Luogu P1186 Marika

Portal

Thinking of solving problems

Topic description Touching. Explain that this is the person with the system game, the system cut a side, and then make this person to go the shortest path longest. The procedure is more natural, first run the shortest path, and then enumerate the path to run the quickest way, the maximum value. This complexity is $o (N^2log (m) $) because the shortest circuit is impossible to ring.

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>using namespacestd;Const intMAXN =1005;Const intMAXM =1000005; inlineintRd () {intx=0, f=1;CharCh=GetChar ();  while(!isdigit (CH)) {f=ch=='-'?0:1; ch=GetChar ();}  while(IsDigit (CH)) {x= (x<<1) + (x<<3) +ch-'0'; ch=GetChar ();} returnf?x:-x;}intN,m,head[maxn],cnt=1, Fa[maxn],dis[maxn],ans;BOOLCUT[MAXM],VIS[MAXN];intto[maxm<<1],nxt[maxm<<1],val[maxm<<1];structnode{intW,id; FriendBOOL operator< (ConstNode A,ConstNode B) {        returnA.w>B.W; }};p Riority_queue<Node>Q;inlinevoidAddintBgintEdintW) {to[++cnt]=ed,nxt[cnt]=head[bg],val[cnt]=w,head[bg]=CNT;}voidDijkstraBOOLflag) {memset (DIS,0x3f,sizeof(DIS)); memset (Vis,false,sizeof(VIS)); Node now;now.id=1; now.w=0;d is[1]=0;    Q.push (now);  while(!Q.empty ()) {Node zz=q.top (); Q.pop ();intx=zz.id; if(DIS[X]!=ZZ.W | | vis[x])Continue; vis[x]=1;  for(RegisterintI=head[x];i;i=Nxt[i]) {            if(Cut[i])Continue;intu=To[i]; if(Dis[u]<=dis[x]+val[i])Continue; if(flag) fa[u]=x; Dis[u]=dis[x]+Val[i]; Now.id=u;now.w=Dis[u];        Q.push (now); }    }        if(flag)return; Ans=Max (Ans,dis[n]);}intMain () {n=rd (), M=rd ();intx, Y, Z  for(RegisterintI=1; i<=m;i++) {x=rd (), Y=rd (), z=Rd ();    Add (x, Y, z), add (y,x,z); } Dijkstra (1); intnow=n,k;  while(Fa[now]) { for(RegisterintI=head[now];i;i=Nxt[i])if(To[i]==fa[now]) {cut[i]=cut[i^1]=1; k=i; Break;} Dijkstra (0); now=fa[now];cut[k]=cut[k^1]=0; } printf ("%d", ans); return 0;}

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Luogu P1186 Marika