UVa 10806 & 费用流+意识流...

题意:

  一张无向图,求两条没有重复的从S到T的路径.

SOL:

  网络流为什么屌呢..因为网络流的容量,流量,费用能对许许多多的问题进行相应的转化,然后它就非常的屌.

  对于这道题呢,不是要没有重复吗?不是一条边只能走一次吗?那么容量上界就是1.不是要有两条吗?那么总流量就是2.不是带权吗?那么加个费用.

  

  WA得惨不忍睹,最后发现边从0开始记异或以后会改变一些非常奇异的边...真是丝帛= =

Code

  

/*==========================================================================
# Last modified: 2016-03-07 14:07
# Filename: uva10806.cpp
# Description:
==========================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <cmath>
#include <ctime>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm> #include <set>
#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,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1] #define maxn 1000
#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);
}
}
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