传送门：
思路：把每天当作一层，一层包含n个点，每层向下一层在原图中有边相连的点连边，表示一天能走一条边，每天的n点向汇连边
枚举天数，每次加一层，满流即输出答案
#include#include #include #include const int maxn=10010,inf=1e9,maxm=1000010;using namespace std;int n,m,S=maxn-1,T=maxn-2,cnt,q[maxn+10],head,tail,pre[maxm],now[maxn],son[maxm],val[maxm],tot=1,dis[maxn];struct Edge{int x,y,z;}E[maxm];void add(int a,int b,int c){pre[++tot]=now[a],now[a]=tot,son[tot]=b,val[tot]=c;}void ins(int a,int b,int c){add(a,b,c),add(b,a,0);}bool bfs(){ memset(dis,-1,sizeof(dis)); q[tail=1]=S,dis[S]=head=0; while (head!=tail){ if (++head>maxn) head=1; int x=q[head]; for (int y=now[x];y;y=pre[y]) if (dis[son[y]]==-1&&val[y]){ if (++tail>maxn) tail=1; dis[son[y]]=dis[x]+1,q[tail]=son[y]; } } return dis[T]>0;}int find(int x,int low){ if (x==T) return low; int y,res=0; for (y=now[x];y;y=pre[y]){ if (dis[son[y]]!=dis[x]+1||!val[y]) continue; int tmp=find(son[y],min(low,val[y])); val[y]-=tmp,val[y^1]+=tmp,res+=tmp,low-=tmp; if (!low) break; } if (!y) dis[x]=-1; return res;}void rebuild(){ for (int i=2;i<=tot;i+=2) val[i]+=val[i^1],val[i^1]=0;}int main(){ scanf("%d%d%d",&n,&m,&cnt); for (int i=1,x,y,z;i<=m;i++) scanf("%d%d%d",&x,&y,&z),E[i]=(Edge){x,y,z}; ins(S,1,cnt); for (int i=1;i<=n+cnt;i++){ rebuild(); for (int j=1;j<=m;j++) ins((i-1)*n+E[j].x,i*n+E[j].y,E[j].z); for (int j=1;j<=n;j++) ins((i-1)*n+j,i*n+j,inf); ins(i*n+n,T,inf); int res=0; while (bfs()) res+=find(S,inf); if (res==cnt) return printf("%d\n",i),0; } return 0;}