from,tp.cap=0,tp.flow=0,tp.cost=-cost;
edges.push_back(tp);
int g=edges.size();
G[from].push_back(g-2);
G[to].push_back(g-1);
}
int BellmanFord(int s,int t,int &flow, int &cost)
{
int i,j,u;
for(i=0; i<N; i++) d[i]=inf;
memset(inq,0,sizeof(inq));
d[s]=0;
inq[s]=1;
p[s]=1;
a[s]=inf;
queue<int>Q;
Q.push(s);
while(!Q.empty())
{
u=Q.front();
Q.pop();
inq[u]=0;
for(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 0;
flow+=a[t];
cost+=d[t]*a[t];
u=t;
while(u!=s)
{
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
u=edges[p[u]].from;
}
return 1;
}
int Mincost(int s,int t)
{
int flow=0,cost=0;
BellmanFord(s,t,flow,cost);
return cost;
}
int main()
{
int i,u,v,c;
while(~scanf("%d%d",&n,&m))
{
for(i=0; i<N; i++) G[i].clear();
edges.clear();
for(i=2; i<n; i++)
{
AddEdge(i,i+n,1,0);
}
for(i=0; i<m; i++)
{
scanf("%d%d%d",&u,&v,&c);
if(u==1||u==n) AddEdge(u,v,1,c);
else AddEdge(u+n,v,1,c);
}
int ans=0;
ans+=Mincost(1,n);
ans+=Mincost(1,n);
printf("%d\n",ans);
}
return 0;
}
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时间: 2024-10-13 21:33:57