CF538H. Summer Dichotomy

技术分享4年前 (2020)更新 技术分享
1,877 0 0

题目大意

CF538H. Summer Dichotomy

题解

看CF上标签的意思应该是2-sat+数据结构优化建图难怪是H题

看了题解,其实想想(也许)能够想出来

设两组的最大l和最小r为l1r1l2r2,则满足l1+l2<=t2且r1+r2>=t1

根据max(l)和min(r)是否在同一区间分类讨论

判断很简单,设代表1和2区间的点,两点之间有边即不能在同一区间

然后判是否是二分图,以及新建的点颜色是否不同即可

最后具体分的人数要考虑一下,还要特判一下全在某个集合的和max(l)>t2的

大概是这样,有一点细节

code

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define inf 2133333333
#define ll long long
//#define file
using namespace std;

int a[600001][2],ls[100003],L[100001],R[100001],b[100001][2],f[100003],len,t1,t2,n,m,i,j,k,l,ml,mr,l1,r1,l2,r2;
bool bz;

void New(int x,int y) {++len;a[len][0]=y;a[len][1]=ls[x];ls[x]=len;}
void NEW(int x,int y) {New(x,y);New(y,x);}

void dfs(int t)
{
	int i;
	if (!bz) return;
	
	for (i=ls[t]; i; i=a[i][1])
	if (!f[a[i][0]])
	{f[a[i][0]]=3-f[t];dfs(a[i][0]);}
	else
	if (f[a[i][0]]==f[t]) {bz=0;return;}
}

void work(bool tp)
{
	int i,j,k,l;
	
	memset(ls,0,sizeof(ls));len=0;
	memset(f,0,sizeof(f));
	fo(i,1,m) NEW(b[i][0],b[i][1]);
	fo(i,1,n)
	if (!tp)
	{
		if (R[i]<t1-mr || R[i]<ml) NEW(n+1,i);
		if (L[i]>t2-ml || mr<L[i]) NEW(n+2,i);
	}
	else
	if (!(L[i]<=t2-ml && R[i]>=t1-mr)) NEW(n+2,i);
	
	bz=1;
	f[n+1]=1;dfs(n+1);
	if (f[n+2]==1 || !bz) return;
	f[n+2]=2;dfs(n+2);
	if (bz)
	{
		fo(i,1,n)
		if (!f[i])
		{
			f[i]=1;dfs(i);
			if (!bz) return;
		}
	}
	
	if (bz)
	{
		printf("POSSIBLE\n");
		
		l1=l2=-inf,r1=r2=inf;
		if (!tp) l1=ml,r2=mr; else l1=ml,r1=mr;
		fo(i,1,n)
		if (f[i]==1)
		l1=max(l1,L[i]),r1=min(r1,R[i]);
		else
		l2=max(l2,L[i]),r2=min(r2,R[i]);
		
		fo(i,1,n) if (f[i]==2) break;
		if (i>n)
		{printf("%d %d\n",l1,max(t1-l1,0));}
		else
		{k=max(l1+l2,t1);printf("%d %d\n",l1+max((k-l1)-r2,0),k-(l1+max((k-l1)-r2,0)));}
		fo(i,1,n) printf("%d",f[i]);
		printf("\n");
		exit(0);
	}
}

int main()
{
	#ifdef file
	freopen("CF538H.in","r",stdin);
	#endif
	
	scanf("%d%d",&t1,&t2);
	scanf("%d%d",&n,&m);
	ml=-inf;mr=inf;
	fo(i,1,n) scanf("%d%d",&L[i],&R[i]),ml=max(ml,L[i]),mr=min(mr,R[i]);
	fo(i,1,m) scanf("%d%d",&b[i][0],&b[i][1]);
	if (ml>t2) {printf("IMPOSSIBLE\n");return 0;}
	
	if (n==1)
	{
		if (L[1]<=t2 && t1<=R[1]) {printf("POSSIBLE\n");printf("%d %d\n",max(L[1],t1),0);printf("1\n");}
		else printf("IMPOSSIBLE\n");
		return 0;
	}
	
	if (ml>mr) {fo(i,1,n) if (mr<L[i] && R[i]<ml) {printf("IMPOSSIBLE\n");return 0;}}
	
	work(0);
	if (ml<=mr)
	work(1);
	
	printf("IMPOSSIBLE\n");
	
	fclose(stdin);
	fclose(stdout);
	
	return 0;
}
© 版权声明

相关文章

暂无评论

您必须登录才能参与评论!
立即登录
暂无评论...