http://poj.org/problem?id=1637
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Sightseeing tour
Description The city executive board in Lund wants to construct a sightseeing tour by bus in Lund, so that tourists can see every corner of the beautiful city. They want to construct the tour so that every street in the city is visited exactly once. The bus should also Input On the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each scenario begins with a line containing two positive integers m and s, 1 <= m <= 200,1 <= s <= 1000 being the number of junctions and streets, Output For each scenario, output one line containing the text "possible" or "impossible", whether or not it‘s possible to construct a sightseeing tour. Sample Input 4 5 8 2 1 0 1 3 0 4 1 1 1 5 0 5 4 1 3 4 0 4 2 1 2 2 0 4 4 1 2 1 2 3 0 3 4 0 1 4 1 3 3 1 2 0 2 3 0 3 2 0 3 4 1 2 0 2 3 1 1 2 0 3 2 0 Sample Output possible impossible impossible possible Source |
混合图的欧拉回路求解方法为：
任意给无向边加方向，然后统计图中点的入度出度，如果所有点的入度和出度的差为偶数则有可能构成欧拉回路;
删除原来的有向边，加方向的无向边容量为1;
增加源点和汇点，如果某点入度大于出度，则和汇点相连，容量为(入度-出度)/2;如果某点出度大于入度，则源点与其相连，容量为(出度-入度)/2;
在新图中跑最大流，如果满流则有可能构成欧拉回路。
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm 4004
#define maxn 404
using namespace std;
int fir[maxn];
int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm];
int e_max;
int iter[maxn],q[maxn],lv[maxn];
void add_edge(int _u,int _v,int _w)
{
int e;
e=e_max++;
u[e]=_u;v[e]=_v;cap[e]=_w;
nex[e]=fir[u[e]];fir[u[e]]=e;
e=e_max++;
u[e]=_v;v[e]=_u;cap[e]=0;
nex[e]=fir[u[e]];fir[u[e]]=e;
}
void dinic_bfs(int s)
{
int f,r;
memset(lv,-1,sizeof lv);
q[f=r=0]=s;
lv[s]=0;
while(f<=r)
{
int x=q[f++];
for (int e=fir[x];~e;e=nex[e])
{
if (cap[e]>flow[e] && lv[v[e]]<0)
{
lv[v[e]]=lv[u[e]]+1;
q[++r]=v[e];
}
}
}
}
int dinic_dfs(int _u,int t,int _f)
{
if (_u==t) return _f;
for (int &e=iter[_u];~e;e=nex[e])
{
if (cap[e]>flow[e] && lv[_u]<lv[v[e]])
{
int _d=dinic_dfs(v[e],t,min(_f,cap[e]-flow[e]));
if (_d>0)
{
flow[e]+=_d;
flow[e^1]-=_d;
return _d;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
memset(flow,0,sizeof flow);
int total_flow=0;
for (;;)
{
dinic_bfs(s);
if (lv[t]<0) return total_flow;
memcpy(iter,fir,sizeof iter);
int _f;
while ((_f=dinic_dfs(s,t,INF))>0)
total_flow+=_f;
}
return total_flow;
}
struct EDGE
{
int u,v,w;
}edge[1111];
int in[222],out[222];
int main()
{
#ifndef ONLINE_JUDGE
freopen("/home/fcbruce/文档/code/t","r",stdin);
#endif // ONLINE_JUDGE
int T_T;
unsigned long long seed=~0ULL;
scanf("%d",&T_T);
while (T_T--)
{
int n,m,s,t,full_flow=0;
e_max=0;
memset(fir,-1,sizeof fir);
memset(in,0,sizeof in);
memset(out,0,sizeof out);
scanf("%d%d",&n,&m);
seed=seed^T_T^n^m^(~0ULL);
srand(seed);
s=0;t=n+1;
for (int i=0;i<m;i++)
{
scanf("%d %d %d",&edge[i].u,&edge[i].v,&edge[i].w);
if (edge[i].w==1)
{
out[edge[i].u]++;
in[edge[i].v]++;
edge[i].w=0;
continue;
}
if (rand()&1)
{
edge[i].w=1;
out[edge[i].u]++;
in[edge[i].v]++;
add_edge(edge[i].u,edge[i].v,1);
}
else
{
edge[i].w=-1;
out[edge[i].v]++;
in[edge[i].u]++;
add_edge(edge[i].v,edge[i].u,1);
}
}
bool flag=true;
for (int i=1;i<=n;i++)
{
if (abs(in[i]-out[i])&1)
{
flag=false;
break;
}
else
{
if (in[i]==out[i]) continue;
if (in[i]>out[i])
add_edge(i,t,in[i]-out[i]>>1);
else
add_edge(s,i,out[i]-in[i]>>1),full_flow+=out[i]-in[i]>>1;
}
}
if (!flag)
{
puts("impossible");
continue;
}
else
{
if (max_flow(s,t)==full_flow)
puts("possible");
else
puts("impossible");
}
}
return 0;
}
POJ 1637 Sightseeing tour (混合图欧拉回路,网络最大流)