hdu 4280 网络流

裸的网络流，递归的dinic会爆栈，在第一行加一句就行了

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cstring>
#include <algorithm>
#include <vector>
#define Maxn 120010
#define Maxm 210000
#define LL int
#define inf 100000000
#define Abs(a) (a)>0?(a):(-a)
using namespace std;
struct Edge{
    int from,to,next;
    LL val;
}edge[Maxm];
const double eps=1e-;
LL value[Maxn];
int head[Maxn],work[Maxn],dis[Maxn],q[Maxn],e,vi[Maxn];
void init()
{
    e=;
    memset(head,-,sizeof(head));
}
void add(int u,int v,LL c)//有向边
{
    edge[e].to=v;edge[e].val=c;edge[e].next=head[u];head[u]=e++;
    edge[e].to=u;edge[e].val=c;edge[e].next=head[v];head[v]=e++;
}
int bfs(int S,int T)
{
    int rear=;
    memset(dis,-,sizeof(dis));
    dis[S]=;q[rear++]=S;
    for(int i=;i<rear;i++)
    {
        for(int j=head[q[i]];j!=-;j=edge[j].next)
        {
            if(edge[j].val&&dis[edge[j].to]==-)
            {
                dis[edge[j].to]=dis[q[i]]+;
                q[rear++]=edge[j].to;
                if(edge[j].to==T) return ;
            }
        }
    }
    return ;
}
LL dfs(int cur,LL a,int T)
{
    if(cur==T) return a;
    for(int &i=work[cur];i!=-;i=edge[i].next)
    {
        if(edge[i].val&&dis[edge[i].to]==dis[cur]+)
        {
            LL t=dfs(edge[i].to,min(a,edge[i].val),T);
            if(t)
            {
                edge[i].val-=t;
                edge[i^].val+=t;
                return t;
            }
        }
    }
    return ;
}
LL Dinic(int S,int T)
{
    LL ans=;
    while(bfs(S,T))
    {
        memcpy(work,head,sizeof(head));
        while(LL t=dfs(S,inf,T)) ans+=t;
    }
    return ans;
}
int main()
{
    int n,m,i,j,num=,t,a,b,west=,east=-,S,T;
    LL c;
    scanf("%d",&t);
    while(t--)
    {
        init();
        west=,east=-;
        scanf("%d%d",&n,&m);
        for(i=;i<=n;i++)
        {
            scanf("%d%d",&a,&b);
            if(a<west)
                west=a,S=i;
            if(a>east)
                east=a,T=i;
        }
        for(i=;i<=m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
        }
        LL ans=Dinic(S,T);
        printf("%d\n",ans);
    }
    return ;
}