POJ3068: "Shortest" pair of paths 题解

注意每个点不能重复，所以要拆点

然后跑一个流量为2的费用流就可以了

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <cstdlib>#include <utility>#include <map>#include <stack>#include <set>#include <vector>#include <queue>#include <deque>#define x first#define y second#define mp make_pair#define pb push_back#define LL long long#define Pair pair<int,int>#define LOWBIT(x) x & (-x)using namespace std;const int MOD=1e9+7;const int INF=1e9;const int magic=348;int dist[200048],h[200048];priority_queue<Pair> q;int prevv[200048],preve[200048];int t,tot=1,head[200048],nxt[200048],to[200048],f[200048],w[200048];inline void addedge(int s,int t,int cap,int cost){to[++tot]=t;nxt[tot]=head[s];head[s]=tot;f[tot]=cap;w[tot]=cost;to[++tot]=s;nxt[tot]=head[t];head[t]=tot;f[tot]=0;w[tot]=-cost;}int n,e;int ca=0;bool dijkstra(){int i,x,y,dd;for (i=1;i<=n*2;i++) dist[i]=INF;dist[n+1]=0;q.push(mp(0,n+1));while (!q.empty()){x=q.top().y;dd=-q.top().x;q.pop();if (dd>dist[x]) continue;for (i=head[x];i;i=nxt[i]){y=to[i];if (f[i] && dist[y]>dist[x]+w[i]+h[x]-h[y]){dist[y]=dist[x]+w[i]+h[x]-h[y];prevv[y]=x;preve[y]=i;q.push(mp(-dist[y],y));}}}if (dist[n]>=INF) return false; else return true;}int min_cost_flow(){int i,u,minf;for (i=1;i<=n*2;i++) h[i]+=dist[i];for (u=n;u!=n+1;u=prevv[u]){f[preve[u]]--;f[preve[u]^1]++;}return h[n];}int main (){int i,j,x,y,c;while (scanf("%d%d",&n,&e) && n){tot=1;for (i=1;i<=n*2;i++){head[i]=0;h[i]=0;}for (i=1;i<=n;i++) addedge(i,n+i,1,0);for (i=1;i<=e;i++){scanf("%d%d%d",&x,&y,&c);x++;y++;addedge(n+x,y,1,c);}int ans=0;for (j=1;j<=2;j++){if (!dijkstra()) break;ans+=min_cost_flow();}printf("Instance #%d: ",++ca);if (j<=2)printf("Not possible\n");elseprintf("%d\n",ans);}return 0;}