Coj 0252 HDNOIP201304 Blocking infection

HDNOIP201304 Blocking infection

Difficulty level: A; programming language: Unlimited; run time limit: 1000ms; run space limit: 51200KB; code length limit: 2000000B

Question Description

H State N cities, N cities with n-1 two-way road interconnected to form a tree, city 1th is the capital, is also the root node in the tree. A highly hazardous infectious disease has erupted in the capital of H country. The authorities have decided to set up quarantine checkpoints on some highways in order to prevent outbreaks from spreading to the border cities (the cities indicated by leaf nodes), but the costs of multiplication on different roads may vary, and in order to avoid contagion in border cities, there must be at least one quarantine checkpoint on the route from the capital to each frontier city, Ask you to program the minimum multiplication total cost to stop the infection.

Input

The first line is an integer n, which indicates the number of cities. The next line of n-1, with 3 integers per line, u, V, W, separated by a space between every two integers, represents a two-way road from City u to City v, with a multiplication cost of W. The data guarantees that a tree is entered, and the root node number is 1.

Output

A total of one row, containing an integer, represents the minimum multiplication total cost required to control the outbreak.

Input example

9 2 1 888 1 3 88 2 4 6 5 2 8 6 3 100 3 7 10 8 3 50 7 9 1

Output example

102

Other Notes

For 60% of the data, 2<n<10, for 80% of the data, 2<n<100, for 100% of the data, 2<n<100000

The puzzle: Write a minimal cut and go straight away ...

WA took two rounds because BFs forgot to set vis[s]=true ... Qaq ...

1#include <iostream>2#include <cstdio>3#include <cmath>4#include <algorithm>5#include <queue>6#include <cstring>7 #definePAU Putchar (")8 #defineENT Putchar (' \ n ')9 using namespacestd;Ten Const intmaxn=100000+Ten, maxm=200000+Ten, inf=-1u>>1; One int  out[MAXN]; A structisap{ -     structted{intX,y,w;ted*nxt,*re;} adj[maxm],*fch[maxn],*ms,*cur[maxn],*RET[MAXN]; -     intd[maxn],gap[maxn],n,s,t; the     voidInitintN) { This->n=n;memset (d,-1,sizeof(d)); Ms=adj;return;} -     voidAddintXintYintW) { -*ms= (Ted) {x,y,w,fch[x],ms+1};fch[x]=ms++; -*ms= (Ted) {y,x,0, fch[y],ms-1};fch[y]=ms++; +         return; -     } +     voidBFs () { Aqueue<int>Q; Q.push (T);d [t]=0; at          while(!Q.empty ()) { -             intu=Q.front (); Q.pop (); -              for(ted*e=fch[u];e;e=e->NXT) { -                 intv=e->y;if(d[v]<0) d[v]=d[u]+1, Q.push (v); -             } -}return; in     } -     intMxflow (intSintT) { to          This->s=s; This->t=t;bfs (); ted*e;intk=s,flow=0; +          for(intI=1; i<=n;i++) gap[d[i]]++,cur[i]=Fch[i]; -          while(d[s]<N) { the             if(k==T) { *                 intMi=inf,pos; for(intI=s;i!=t;i=cur[i]->y)if(CUR[I]-&GT;W&LT;MI) mi=cur[i]->w,pos=i; $                  for(intI=s;i!=t;i=cur[i]->y) cur[i]->w-=mi,cur[i]->re->w+=mi;flow+=mi;k=Pos;Panax Notoginseng} for(E=CUR[K];E;E=E-&GT;NXT)if(e->w&&d[k]==d[e->y]+1) Break; -             if(e) cur[k]=e,ret[e->y]=e->re,k=e->y; the             Else{if(--gap[d[k]]==0) Break; cur[k]=fch[k];intMi=N; +                  for(E=FCH[K];E;E=E-&GT;NXT)if(E-&GT;W&AMP;&AMP;D[E-&GT;Y]&LT;MI) mi=d[e->y]; Ad[k]=mi+1; gap[d[k]]++;if(k!=s) k=ret[k]->y; the             } +}returnflow; -     } $ }sol; $ structted{intX,Y,W;TED*NXT;} adj[maxm],*fch[maxn],*ms=adj; - voidAddintXintYintW) { -*ms= (Ted) {x,y,w,fch[x]};fch[x]=ms++; the*ms= (Ted) {y,x,w,fch[y]};fch[y]=ms++; -     return;Wuyi } the BOOLVIS[MAXN]; - voidBFsintS) { Wuqueue<int>Q; Q.push (S); vis[s]=true; -      while(!Q.empty ()) { About         intu=Q.front (); Q.pop (); $          for(ted*e=fch[u];e;e=e->NXT) { -             intv=e->y;if(!Vis[v]) { -vis[v]=true; out[U]++;sol.add (U,V,E-&GT;W); Q.push (v);//printf ("add:%d->%d%d\n", u,v,e->w); System ("pause"); -             } A         } +}return; the } - intn,m;queue<int>Tset; $ voidMakeg () { the      for(intI=1; i<=n;i++)if(! out[i]) Tset.push (i);//printf ("Tset:"); the      while(!Tset.empty ()) { the         intX=tset.front (); Tset.pop (); Sol.add (x,n+1, 1e9);//printf ("%d", x); the}return; - } inInlineintRead () { the     intx=0, sig=1;CharCh=GetChar (); the      while(!isdigit (CH)) {if(ch=='-') sig=-1; ch=GetChar ();} About      while(IsDigit (CH)) x=Ten*x+ch-'0', ch=GetChar (); the     returnx*=Sig; the } theInlinevoidWriteintx) { +     if(x==0) {Putchar ('0');return;}if(x<0) Putchar ('-'), x=-x; -     intlen=0, buf[ the]; while(x) buf[len++]=x%Ten, x/=Ten; the      for(inti=len-1; i>=0; i--) Putchar (buf[i]+'0');return;Bayi } the voidinit () { theN=read (); M=~-n;sol.init (n+1);intx,y,w; -      for(intI=1; i<=m;i++){ -X=read (); Y=read (); w=read (); Add (x,y,w); the}BFS (1); Makeg (); Write (Sol.mxflow (1, n+1)); the     return; the } the voidWork () { -     return; the } the voidprint () { the     return;94 } the intMain () { theInit (); work ();p rint ();return 0; the}

Coj 0252 HDNOIP201304 Blocking infection