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<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>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;}