UVa1658: Admiral 题解

来源：互联网发布：阿里云大数据编辑：程序博客网时间：2024/05/21 18:35

从起点到终点找两条没有公共端点的路径，使得权值和最小，这是老套路了

保证每个点只经过一次，可以拆点

然后跑流量为2的费用流

#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>#include <sstream>#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=0x7ffffff;const int magic=348;int n,e;int start,ed;int prevv[100048],preve[100048],h[100048];int t,tot,head[100048],to[200048],f[200048],nxt[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 dist[100048];priority_queue<Pair> q;void dijkstra(){int i,x,y,dd;for (i=1;i<=t;i++) dist[i]=INF;dist[start]=0;q.push(mp(0,start));while (!q.empty()){x=q.top().y;dd=-q.top().x;q.pop();if (dist[x]<dd) 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));}}}}int min_cost_flow(){int i,minf,u,res=0;for (i=1;i<=t;i++) h[i]+=dist[i];minf=INF;for (u=ed;u!=start;u=prevv[u]) minf=min(minf,f[preve[u]]);res=minf*h[ed];for (u=ed;u!=start;u=prevv[u]){f[preve[u]]-=minf;f[preve[u]^1]+=minf;}return res;}int main (){int i,j,x,y,c;while (scanf("%d%d",&n,&e)!=EOF){t=n*2;tot=1;for (i=1;i<=t;i++) head[i]=0;for (i=1;i<=t;i++) 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);addedge(n+x,y,1,c);}int ans=0;start=n+1;ed=n;for (i=1;i<=2;i++){dijkstra();ans+=min_cost_flow();}printf("%d\n",ans);}return 0;}

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