UVa 10806 & cost stream + stream of consciousness ...

Test instructions

An no-go graph, which asks for two paths from S to t that are not duplicated.

SOL:

Network Flow why Dick? Because of the capacity of the network flow, traffic, the cost can be a lot of problems to the corresponding conversion, and then it is very dick.

For this problem, is not to not repeat it? is not a side can only walk once? Then the upper bound of the capacity is 1. Is there a two? Then the total traffic is 2. Not with the right? So add a fee.

　　

WA was miserable, and finally found that the side from 0 began to remember the difference or later will change some very strange side ... It's silk.

Code

　　

/*==========================================================================# last modified:2016-03-07 14:07# filename:uva10806.cpp# Description: ==========================================================================*/ #define ME Acrossthesky #include <cstdio> #include <cmath> #include <ctime> #include <string> #in Clude <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <set&gt ; #include <map> #include <stack> #include <queue> #include <vector> #define LOWBIT (x) (x) & (-X ) #define for (I,A,B) for ((i) = (a);(i) <= (b);(i) + +) #define FORP (I,A,B) for (int i= (a); i<= (b); i++) #define FORM (i,a,b ) for (int i= (a); i>= (b); i--) #define LS (A, a) (((a) + (b)) << 1) #define RS (A, B) (((a) + (c)) >> 1) #define GETLC (a) ch[(a)][0] #define GETRC (a) ch[(a)][1] #define MAXN #define MAXM 100000 #define PI 3.1415926535898 #define _E 2. 718281828459 #define INF 1070000000 using NAMESPAce STD; typedef long Long LL;  typedef unsigned long long ull;     Template<class t> inline void read (t& num) {bool start=false,neg=false;     char c;     num=0;         while ((C=getchar ())!=eof) {if (c== '-') start=neg=true;             else if (c>= ' 0 ' && c<= ' 9 ') {start=true;         num=num*10+c-' 0 ';     } else if (start) break; } if (neg) num=-num; }/*==================split line==================*/struct edge{int from,to,w,c;} E[maxm];int sume,flow,ans,n,m;int d[maxn],first[maxn],next[maxm],from[maxn];bool inq[maxn];void addedge (int x,int y, int cap,int cost) {sume++; e[sume].from=x; e[sume].to=y; e[sume].w=cap; e[sume].c=cost;next[sume]=first[x]; first[x]= sume;sume++; E[sume].from=y; E[sume].to=x; e[sume].w=0; E[sume].c=-cost;next[sume]=first[y]; First[y]=sume;} BOOL SPFA () {queue<int> q; Forp (i,0,n) D[i]=inf;memset (inq,false,sizeof (INQ)); memset (From,0,sizeof (from)); Q.push (0); d[0]=0; Inq[0]=true;while (!q.empty ()) {int Now=q.fronT (); Q.pop (); inq[now]=false;for (int i=first[now];i!=-1;i=next[i]) if (E[I].W && d[now]+e[i].c<d[e[i].to]) {d[e[i].to]= D[now]+e[i].c;from[e[i].to]=i;//q.push (e[i].to); if (!inq[e[i].to]) {Inq[e[i].to]=true;q.push (e[i].to);}}} if (D[n]==inf) return False;return true;} void Mincost () {int X=inf,i=from[n];while (i) {x=min (e[i].w,x); i=from[e[i].from];} Flow+=x;i=from[n];while (i) {e[i].w-=x;e[i^1].w+=x;//ans+= (X*E[I].C); i=from[e[i].from];} Ans+=d[n]*x;} void Reset () {sume=1; flow=0; Ans=0;memset (E,0,sizeof (e)); Forp (I,0,MAXN) First[i]=-1;memset (next,0,sizeof (next));} int main () {while (scanf ("%d", &n)!=eof) {reset (); if (n==0) return 0; Read (m); Forp (i,1,m) {int x,y,w; read (x); Read (y); Read (w); Addedge (x,y,1,w); Addedge (y,x,1,w);} addedge (0,1,2,0); Addedge (n,n+1,2,0); n++; while (SPFA ()) mincost (); if (flow<2) printf ("Back to jail\n"); else printf ("%d\n", ans);}}

UVa 10806 & cost stream + stream of consciousness ...