BZOJ3218:a + b Problem 网络流+线段树

思路:

首先最小割的建模是显然的.

令\(S\)集合表示选择白色,\(T\)集合表示选择黑色.

则有:

\[S->i~c=w_i\]表示割掉这条边之后,\(i\)在\(T\)集合,颜色为黑色,失去了白色的价值.

同理\[i->T~c=b_i\].

接下来假设有一堆点,若这些点有一个为白色,且\(i\)为黑色,则损失价值\(p_i\).

这个怎么搞?

设一个辅助节点\(x\),让所有对\(i\)有影响的点\(j\)都连\(j->x~c=INF\).然后再连接\(x->i~c=p_i\).

这样的话,如果有某个点\(x\)选了白色(属于\(S\)集),而节点\(i\)为黑色(属于\(T\)集),那么这条\(p_i\)的边就必定需要割掉.

可是这样边数爆表.

注意到一堆\(INF\)的边很是浪费,于是我们用线段树优化.

首先不考虑标号的限制,只考虑\(l\leq{a}\leq{r}\)的限制.

离散化后,我们用线段树上的\(O(logn)\)个节点来连向辅助节点.

接下来是标号的限制...

我们用一颗可持久化线段树来维护连边即可.

(关于可持久化线段树的姿势可以看代码)

(内附良心样例)

#include<cstdio>
#include<cstring>
#include<cctype>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<map>
using namespace std;
  
#define INF 0x3f3f3f3f
  
#define N 5010
int a[N],b[N],w[N],l[N],r[N],p[N];
  
int seq[N*3],id;
  
struct MaxflowSolver{
    int head[300010],next[1000010],end[1000010],flow[1000010],ind;
    int d[300010],gap[300010],cur[300010],stk[300010],top;
    inline void reset(){ind=top=0,memset(head,-1,sizeof head);}
    inline void addedge(int a,int b,int f){int q=ind++;end[q]=b,next[q]=head[a],head[a]=q,flow[q]=f;}
    inline void make(int a,int b,int f){/*printf("%d %d %d\n",a,b,f);*/addedge(a,b,f),addedge(b,a,0);}
    inline void bfs(int T){
        static int q[300010];int fr=0,ta=0;
        memset(d,-1,sizeof d),memset(gap,0,sizeof gap),++gap[d[T]=0],q[ta++]=T;
        while(fr^ta){
            int i=q[fr++];for(int j=head[i];j!=-1;j=next[j])if(d[end[j]]==-1)++gap[d[end[j]]=d[i]+1],q[ta++]=end[j];
        }
    }
    inline int Maxflow(int S,int T,int cnt){
        int res=0,i,Min,ins,u=S;bfs(T),memcpy(cur,head,sizeof cur);
        while(d[S]<cnt){
            if(u==T){
                for(Min=INF,i=0;i<top;++i)if(Min>flow[stk[i]])Min=flow[stk[i]],ins=i;
                for(res+=Min,i=0;i<top;++i)flow[stk[i]]-=Min,flow[stk[i]^1]+=Min;
                u=end[stk[top=ins]^1];
            }
            if(u!=T&&!gap[d[u]-1])break;
            bool find=0;
            for(int&j=cur[u];j!=-1;j=next[j])if(flow[j]&&d[u]==d[end[j]]+1){find=1,ins=j;break;}
            if(find){cur[u]=stk[top++]=ins,u=end[ins];}
            else{
                Min=cnt;for(int j=head[u];j!=-1;j=next[j])if(flow[j]&&d[end[j]]<Min)Min=d[end[j]],cur[u]=j;
                if(!--gap[d[u]])break;++gap[d[u]=Min+1];
                if(u!=S)u=end[stk[--top]^1];
            }
        }return res;
    }
}SteinsGate;
  
struct Node{
    int l,r;
}S[300010];int cnt;
  
int root[5010];
int Newversion(int last,int tl,int tr,int ins){
    int q=++cnt;S[q]=S[last];if(last)SteinsGate.make(last,q,INF);if(tl==tr)return q;
    int mid=(tl+tr)>>1;
    if(ins<=mid)S[q].l=Newversion(S[last].l,tl,mid,ins);else S[q].r=Newversion(S[last].r,mid+1,tr,ins);
    return q;
}
  
void Addedge(int q,int tl,int tr,int dl,int dr,int topoint){
    if(dl<=tl&&tr<=dr){if(q)SteinsGate.make(q,topoint,INF);return;}
    int mid=(tl+tr)>>1;
    if(dr<=mid)Addedge(S[q].l,tl,mid,dl,dr,topoint);
    else if(dl>mid)Addedge(S[q].r,mid+1,tr,dl,dr,topoint);
    else Addedge(S[q].l,tl,mid,dl,mid,topoint),Addedge(S[q].r,mid+1,tr,mid+1,dr,topoint);
}
  
int main(){
    int n;scanf("%d",&n);register int i,j;
    for(i=1;i<=n;++i)scanf("%d%d%d%d%d%d",&a[i],&b[i],&w[i],&l[i],&r[i],&p[i]);
  
    for(i=1;i<=n;++i)seq[++id]=a[i],seq[++id]=l[i],seq[++id]=r[i];sort(seq+1,seq+id+1);
    map<int,int>M;int num=0;for(seq[0]=-1<<30,i=1;i<=id;++i)if(seq[i]!=seq[i-1])M[seq[i]]=++num;
    for(i=1;i<=n;++i)a[i]=M[a[i]],l[i]=M[l[i]],r[i]=M[r[i]];
  
    SteinsGate.reset();
    for(i=1;i<=n;++i)root[i]=Newversion(root[i-1],1,num,a[i]),SteinsGate.make(250000+i,cnt,INF);
    for(i=1;i<=cnt;++i){
        if(S[i].l)SteinsGate.make(S[i].l,i,INF);
        if(S[i].r)SteinsGate.make(S[i].r,i,INF);
    }
  
    int S=0,T=++cnt,res=0;
    for(i=1;i<=n;++i)res+=b[i]+w[i],SteinsGate.make(S,250000+i,w[i]),SteinsGate.make(250000+i,T,b[i]);
    for(i=1;i<=n;++i)if(i)Addedge(root[i-1],1,num,l[i],r[i],++cnt),SteinsGate.make(cnt,250000+i,p[i]);
  
    printf("%d",res-SteinsGate.Maxflow(S,T,cnt+1+n));
  
#ifndef ONLINE_JUDGE
    system("pause");
#endif
  
    return 0;
}
/*10
0 1 7 3 9 2
7 4 0 9 10 5
1 0 4 2 10 2
7 9 1 5 7 2
6 3 5 3 6 2
6 6 4 1 8 1
6 1 6 0 6 5
2 2 5 0 9 3
5 1 3 0 2 5
5 6 7 1 1 2*/