【bzoj2330】[SCOI2011]糖果

裸的差分约束系统.建图很好建，跑一个最长路就可以了

#include
#include
#include
#include
#include
#include
using namespace std;
const int N=100010;
struct edge{
    int v,w,next;
}e[300010];
long long ans;
int te,k,n;
int head[N],d[N],inq[N],vis[N];
queue<int>q;
inline int F()
{
    register int aa,bb;register char ch;
    while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');ch=='-'?aa=bb=0:(aa=ch-'0',bb=1);
    while(ch=getchar(),ch>='0'&&ch<='9')aa=(aa<<3)+(aa<<1)+ch-'0';return bb?aa:-aa;
}
void add(int u,int v,int w)
{
    e[++te].v=v;
    e[te].w=w;
    e[te].next=head[u];
    head[u]=te;
}
int spfa(int s)
{
    memset(inq,0,sizeof(inq));
    memset(vis,0,sizeof(vis));
    memset(d,128,sizeof(d));
    while(!q.empty())q.pop();
    inq[s]=1;vis[s]=1;q.push(s);d[s]=0;
    while(!q.empty())
    {
        int u=q.front();q.pop();vis[u]=0;
        for (int i=head[u];i;i=e[i].next)
        {
            int v=e[i].v;
            if (d[v]if (!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                    inq[v]++;
                    if (inq[v]>n)return 0;
                }
            }
        }
    }
    return 1;
}

int main()
{
    te=0;
    memset(head,0,sizeof(head));
    int op,u,v;
    scanf("%d%d",&n,&k);
    for (int i=1;i<=k;i++)
    {
        scanf("%d%d%d",&op,&u,&v);
        if (op==1)
        add(u,v,0),add(v,u,0);
        else if (op==2) 
        add(u,v,1);
        else if (op==3) 
        add(v,u,0);
        else if (op==4)
        add(v,u,1);
        else add(u,v,0);
        if ((op==2||op==4)&&u==v){cout<<-1<return 0;} 
    }
    for (int i=n;i>=1;--i)
    add(0,i,1);
    if (!spfa(0))
    cout<<-1;
    else 
    {
        for (int i=1;i<=n;++i)
        ans+=d[i];
        cout<