CF434D Nanami's Power Plant

就是切糕那道题，首先对每个函数连一串，然后\(x_u\leq x_v+d\)这个条件就是\(u\)函数\(i\)取值连向\(v\)函数\(i-d\)取值边权为inf，然后答案就是最小割了。

#include<bits/stdc++.h>
#define il inline//
#define vd void
typedef long long ll;
il int gi(){
    int x=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int a[51],b[51],c[51],l[51],r[51];
int num[51][202],cnt,S,T;
int fir[11000],head[10010],dep[10010],dis[10000010],nxt[10000010],id=1;
ll w[10000010];
il vd link(int a,int b,ll c){
    nxt[++id]=fir[a],fir[a]=id,dis[id]=b,w[id]=c;
    nxt[++id]=fir[b],fir[b]=id,dis[id]=a,w[id]=0;
}
il bool BFS(){
    static int que[11000],hd,tl;
    memset(dep,0,sizeof dep);
    hd=tl=0;que[tl++]=S;dep[S]=1;
    while(hd^tl){
        int x=que[hd++];
        for(int i=fir[x];i;i=nxt[i])
            if(!dep[dis[i]]&&w[i])
                dep[dis[i]]=dep[x]+1,que[tl++]=dis[i];
    }
    return dep[T];
}
il ll Dinic(int x,ll maxflow){
    if(x==T)return maxflow;
    ll ret=0;
    for(int&i=head[x];i;i=nxt[i])
        if(w[i]&&dep[dis[i]]==dep[x]+1){
            ll d=Dinic(dis[i],std::min(maxflow-ret,w[i]));
            w[i]-=d,w[i^1]+=d,ret+=d;
            if(ret==maxflow)break;
        }
     return ret;
}
int main(){
    int n=gi(),m=gi();S=++cnt,T=++cnt;
    for(int i=1;i<=n;++i)a[i]=gi(),b[i]=gi(),c[i]=gi();
    for(int i=1;i<=n;++i)l[i]=gi(),r[i]=gi();
    for(int i=1;i<=n;++i){
        for(int j=l[i];j<=r[i]+1;++j)num[i][j+100]=++cnt;
        link(S,num[i][l[i]+100],1e18);
        for(int j=l[i];j<=r[i];++j)link(num[i][j+100],num[i][j+101],(1<<25)-(a[i]*j*j+b[i]*j+c[i]));
        link(num[i][r[i]+101],T,1e18);
    }
    int u,v,d;
    while(m--){
        u=gi(),v=gi(),d=gi();
        for(int i=std::max(l[u]-d,l[v])+100;i<=std::min(r[u]-d,r[v])+101;++i)link(num[u][i+d],num[v][i],1e18);
    }
    ll ans=(1ll<<25)*n;
    while(BFS())memcpy(head,fir,sizeof head),ans-=Dinic(S,1e18);
    printf("%lld\n",ans);
    return 0;
}