poj1637 Sightseeing tour

Sightseeing tour

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8859		Accepted: 3728

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 start and end at the same junction. As in any city, the streets are either one-way or two-way, traffic rules that must be obeyed by the tour bus. Help the executive board and determine if it's possible to construct a sightseeing tour under these constraints.

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, respectively. The following s lines contain the streets. Each street is described with three integers, xi, yi, and di, 1 <= xi,yi <= m, 0 <= di <= 1, where xi and yi are the junctions connected by a street. If di=1, then the street is a one-way street (going from xi to yi), otherwise it's a two-way street. You may assume that there exists a junction from where all other junctions can be reached.

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

Northwestern Europe 2003

混合欧拉回路，用最大流求解。

【建模方法】
把该图的无向边随便定向，计算每个点的入度和出度。如果有某个点出入度
之差为奇数，那么肯定不存在欧拉回路。因为欧拉回路要求每点入度 = 出度，
也就是总度数为偶数，存在奇数度点必不能有欧拉回路。
好了，现在每个点入度和出度之差均为偶数。那么将这个偶数除以2，得x。
也就是说，对于每一个点，只要将x 条边改变方向（入>出就是变入，出>入就是变出），就能保证出=入。如果每个点都是出=入，那么很明显，该图就存在欧拉回路。
现在的问题就变成了：我该改变哪些边，可以让每个点出=入？构造网络流模型。首先，有向边是不能改变方向的，要之无用，删。一开始不是把无向边定向了吗？定的是什么向，就把网络构建成什么样，边长容量上限1。另新建s和t。对于入>出的点u，连接边(u, t)、容量为x，对于出>入的点v，连接边(s, v)，容量为x（注意对不同的点x不同）。之后，察看是否有满流的分配。有就是能有欧拉回路，没有就是没有。欧拉回路是哪个？察看流值分配，将所有流量非 0（上限是1，流值不是0就是1）的边反向，就能得到每点入度=出度的欧拉图。
由于是满流，所以每个入>出的点，都有x条边进来，将这些进来的边反向，OK，入=出了。对于出>入的点亦然。那么，没和s、t连接的点怎么办？和s连接的条件是出>入，和t连接的条件是入>出，那么这个既没和s也没和t连接的点，自然早在开始就已经满足入=出了。那么在网络流过程中，这些点属于“中间点”。我们知道中间点流量不允许有累积的，这样，进去多少就出来多少，反向之后，自然仍保持平衡。
所以，就这样，混合图欧拉回路问题，解了。（来自《网络流建模汇总   by Edelweiss》）

15716369    ksq2013    1637    Accepted    932K    32MS    G++    2812B    2016-07-13 13:18:25

#include<cstdio>
#include<cstring>
#include<iostream>
#define INF 0x3f3f3f3f
using namespace std;
int n,m,s,t,ecnt,first[250],nxt[25000],du[250],res;
struct Edge{int u,v,cap,flow;}e[25000];
bool vis[250];
int q[25000],d[250],cur[250],num[250],p[25000];
bool judge()
{
    for(int i=1;i<=n;i++)
        if(du[i]&1)return false;
    return true;
}
void Link(int x,int y,int z)
{
    e[++ecnt].u=x,e[ecnt].v=y,e[ecnt].cap=z,e[ecnt].flow=0;
    nxt[ecnt]=first[x];
    first[x]=ecnt;
    e[++ecnt].u=y,e[ecnt].v=x,e[ecnt].cap=0,e[ecnt].flow=0;
    nxt[ecnt]=first[y];
    first[y]=ecnt;
}
void design()
{
    res=0;
    for(int i=1;i<=n;i++){
        if(du[i]<0)Link(i,t,-du[i]/2);
        if(du[i]>0)Link(s,i,du[i]/2),res+=(du[i]/2);
    }
}
void bfs()
{
    memset(vis,false,sizeof(vis));
    int head=0,tail=1;
    q[0]=t;d[t]=0;vis[t]=true;
    while(head^tail){
        int now=q[head++];
        for(int i=first[now];i;i=nxt[i])
        if(!vis[e[i].u]&&e[i].cap>e[i].flow){
            vis[e[i].u]=true;
            d[e[i].u]=d[now]+1;
            q[tail++]=e[i].u;
        }
    }
}
int Agument()
{
    int x=t,a=INF;
    while(x^s){
        a=min(a,e[p[x]].cap-e[p[x]].flow);
        x=e[p[x]].u;
    }
    x=t;
    while(x^s){
        e[p[x]].flow+=a;
        e[p[x]^1].flow-=a;
        x=e[p[x]].u;
    }
    return a;
}
int ISAP()
{
    int flow=0;
    bfs();
    memset(num,0,sizeof(num));
    for(int i=0;i<=n+1;i++)num[d[i]]++;
    int x=s;
    for(int i=0;i<=n+1;i++)cur[i]=first[i];
    while(d[s]<n+2){
        if(!(x^t)){
            flow+=Agument();
            x=s;
        }
        bool advanced=false;
        for(int i=cur[x];i;i=nxt[i])
            if(e[i].cap>e[i].flow&&d[x]==d[e[i].v]+1){
                advanced=true;
                p[e[i].v]=i;
                cur[x]=i;
                x=e[i].v;
                break;
            }
        if(!advanced){
            int mn=n+1;
            for(int i=first[x];i;i=nxt[i])
                if(e[i].cap>e[i].flow)
                    mn=min(mn,d[e[i].v]);
            if(--num[d[x]]==0)break;
            num[d[x]=mn+1]++;
            cur[x]=first[x];
            if(x^s)x=e[p[x]].u;
        }
    }
    return flow;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(;T;T--){
        scanf("%d%d",&n,&m);
        s=0,t=n+1,ecnt=1;
        memset(d,0,sizeof(d));
        memset(p,0,sizeof(p));
        memset(nxt,0,sizeof(nxt));
        memset(first,0,sizeof(first));
        memset(du,0,sizeof(du));
        for(int x,y,z,i=1;i<=m;i++){
            scanf("%d%d%d",&x,&y,&z);
            du[x]++,du[y]--;
            if(!z)Link(x,y,1);
        }
        if(!judge()){puts("impossible");continue;}
        design();
        int tmp=ISAP();
        if(tmp^res)puts("impossible");
        else puts("possible");
    }
    return 0;
}