poj 1637 （浅谈最大流在解决混合图欧拉回路中的应用）

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题解：
本人的欧拉路径专题总结☜戳这里。

#include
#include
#include
#include
#include
using namespace std;
const int MAXN=202,MAXM=1002,INF=0x3f3f3f3f;
int n,m;
int head[MAXN],etot;
struct EDGE {
    int v,nxt,r;
}e[MAXM<<1];
int in[MAXN],out[MAXN],dis[MAXN],cur[MAXN],S,T,maxflow,sumflow;
inline int read() {
    int x=0;char c=getchar();
    while (c<'0'||c>'9') c=getchar();
    while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
    return x;
}
inline void adde(int u,int v,int r) {
    e[etot].nxt=head[u],e[etot].v=v,e[etot].r=r,head[u]=etot++;
    e[etot].nxt=head[v],e[etot].v=u,e[etot].r=0,head[v]=etot++;
}
inline bool bfs() {
    queue<int > q;
    memset(dis,-1,sizeof(dis));
    dis[S]=0,q.push(S);
    while (!q.empty()) {
        int p=q.front();
        q.pop();
        for (int i=head[p];~i;i=e[i].nxt) {
            int v=e[i].v;
            if (dis[v]==-1&&e[i].r>0)
                dis[v]=dis[p]+1,q.push(v);
        }
    }
    return ~dis[T];
}
int dfs(int p,int low) {
    if (p==T||low==0) return low;
    int cost=0,flow;
    for (int &i=cur[p];~i;i=e[i].nxt) {
        int v=e[i].v;
        if (dis[v]==dis[p]+1&&e[i].r>0&&(flow=dfs(v,min(e[i].r,low)))) {
            e[i].r-=flow;
            e[i^1].r+=flow;
            low-=flow;
            cost+=flow;
            if (low==0) return cost;
        }
    }
    if (low>0) dis[p]=-1;
    return cost;
}
int main() {
//  freopen("poj 1637.in","r",stdin);
    int kase=read();
    while (kase--) {
        etot=maxflow=sumflow=0;
        memset(head,-1,sizeof(head));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        n=read(),m=read(),S=0,T=n+1;
        while (m--) {
            int u=read(),v=read(),type=read();
            ++in[v],++out[u];
            if (!type) adde(u,v,1);//manually set the direction
        }
        bool euler=true;
        for (int i=1;i<=n;++i) {
            int temp=out[i]-in[i];
            if (temp&1) {euler=false;break;}
            if (temp>0) adde(S,i,temp>>1),sumflow+=(temp>>1);
            else if (temp<0) adde(i,T,(-temp)>>1);
        }
        if (!euler) {puts("impossible");continue;}
        while (bfs()) {//Dinic-maxflow
            for (int i=S;i<=T;++i) cur[i]=head[i];
            int a;
            while (a=dfs(S,INF)) maxflow+=a;
        }
        puts(sumflow^maxflow?"impossible":"possible");
    }
    return 0;
}