兰州大学第一届『飞马杯』程序设计竞赛 - ★★体育课排队★★(二分+最大流)

Posted 2021-06-13 Frozen_Guardian

题目链接：点击查看

题目大意：给出 $n$ 个人，再给出 $n$ 个指定位置，每个人每秒钟可以向上下左右四个方向移动一个单位，问最少需要多长时间，才能使所有人都到达指定位置

题目分析：数据比较小，可以直接二分时间然后跑最大流 $check$，输出路径也是中规中矩的，注意行末空格就好啦

代码：

// Problem: ★★体育课排队★★
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/16520/B
// Memory Limit: 524288 MB
// Time Limit: 8000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) x&-x
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;
}
template<typename T>
inline void write(T x)
{
	if(x<0){x=~(x-1);putchar('-');}
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=2e3+100;
struct Edge
{
	int to,w,next;
}edge[N*N*2];//边数
int head[N],cnt;
void addedge(int u,int v,int w)
{
	edge[cnt].to=v;
	edge[cnt].w=w;
	edge[cnt].next=head[u];
	head[u]=cnt++;
	edge[cnt].to=u;
	edge[cnt].w=0;//反向边边权设置为0
	edge[cnt].next=head[v];
	head[v]=cnt++;
}
int d[N],now[N];//深度 当前弧优化
bool bfs(int s,int t)//寻找增广路
{
	memset(d,0,sizeof(d));
	queue<int>q;
	q.push(s);
	now[s]=head[s];
	d[s]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			int v=edge[i].to;
			int w=edge[i].w;
			if(d[v])
				continue;
			if(!w)
				continue;
			d[v]=d[u]+1;
			now[v]=head[v];
			q.push(v);
			if(v==t)
				return true;
		}
	}
	return false;
}
int dinic(int x,int t,int flow)//更新答案
{
	if(x==t)
		return flow;
	int rest=flow,i;
	for(i=now[x];i!=-1&&rest;i=edge[i].next)
	{
		int v=edge[i].to;
		int w=edge[i].w;
		if(w&&d[v]==d[x]+1)
		{
			int k=dinic(v,t,min(rest,w));
			if(!k)
				d[v]=0;
			edge[i].w-=k;
			edge[i^1].w+=k;
			rest-=k;
		}
	}
	now[x]=i;
	return flow-rest;
}
void init()
{
    memset(now,0,sizeof(now));
	memset(head,-1,sizeof(head));
	cnt=0;
}
int solve(int st,int ed)
{
	int ans=0,flow;
	while(bfs(st,ed))
		while(flow=dinic(st,ed,inf))
			ans+=flow;
	return ans;
}
int n,m;
int x[N],y[N],s[N],xx[N],yy[N];
bool check(int mid) {
	init();
	int st=N-1,ed=st-1;
	for(int i=1;i<=n;i++) {
		addedge(st,i,1);
		addedge(i+n,ed,1);
	}
	int tot;
	for(int i=1;i<=n;i++) {
		tot=0;
		for(int j=1;j<=m;j++) {
			for(int k=0;k<s[j];k++) {
				tot++;
				if(abs(x[i]-(xx[j]+k))+abs(y[i]-yy[j])<=mid) {
					addedge(i,tot+n,1);
				}
			}
		}
	}
	return solve(st,ed)==n;
}
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	int w;
	cin>>w;
	while(w--) {
		read(n),read(m);
		for(int i=1;i<=n;i++) {
			read(x[i]),read(y[i]);
		}
		for(int i=1;i<=m;i++) {
			read(s[i]),read(xx[i]),read(yy[i]);
		}
		int l=0,r=5000,ans=0;
		while(l<=r) {
			int mid=(l+r)>>1;
			if(check(mid)) {
				ans=mid;
				r=mid-1;
			} else {
				l=mid+1;
			}
		}
		check(ans);
		cout<<ans<<endl;
		int tot=0;
		for(int j=1;j<=m;j++) {
			for(int k=0;k<s[j];k++) {
				tot++;
				for(int i=head[n+tot];i!=-1;i=edge[i].next) {
					int to=edge[i].to;
					if(to>=1&&to<=n&&edge[i].w==1) {
						printf("%d%c",to,k==s[j]-1?'\\n':' ');
						break;
					}
				}
			}
		}
	}
	return 0;
}