[POI2015][bzoj4383] Pustynia [线段树优化建图+拓扑排序]

Posted 2021-01-04 dedicatus545

题面

bzoj权限题传送门

luogu传送门

思路

首先，这个题目显然可以从所有小的点往大的连边，然后如果没环就一定可行，从起点（入读为0）开始构造就好了

但是问题来了，如果每个都连的话，本题中边数是$O(n^2)$级别的，显然会挂

发现两条性质：

1.所有的限制条件中，给定的总点数不超过3e5个

2.是一个点比一段区间大

第二个条件决定了我们可以利用线段树优化建图，而第一个条件告诉了我们，本题的总边数应该是$sumkastlog_2n$级别的

那么就做完了

注意拓扑排序的时候有个技巧，把连向实际的点的有向边边权标1，其他标0，这样方便处理

Code

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cassert>
#define ll long long
using namespace std;
inline int read(){
????int re=0,flag=1;char ch=getchar();
????while(ch>'9'||ch<'0'){
????????if(ch=='-') flag=-1;
????????ch=getchar();
????}
????while(ch>='0'&&ch<='9') re=(re<<1)+(re<<3)+ch-'0',ch=getchar();
????return re*flag;
}
int n,m,s,first[1000010],ans[1000010],cnte,pos[1000010],cnt=0,in[1000010],is[1000010],isn[1000010];
struct edge{
????int to,next,w;
}a[8000010];
inline void add(int u,int v,int w){
????in[v]++;
????a[++cnte]=(edge){v,first[u],w};first[u]=cnte;
}
int q[1000010],dp[1000010];
namespace seg{
????int ch[400010][2];
????int build(int l,int r){
????????int cur=++cnt;is[cur]=1;
????????if(l==r){
????????????isn[cur]=1;
????????????return pos[l]=cur;
????????}
????????int mid=(l+r)>>1;
????????ch[cur][0]=build(l,mid);
????????ch[cur][1]=build(mid+1,r);
????????add(ch[cur][0],cur,0);
????????add(ch[cur][1],cur,0);
????????return cur;
????}
????void change(int l,int r,int ql,int qr,int cur,int ori){
????????if(ql>qr) return;
????????if(l>=ql&&r<=qr){
????????????add(cur,ori,1);return;
????????}
????????int mid=(l+r)>>1;
????????if(mid>=ql) change(l,mid,ql,qr,ch[cur][0],ori);
????????if(mid<qr) change(mid+1,r,ql,qr,ch[cur][1],ori);
????}
}
void topo(){
????int i,u,v,head=0,tail=0,tot=0;
????for(i=1;i<=cnt;i++) if(!in[i]){
????????q[tail++]=i;dp[i]=1;
????}
????while(head<tail){
????????u=q[head++];tot+=isn[u];
????????if(dp[u]>1e9){
????????????puts("NIE");return;
????????}
????????if((!is[u])&&(dp[u]>ans[u])){
????????????puts("NIE");return;
????????}
????????if((!is[u])&&(dp[u]<ans[u])) dp[u]=ans[u];
????????for(i=first[u];~i;i=a[i].next){
????????????v=a[i].to;
????????????if(dp[v]<dp[u]+a[i].w){
????????????????dp[v]=dp[u]+a[i].w;
????????????}
????????????if(!(--in[v])) q[tail++]=v;
????????}
????}
????if(tot<n){
????????puts("NIE");return;
????}
????puts("TAK");
????for(i=1;i<=n;i++) printf("%d ",dp[pos[i]]);
}
int main(){
????memset(first,-1,sizeof(first));
????n=read();s=read();m=read();int i,j,t1,t2,t3;
????memset(ans,63,sizeof(ans));
????int root=seg::build(1,n);
????for(i=1;i<=s;i++){
????????t1=read();t2=read();
????????ans[pos[t1]]=t2;is[pos[t1]]=0;
????}
????for(i=1;i<=m;i++){
????????t1=read();t2=read();t3=read();
????????cnt++;is[cnt]=1;
????????for(j=1;j<=t3;j++) q[j]=read(),add(cnt,pos[q[j]],0);
????????seg::change(1,n,t1,q[1]-1,root,cnt);
????????seg::change(1,n,q[t3]+1,t2,root,cnt);
????????for(j=1;j<t3;j++) seg::change(1,n,q[j]+1,q[j+1]-1,root,cnt);
????}
????topo();
}