UVALive - 6889[并查集+STL]

题目链接:https://vjudge.net/contest/301219#problem/F

解题思路:

枚举每个矩形的时候，看它是否需要和其他人合并只需要查看它的外形边框是否又被标记，这个可以直接用离散化，然后set存一下每个矩形四个格子，就可以用log(n)找到合并的矩形，然后后并查集并一下就好了。

#include <bits/stdc++.h>
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair <int,int> pa;
const int mx = 1e5 + 10;
int n,m,num[mx],a[mx],b[mx]; 
map<int,set<pa>> r,c; 
int fa[mx];
ll sum[mx];
struct node
{
	int x1,y1;
	int x2,y2;
}s[mx];
int find(int x){
	return x==fa[x]? x:fa[x]=find(fa[x]); 
}
void merge(int x,set<pa>& st,int l,int r)
{
	auto it = st.lower_bound(pa(l,0));
	x = find(x);
	while(it!=st.end()&&(it->fi)<=r)
	{
		l = find(it->se);
		if(l!=x) sum[x] += sum[l],fa[l] = x; 
		it++;
	}
}
int main()
{
	while(~scanf("%d",&n)){
		int cx = 0,cy = 0;
		r.clear(),c.clear();
		for(int i=1;i<=n;i++){
			scanf("%d%d%d%d",&s[i].x1,&s[i].y1,&s[i].x2,&s[i].y2);
			sum[i] = s[i].x2*s[i].y2;
			s[i].x2 += s[i].x1 - 1;
			s[i].y2 += s[i].y1 - 1;
			fa[i] = i;
		}
		for(int i=1;i<=n;i++){
			int x1 = s[i].x1;
			int x2 = s[i].x2;
			int y1 = s[i].y1;
			int y2 = s[i].y2;
			r[x1].insert(pa(y1,i));r[x1].insert(pa(y2,i));
			r[x2].insert(pa(y1,i));r[x2].insert(pa(y2,i));
			c[y1].insert(pa(x1,i));c[y1].insert(pa(x2,i));
			c[y2].insert(pa(x1,i));c[y2].insert(pa(x2,i));
		}
		for(int i=1;i<=n;i++){
			if(r.count(s[i].x1-1)) 
			merge(i,r[s[i].x1-1],s[i].y1-1,s[i].y2+1);
			if(r.count(s[i].x2+1)) 
			merge(i,r[s[i].x2+1],s[i].y1-1,s[i].y2+1);
			if(c.count(s[i].y1-1)) 
			merge(i,c[s[i].y1-1],s[i].x1-1,s[i].x2+1);
			if(c.count(s[i].y2+1)) 
			merge(i,c[s[i].y2+1],s[i].x1-1,s[i].x2+1);
		}
		ll ans = 0;
		for(int i=1;i<=n;i++) if(find(i)==i)
		ans = max(ans,sum[i]);
		printf("%d\n",ans);
	}
    return 0;
}