HDU 6166 Senior Pan

来源:互联网 发布:邯郸煤炭软件 编辑:程序博客网 时间:2024/06/04 20:03
多源最短路次短路,还要保证最短路和次短路的起点不同
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>using namespace std;typedef long long ll;const int MAXN =(int)1e5+10;const ll INF =(ll)1e11;struct edge2{int to,nxt;ll w;}ed2[MAXN];struct edge1{int to,nxt;ll w;}ed1[MAXN];int h1[MAXN],c1,h2[MAXN],c2;void addedge(int u,int v,ll w){ed1[c1].to=v;ed1[c1].nxt=h1[u];ed1[c1].w=w;h1[u]=c1++;ed2[c2].to=u;ed2[c2].nxt=h2[v];ed2[c2].w=w;h2[v]=c2++;}struct node{int s,u;ll dis;node(){}node(int _s,int _u,ll _dis){s=_s;u=_u;dis=_dis;}bool operator < (const node &a)const{return dis>a.dis;}};ll d1[MAXN],d2[MAXN],d3[MAXN],d4[MAXN];int m1[MAXN],m2[MAXN],m3[MAXN],m4[MAXN];priority_queue<node>q1,q2;void init(int n){for(int i=1;i<=n;i++)h1[i]=h2[i]=-1,m1[i]=m2[i]=m3[i]=m4[i]=0,d1[i]=d2[i]=d3[i]=d4[i]=INF;c1=c2=0;}void dij1(){while(!q1.empty()){node now=q1.top();q1.pop();int u=now.u,s=now.s;ll dis=now.dis;if(m2[u]!=0)continue;if(m1[u]==0){m1[u]=s;d1[u]=dis;}else {m2[u]=s;d2[u]=dis;}for(int i=h1[u];i!=-1;i=ed1[i].nxt){int v=ed1[i].to;if(m2[v]!=0||m1[v]==s)continue;if(m1[v]==0) if(d1[v]>dis+ed1[i].w)q1.push(node(s,v,dis+ed1[i].w));else if(d2[v]>dis+ed1[i].w)q1.push(node(s,v,dis+ed1[i].w));}}}void dij2(){while(!q2.empty()){node now=q2.top();q2.pop();int u=now.u,s=now.s;ll dis=now.dis;if(m4[u]!=0)continue;if(m3[u]==0){m3[u]=s;d3[u]=dis;}else {m4[u]=s;d4[u]=dis;}for(int i=h2[u];i!=-1;i=ed2[i].nxt){int v=ed2[i].to;if(m4[v]!=0||m3[v]==s)continue;if(m3[v]==0) if(d3[v]>dis+ed2[i].w)q2.push(node(s,v,dis+ed2[i].w));else if(d4[v]>dis+ed2[i].w)q2.push(node(s,v,dis+ed2[i].w));}}}int main(){int t;scanf("%d",&t);for(int ca=1;ca<=t;ca++){int n,m;scanf("%d%d",&n,&m);init(n);for(int i=1;i<=m;i++){int u,v;ll w;scanf("%d%d%lld",&u,&v,&w);addedge(u,v,w);}int k;scanf("%d",&k);for(int i=1;i<=k;i++){int x;scanf("%d",&x);q1.push(node(x,x,0));q2.push(node(x,x,0));}dij1();dij2();ll ans=INF;for(int i=1;i<=n;i++){if(m1[i]==m3[i])ans=min(ans,d1[i]+d4[i]),ans=min(ans,d2[i]+d3[i]);else ans=min(ans,d1[i]+d3[i]);}printf("Case #%d: %lld\n",ca,ans);}return 0;}