POJ1637_Sightseeing tour

给一个联通图,有的是单向边,有的是双向边,问是否存在欧拉回路。

乍一看毫无思路,可以这样来搞,对于每条无向边,我们随便指定一个方向,看看是否能够做到所有点的度数之和为偶数。

接下来,对于我们指定的边,假设指定的是U->V,那么我们也同时在网络中设置一条同样的边,使得流量为1,最后如果某点的出入度只差不为0,那么我们把那个差除以2,这表示我们在这个点处至少需要改变多少条无向边的方向。对于出大于入的点,我们连源点,对于如大于出的点,我们连汇点,最后跑最大流看看时候所有的与源点和汇点的边能否满流即可。

一开始的想法是,不指定方向,直接跑最大流,这样是错的,因为无法保证每条无向边最终都被指定了方向,也就是对应回题目里面,不一定无向边都遍历到了,而通过首先指定一个方向,然后再改变方向的方法可以保证这一点。

召唤代码君:

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 2220
#define maxm 22220
using namespace std;

int to[maxm],next[maxm],first[maxn],c[maxm],edge;
int d[maxn],tag[maxn],can[maxn],dgr[maxn],TAG=520;
int n,m,T,s,t,sum,ans;
int Q[maxn],bot,top;
const int inf=~0U>>1;

void _init()
{
    sum=ans=s=0,t=n+1,edge=-1;
    for (int i=s; i<=t; i++) first[i]=-1,dgr[i]=0;
}

bool check()
{
    for (int i=1; i<=n; i++)
        if (dgr[i]&1) return false;
    return true;
}

void addedge(int U,int V,int W)
{
    edge++;
    to[edge]=V,c[edge]=W,next[edge]=first[U],first[U]=edge;
    edge++;
    to[edge]=U,c[edge]=0,next[edge]=first[V],first[V]=edge;
}

bool bfs()
{
    Q[bot=top=1]=t,d[t]=0,tag[t]=++TAG;
    while (bot<=top)
    {
        int cur=Q[bot++];
        for (int i=first[cur]; i!=-1; i=next[i])
            if (c[i^1]>0 && tag[to[i]]!=TAG)
            {
                tag[to[i]]=TAG,d[to[i]]=d[cur]+1;
                can[to[i]]=false,Q[++top]=to[i];
                if (to[i]==s) return true;
            }
    }
    return false;
}

int dfs(int cur,int num)
{
    if (cur==t) return num;
    int tmp=num,k;
    for (int i=first[cur]; i!=-1; i=next[i])
        if (c[i]>0 && tag[to[i]]==TAG && d[to[i]]==d[cur]-1 && !can[to[i]])
        {
            k=dfs(to[i],min(num,c[i]));
            if (k) num-=k,c[i]-=k,c[i^1]+=k;
            if (num==0) break;
        }
    if (num>0) can[cur]=true;
    return tmp-num;
}

int main()
{
    int U,V,W;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        _init();
        while (m--)
        {
            scanf("%d%d%d",&U,&V,&W);
            dgr[U]--,dgr[V]++;
            if (W==0) addedge(U,V,1);
        }
        if (!check())
        {
            puts("impossible");
            continue;
        }
        for (int i=1; i<=n; i++)
        {
            if (dgr[i]==0) continue;
            if (dgr[i]<0) addedge(s,i,-dgr[i]/2);
            else
            {
                sum+=dgr[i]/2;
                addedge(i,t,dgr[i]/2);
            }
        }
        while (bfs()) ans+=dfs(s,inf);
        if (ans==sum) puts("possible");
            else puts("impossible");
    }
    return 0;
}

POJ1637_Sightseeing tour

时间: 2024-11-01 14:24:57

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