【BZOJ3218】a+b problem (最小割 + 主席树)

继续优化：把a[ ]离散化

#include
using namespace std;
const int inf=1000000007;
const int N=200010;
const int M=1000010;
 
struct Edge{
    int u,v,f,next;
}edge[M];
int head[N],cnt;
int s,t,flow,level[N];
 
struct Node{
    int x,id;
}e[5010];
int L[5010],R[5010],num,key[5010],seq[5010];
 
int tot,ls[N],rs[N],rt[N],Cnt[N];
 
int n,ans;
 
void add(int u,int v,int f){
    edge[cnt].u=u;edge[cnt].v=v;edge[cnt].f=f;edge[cnt].next=head[u];head[u]=cnt++;
    edge[cnt].u=v;edge[cnt].v=u;edge[cnt].f=0;edge[cnt].next=head[v];head[v]=cnt++;
} 
bool bfs(){
    queue<int> q;
    memset(level,-1,sizeof(level));
    level[s]=0;
    q.push(s);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            int f=edge[i].f;
            if(f>0&&level[v]<0){
                level[v]=level[u]+1;
                q.push(v);
            }
        }
    }
    return level[t]>0;
}
int dfs(int u,int mx){
    if(u==t||!mx) return mx;
    int tmp,res=0;
    for(int i=head[u];i!=-1&&mx;i=edge[i].next){
        int v=edge[i].v;
        int f=edge[i].f;
        if(level[v]==level[u]+1&&(tmp=dfs(v,min(mx,f)))){
            edge[i].f-=tmp;
            edge[i^1].f+=tmp;
            res+=tmp;
            mx-=tmp;
        }
    }
    if(!res)
        level[u]=-1;
    return res;
}
int dinic(){
    int maxflow=0;
    while(bfs()){
        maxflow+=dfs(s,inf);
    }
    return maxflow;
}
 
bool cmp(Node x,Node y){
    return x.x<y.x;
}
int findl(int x){
    int l=1,r=num,ans=num;
    while(l<=r){
        int mid=(l+r)/2;
        if(seq[mid]>=x) ans=mid,r=mid-1; 
        else l=mid+1;
    }
    return ans;
}
int findr(int x){
    int l=1,r=num,ans=1;
    while(l<=r){
        int mid=(l+r)/2;
        if(seq[mid]<=x) ans=mid,l=mid+1; 
        else r=mid-1;
    }
    return ans;
}
int modify(int pre,int l,int r,int i){
    int now=++tot;
    if(l==r){
        ls[now]=rs[now]=0;
        Cnt[now]=Cnt[pre]+1;
    }
    else{
        int mid=(l+r)/2;
        if(key[i]<=mid){
            ls[now]=modify(ls[pre],l,mid,i);
            rs[now]=rs[pre];
        }
        else{
            rs[now]=modify(rs[pre],mid+1,r,i);
            ls[now]=ls[pre];
        }
        Cnt[now]=Cnt[ls[now]]+Cnt[rs[now]];
    }
    add(i,2*n+1+now,inf);
    add(2*n+1+pre,2*n+1+now,inf);
    return now;
}
void query(int o,int l,int r,int i){
    if (!Cnt[o]) return;
    if (L[i]<=l&&r<=R[i]){
        add(2*n+1+o,n+i,inf);
        return;
    }
    int mid=(l+r)/2;
    if(L[i]<=mid) query(ls[o],l,mid,i);
    if(mid<R[i]) query(rs[o],mid+1,r,i);
}
int main(){
    cnt=0;memset(head,-1,sizeof(head));
    scanf("%d",&n);
    s=0;t=100000;
    for(int i=1;i<=n;i++){
        int b,w,p;
        scanf("%d%d%d%d%d%d",&e[i].x,&b,&w,&L[i],&R[i],&p);
        e[i].id=i;
        ans+=b+w;
        add(s,i,w);
        add(i,t,b);
        add(n+i,i,p);
    }
    sort(e+1,e+n+1,cmp);
    num=1;key[e[1].id]=1;seq[1]=e[1].x;
    for(int i=2;i<=n;i++)
      if(e[i].x==e[i-1].x) key[e[i].id]=num;
      else key[e[i].id]=++num,seq[num]=e[i].x;
    tot=0;rt[0]=ls[0]=rs[0]=0;
    for(int i=1;i<=n;i++){
        L[i]=findl(L[i]);R[i]=findr(R[i]);
        rt[i]=modify(rt[i-1],1,num,i);
        query(rt[i-1],1,num,i);     
    }
    ans-=dinic();
    printf("%d\n",ans); 
    return 0;
}