[USACO08JAN]Haybale Guessing(LuoguP2898)

The cows, who always have an inferiority complex about their intelligence, have a new guessing game to sharpen their brains.A designated 'Hay Cow' hides behind the barn and creates N (1 ≤ N ≤ 1,000,000) uniquely-sized stacks (conveniently numbered 1..N) of hay bales, each with 1..1,000,000,000 bales of hay.The other cows then ask the Hay Cow a series of Q (1 ≤ Q ≤ 25,000) questions about the the stacks, all having the same form:What is the smallest number of bales of any stack in the range of stack numbers Ql..Qh (1 ≤ Ql ≤ N; Ql ≤ Qh ≤ N)?The Hay Cow answers each of these queries with a single integer A whose truthfulness is not guaranteed.Help the other cows determine if the answers given by the Hay Cow are self-consistent or if certain answers contradict others.
并查集+二分。
有两种产生矛盾的情况。
题目中的两个元素的值没有任何数量是一样的，所以如果两个最小值相同的区间没有交集，则这两个区间产生矛盾。
如果一个区间包含一个最小值比其小的区间，则产生矛盾。
可以二分答案判断到此是否可行。
为方便处理矛盾，二分检查时我们按照预估最小值大到小排序。注意并查集合并区间的细节处理。当区间左右端点相同时，并查集找左右端点是显然会找到相同元素且这合法。并查集合并区间时，是将区间内的元素的父亲指向右端点的后一个元素，如果查询区间的右端点是本身，该点实际并未合并。其余情况如果要检查的区间被完全包含则存在矛盾。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long ll;
#define rint register int

inline void getint(int &x)
{
	char c;
	for(c=getchar();c>'9'||c<'0';c=getchar());
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<1)+(x<<3)+c-'0';
}
const int MAXN=1000000+5;
struct Seg
{
	int l,r,v;
}e[MAXN];
int lim,n;
int srt[MAXN],fa[MAXN];

bool cmp(int a,int b)
{
	if(e[a].v==e[b].v)
		return e[a].l==e[b].l?e[a].r<e[b].r:e[a].l<e[b].l;
	return e[a].v>e[b].v;
}

int find(int x)
{
	return fa[x]==x?x:fa[x]=find(fa[x]);
}

bool check(int mid)
{
	for(rint i=1;i<=mid;++i)
		srt[i]=i;
	for(rint i=1;i<=lim;++i)
		fa[i]=i;
	sort(srt+1,srt+mid+1,cmp);
	int x,y;
	for(int i=1;i<=mid;++i)
	{
		int t=e[srt[i]].v,l=e[srt[i]].l,
                r=e[srt[i]].r,lm=e[srt[i]].l,rm=e[srt[i]].r;
		while(i+1<=mid&&t==e[srt[i+1]].v)
		{
			++i;
			if(e[srt[i]].l>r)
				return 0;
			l=max(l,e[srt[i]].l);
			r=min(r,e[srt[i]].r);
			lm=min(lm,e[srt[i]].l);
			rm=max(rm,e[srt[i]].r);
		}
		int x=find(l),y=find(r);
		if((l!=r&&x==y&&y!=r)||(l==r&&y!=r)) //注意此处判断
			return 0;
		x=lm,y=rm;
		int rg=find(y+1);
		while(rg!=find(x))
		{
			int fx=find(x);
			fa[fx]=fa[fx+1];
		}
	}
	return 1;
}

int main()
{
	scanf("%d%d",&lim,&n);
	for(int i=1;i<=n;++i)
		getint(e[i].l),getint(e[i].r),getint(e[i].v);
	int l=1,r=n+1;
	while(r-l>1)
	{
		int mid=l+r>>1;
		if(check(mid))
			l=mid;
		else r=mid;
	}
	printf("%d",l==n?0:l+1);
	return 0;
}