【BZOJ1822】[JSOI2010]冷冻波（二分，网络流）

题面

BZOJ

洛谷

题解

先预处理每个巫妖可以打到哪些小精灵，然后二分答案，网络流判定即可。

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<vector>

#include<queue>

using namespace std;

#define inf 1000000000

#define ll long long

#define MAXL 100000

#define MAX 500

inline int read()

{

	int x=0;bool t=false;char ch=getchar();

	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();

	if(ch=='-')t=true,ch=getchar();

	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();

	return t?-x:x;

}

struct Line{int v,next,w;}e[MAXL];

int h[MAX],cnt=2;

inline void Add(int u,int v,int w)

{

	e[cnt]=(Line){v,h[u],w};h[u]=cnt++;

	e[cnt]=(Line){u,h[v],0};h[v]=cnt++;

}

int S,T;

int level[MAX];

bool bfs()

{

	memset(level,0,sizeof(level));level[S]=1;

	queue<int> Q;Q.push(S);

	while(!Q.empty())

	{

		int u=Q.front();Q.pop();

		for(int i=h[u];i;i=e[i].next)

			if(e[i].w&&!level[e[i].v])

				level[e[i].v]=level[u]+1,Q.push(e[i].v);

	}

	return level[T];

}

int dfs(int u,int flow)

{

	if(u==T||!flow)return flow;

	int ret=0;

	for(int i=h[u];i;i=e[i].next)

	{

		int v=e[i].v,d;

		if(e[i].w&&level[v]==level[u]+1)

		{

			d=dfs(v,min(flow,e[i].w));

			flow-=d;ret+=d;

			e[i].w-=d;e[i^1].w+=d;

			if(!flow)break;

		}

	}

	if(!ret)level[u]=0;

	return ret;

}

int Dinic()

{

	int ret=0;

	while(bfs())ret+=dfs(S,inf);

	return ret;

}

struct Node{int x,y,r,t;}g[MAX],a[MAX],t[MAX];

ll Sqr(ll x){return x*x;}

ll SqrDis(Node a,Node b){return Sqr(a.x-b.x)+Sqr(a.y-b.y);}

bool match(Node a,Node b,Node c)

{

	if(SqrDis(a,b)>=Sqr(a.r))return false;

	ll A=b.y-a.y,B=a.x-b.x,C=b.x*a.y-a.x*b.y;

	ll d=A*c.x+B*c.y+C,dd=Sqr(A)+Sqr(B);

	if(c.r&&Sqr(d)<Sqr(c.r)*dd)return false;

	return true;

}

bool G[MAX][MAX];

int n,m,K;

bool check(int mid)

{

	memset(h,0,sizeof(h));cnt=2;S=0;T=n+m+1;

	for(int i=1;i<=n;++i)Add(S,i,mid/g[i].t+1);

	for(int i=1;i<=m;++i)Add(i+n,T,1);

	for(int i=1;i<=n;++i)

		for(int j=1;j<=m;++j)

			if(G[i][j])Add(i,j+n,1);

	return Dinic()>=m;

}

int main()

{

	n=read();m=read();K=read();

	for(int i=1;i<=n;++i)g[i].x=read(),g[i].y=read(),g[i].r=read(),g[i].t=read();

	for(int i=1;i<=m;++i)a[i].x=read(),a[i].y=read();

	for(int i=1;i<=K;++i)t[i].x=read(),t[i].y=read(),t[i].r=read();

	for(int i=1;i<=n;++i)

		for(int j=1;j<=m;++j)

		{

			bool fl=true;

			for(int k=0;k<=K;++k)

				if(!match(g[i],a[j],t[k])){fl=false;break;}

			G[i][j]=fl;

		}

	for(int j=1;j<=m;++j)

	{

		bool fl=false;

		for(int k=1;k<=n;++k)

			if(G[k][j])fl=true;

		if(!fl){puts("-1");return 0;}

	}

	int l=0,r=1e4,ret=1e9;

	while(l<=r)

	{

		int mid=(l+r)>>1;

		if(check(mid))r=mid-1,ret=mid;

		else l=mid+1;

	}

	printf("%d\n",ret);

	return 0;

}