題目大意：一個人，如果在一人點沒有出邊時，可以跳轉到任意點，求最小跳轉次數使得走完所有必走邊。
題目思路：有上下界的最小流。
[cpp]
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<queue>
#include<algorithm>
#include<vector>
#include<stack>
#include<list>
#include<iostream>
#include<map>
using namespace std;
#define inf 0x3f3f3f3f
#define Max 550
int max(int a,int b)
{
return a>b?a:b;
}
int min(int a,int b)
{
return a<b?a:b;
}
int dis[Max],gap[Max],pre[Max],cur[Max],p[Max];
int in[Max],out[Max];
int n,m,s,t,eid;
struct node
{
int to,next,c;
}e[40400],ee[40400];
void addedge(int u,int v,int c)
{
ee[eid].to=v;
ee[eid].c=c;
ee[eid].next=p[u];
p[u]=eid++;
}
int ISAP(int st,int ed,int n,int count) ///起點，終點，頂點數
{
memset(dis, 0, sizeof(dis));
memset(gap, 0, sizeof(gap)); gap[0]=n;
memcpy(cur, p, sizeof(p)); ///memcpy!
int i,flag,v,u=pre[st]=st,maxflow=0,aug=inf; //puts("akk");
while(dis[st] < n)
{
for(flag=0,i=cur[u];i!=-1; i=e[i].next) /// cur[u]
if(e[i].c&& dis[u] == dis[e[i].to]+1)
{
flag = 1;
break;
}
if(flag)
{
if(aug > e[i].c)
aug = e[i].c;
v = e[i].to;
pre[v] = u;
cur[u] = i;
u = v;
if(u == ed)
{
for(u=pre[u]; 1;u=pre[u]) ///notice!
{
e[cur[u]].c -= aug;
e[cur[u]^1].c += aug;
if(u==st) break;
}
maxflow += aug;
aug = inf;
}
}
else
{
int minx = n;
for(i=p[u]; i!=-1; i=e[i].next)
if(e[i].c&& dis[e[i].to]<minx)
{
minx = dis[e[i].to];
cur[u] = i;
}
if(--gap[dis[u]] == 0)
break;
dis[u] = minx+1;
gap[dis[u]]++;
u = pre[u];
}
}
return maxflow;
}
int main()
{
int m,n,t,count=1,sum;
int u,v,i,j,k,x,y,tp;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
memset(p,-1,sizeof(p));
eid=0;
sum=0;
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&tp);
if(tp==1)
{
in[v]+=1;
sum+=1;
out[u]+=1;
addedge(u,v,inf-1);
addedge(v,u,0);
}
else
{
addedge(u,v,inf);
addedge(v,u,0);
}
}
for(i=1;i<=n;i++)
{
addedge(n+2,i,in[i]);
addedge(i,n+2,0);
addedge(i,n+3,out[i]);
addedge(n+3,i,0);
addedge(0,i,inf);
addedge(i,0,0);
if(out[i]==0)
{
addedge(i,n+1,inf);
addedge(n+1,i,0);
}
}
addedge(n+1,0,inf);
addedge(0,n+1,0);
int l,r,mid;
l=0;r=sum+2;
while(l<=r)
{
mid=(l+r)>>1;
for(i=0;i<=eid;i++) e[i]=ee[i];
e[p[n+1]].c=mid;
e[p[0]].c=0;
if(ISAP(n+2,n+3,n+4,count)==sum)
r=mid-1;
else
l=mid+1;
}
printf("Case #%d: %d\n",count++,l);
}
}
