计蒜客 菜鸟物流的运输网络

n(≤100)个点，m(≤n(n−1)/2)​ 条边。
输出任意一个a->mid->b不经过重复点的方案。
联想到 SPOJ 962 Intergalactic Map。
最大流

#include<set>#include<cmath>#include<queue>#include<stack>#include<cstdio>#include<bitset>#include<cassert>#include<cstring>#include<complex>#include<iostream>#include<algorithm>#define pi acos(-1)#define inf (1<<30)#define INF (1<<62)#define y1 bflaisfnmasf#define y2 fsafgmalg#define tm afnsjkf#define j1 sfakf#define j2 fasndfkas#define CLR(x,f) memset(x,f,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define prt(x) cout<<#x<<":"<<x<<" "#define prtn(x) cout<<#x<<":"<<x<<endl#define huh(x) printf("--------case(%d)--------\n",x)#define travel(x) for(Edge *e=h[x];e;e=e->n)#define oo 1000000000#define TL#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;typedef long long ll;const int M=305;struct Edge {    int to, cap, next;} edge[M*M];int head[M], dist[M], mark[M], work[M],Q[M];int n,m;int src,dest,tot;void add(int a, int b, int c) {    edge[tot].to = b, edge[tot].cap = c, edge[tot].next = head[a], head[a] = tot++;    edge[tot].to = a, edge[tot].cap = 0, edge[tot].next = head[b], head[b] = tot++;}bool BFS() {    int i, y, k,L, H;    for (i = 0; i <= dest; i++)dist[i] = -1;    dist[src] = 0;    L=H=0;    Q[H++]=src;    while(L<H){        k=Q[L++];        for(i=head[k];i!=-1;i=edge[i].next){            y=edge[i].to;            if(edge[i].cap>0&&dist[y]==-1) {                dist[y]=dist[k]+1;                Q[H++]=y;            }        }    }    return (dist[dest] >= 0);}int frm[M];int DFS(int x, int flow) {    mark[x] = 1;    if (x == dest)return flow;    for (int y, temp, &i = work[x]; i != -1; i = edge[i].next) {        y = edge[i].to;        if (edge[i].cap > 0 && dist[y] == dist[x] + 1 && !mark[y]) {            if ((temp = DFS(y, min(edge[i].cap, flow))) > 0) {                if(!(i&1)&&(y>>1)!=(x>>1))frm[y>>1]=x>>1;//之前忘记!(i&1)了                edge[i].cap -= temp;                edge[i^1].cap += temp;                return temp;            }        }    }    return 0;}int Dinic_flow(){    int i,ans=0,flow;    while(BFS()){        for(i=0;i<=dest;i++)work[i]=head[i];        while(1){            for(i=0;i<=dest; i++)mark[i]=0;            flow=DFS(src,oo);            if(flow==0)break;            ans+=flow;        }    }    return ans;}int a[M],b[M],a0,b0;int st,en,mid;void query(int x){    a[a0++]=x;    if(frm[x]!=mid)query(frm[x]);}void solve(){    CLR(head,-1);    CLR(frm,0);    tot=0;    scanf("%d%d",&n,&m);    scanf("%d%d%d",&st,&en,&mid);    dest=n+1<<1|1;    src=n+1<<1;    add(src,mid<<1|1,2);    add(st<<1|1,dest,1);    add(en<<1|1,dest,1);    for(int i=1;i<=n;i++)        if(i!=mid)add(i<<1,i<<1|1,1);    for(int u,v,i=1;i<=m;i++){        scanf("%d%d",&u,&v);        add(u<<1|1,v<<1,1);        add(v<<1|1,u<<1,1);    }//  huh(Dinic_flow());    if(Dinic_flow()==2);    a0=0;query(st);    b0=a0;for(int i=0;i<a0;i++)b[i]=a[i];    b[b0++]=mid;    a0=0;query(en);    for(int i=a0-1;i>=0;i--)b[b0++]=a[i];    for(int i=0;i+1<b0;i++){        printf("%d ",b[i]);    }printf("%d\n",b[b0-1]);}int main(){    int cas;scanf("%d",&cas);    while(cas--)solve();    return 0;}