sgu438-The_Glorious_Karlutka_River

Description

SGU似乎死了... 题目搬到了Codeforces...

Problem - 99999438 - Codeforces

Solution

动态最大流.

考虑如果不求时间, 只求能否到达对岸, 这显然是一个拆点最大流.

如果加上时间的限制, 考虑按照时间拆点, 然后枚举时间 \(t\), 并且连上 \(t\) 时间能到达的点.

具体的, 设河的这一岸为 \(s\), 对岸为 \(t\). \(t\) 时刻的点 \(p\) 为 \(p_t\), 并拆成两个点 \(p_t'\), \(p_t''\).

对于 \(t\) 时刻, 连边 \((p_t', p_t'', c_p)\). 对于能到达 \(s\) 或 \(t\) 的节点, 分别连边 \((s, p_t',\infty)\), \((p_{t-1}'', t, \infty)\). 对于能互相到达的点 \(u, v\), 连边 \((u_{t-1}, v_t, \infty)\), \((v_{t-1}, u_t, \infty)\).

那么当总最大流 \(\ge m\) 时, 当前枚举的时间即为答案.

Code

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
using namespace std;
#define rep(i,l,r) for(register int i=(l);i<=(r);++i)
#define repdo(i,l,r) for(register int i=(l);i>=(r);--i)
#define il inline
typedef double db;
typedef long long ll;

//---------------------------------------
const int nsz=60,n2sz=3050,psz=14050,esz=1e6+50,ninf=1e9+7;
int n,m,d,w,pos[nsz][3],link[nsz][2];

int pow2(int v){return v*v;}
struct te0{int f,t;}e0[n2sz];
int pe0=0;

struct te{int t,pr,fl;}edge[esz*2];
int hd[psz],pe=1,ss,tt,np;
void adde(int f,int t,int fl){edge[++pe]=(te){t,hd[f],fl};hd[f]=pe;}
void addsg(int f,int t,int fl){adde(f,t,fl),adde(t,f,0);}

int cur[psz],gap[psz],dep[psz];
void init(){
	rep(i,1,np)cur[i]=hd[i],gap[i]=0,dep[i]=0;
	static int que[psz],qh,qt;
	qh=1,qt=0;
	que[++qt]=tt,gap[1]=1,dep[tt]=1;
	while(qh<=qt){
		int u=que[qh++];
		for(int i=hd[u],v;i;i=edge[i].pr){
			v=edge[i].t;
			if(dep[v])continue;
			dep[v]=dep[u]+1,++gap[dep[v]],que[++qt]=v;
		}
	}
}
int dfs(int p,int mi){
	if(p==tt||mi==0)return mi;
	int fl=0,tmp;
	for(int &i=cur[p],v;i;i=edge[i].pr){
		v=edge[i].t;
		if(dep[v]+1!=dep[p])continue;
		tmp=dfs(v,min(mi,edge[i].fl));
		fl+=tmp,mi-=tmp,edge[i].fl-=tmp,edge[i^1].fl+=tmp;
		if(mi==0)return fl;
	}
	if(gap[dep[p]]==1)dep[ss]=np+1;
	--gap[dep[p]],++dep[p],++gap[dep[p]];
	cur[p]=hd[p];
	return fl;
}
int maxfl(){
	init();
	int res=0;
	while(dep[ss]<=np)res+=dfs(ss,ninf);
	return res;
}


int tr(int p,int tm,int fl){//fl==0: from;  fl==1: to
	return (n*(tm-1)+p)*2-1+fl;
}

int sum=0;
int sol(){
	if(d>=w)return 1;
	else if(n==0)return -1;
	ss=(n+m+5)*n*2+1,tt=(n+m+5)*n*2+2,np=tt;
	rep(tm,1,n+m+1){
		rep(i,1,n)if(link[i][0])addsg(ss,tr(i,tm,0),ninf);
		if(tm>1)rep(i,1,n)if(link[i][1])addsg(tr(i,tm-1,1),tt,ninf);
		rep(i,1,n)addsg(tr(i,tm,0),tr(i,tm,1),pos[i][2]);
		if(tm>1){
			rep(i,1,pe0){
				addsg(tr(e0[i].f,tm-1,1),tr(e0[i].t,tm,0),ninf);
				addsg(tr(e0[i].t,tm-1,1),tr(e0[i].f,tm,0),ninf);
			}
		}
		sum+=maxfl();
//		printf("tm=%d sum=%d\n",tm,sum);
		if(sum>=m)return tm;
	}
	return -1;
}

int main(){
	ios::sync_with_stdio(0),cin.tie(0);
	cin>>n>>m>>d>>w;
	rep(i,1,n)cin>>pos[i][0]>>pos[i][1]>>pos[i][2];
	rep(i,1,n){
		if(pos[i][1]<=d)link[i][0]=1;
		if(pos[i][1]>=w-d)link[i][1]=1;
		rep(j,i+1,n){
			if(pow2(pos[i][0]-pos[j][0])+pow2(pos[i][1]-pos[j][1])<=pow2(d))e0[++pe0]=(te0){i,j};
		}
	}
	int ans=sol();
	if(ans==-1)cout<<"IMPOSSIBLE\n";
	else cout<<ans<<'\n';
	return 0;
}