题意：给你一张带权有向图，让你求最大树形图。并在此前提下令n号结点父亲的编号最小。
比赛的时候套了个二分，TLE了。
实际上可以给每个边的权值乘1000，对于n号结点的父边，加上(999-父结点编号)大小的权值，这样即可保证最大树形图的前提下，n号结点父亲的编号最小。
网上找了个朱-刘算法的板子，把边权取负就能跑最大树形图了。
#include <cstdio>
#include <string>
#include <cstring>
#define MAXN 1005
#define MAXM 10005
using namespace std;
struct node
{
int u, v;
int w;
}edge[MAXM];
int pre[MAXN], id[MAXN], vis[MAXN], n, m;
int in[MAXN];
typedef int type;
#define INF 2000000000
type Directed_MST(int root, int V, int E)
{
type ret = 0;
while(true)
{
//1.?????
for(int i = 0; i < V; i++) in[i] = INF;
for(int i = 0; i < E; i++)
{
int u = edge[i].u;
int v = edge[i].v;
if(edge[i].w < in[v] && u != v) {pre[v] = u; in[v] = edge[i].w;}
}
for(int i = 0; i < V; i++)
{
if(i == root) continue;
if(in[i] == INF) return -1;//???????????,???????
}
//2.??
int cnt = 0;
memset(id, -1, sizeof(id));
memset(vis, -1, sizeof(vis));
in[root] = 0;
for(int i = 0; i < V; i++) //?????
{
ret += in[i];
int v = i;
while(vis[v] != i && id[v] == -1 && v != root) //?????????,?????????,???????
{
vis[v] = i;
v = pre[v];
}
if(v != root && id[v] == -1)//??
{
for(int u = pre[v]; u != v; u = pre[u]) id[u] = cnt;
id[v] = cnt++;
}
}
if(cnt == 0) break; //?? ?break
for(int i = 0; i < V; i++)
if(id[i] == -1) id[i] = cnt++;
//3.????
for(int i = 0; i < E; i++)
{
int u = edge[i].u;
int v = edge[i].v;
edge[i].u = id[u];
edge[i].v = id[v];
if(id[u] != id[v]) edge[i].w -= in[v];
}
V = cnt;
root = id[root];
}
return ret;
}
int zu,xs[10005],ys[10005],zs[10005];
int main()
{
// freopen("1009.in","r",stdin);
// freopen("1009.out","w",stdout);
scanf("%d",&zu);
for(;zu;--zu)
{
scanf("%d%d", &n, &m);
for(int i = 0; i < m; i++)
{
scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
edge[i].u--;
edge[i].v--;
edge[i].w*=1000;
if(edge[i].v==n-1){
edge[i].w+=(999-edge[i].u);
}
edge[i].w=-edge[i].w;
xs[i]=edge[i].u;
ys[i]=edge[i].v;
zs[i]=edge[i].w;
if(edge[i].u == edge[i].v){
edge[i].w = INF;
zs[i] = INF;
}
}
int ans = Directed_MST(0, n, m);
printf("%d %d\n",(-ans)/1000,1000-(-ans)%1000);
}
return 0;
}