[USACO08JAN]haybale猜测Haybale Guessing

题目描述

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.

给一段长度为n，每个位置上的数都不同的序列a[1..n]和q和问答，每个问答是(x, y, r)代表RMQ(a, x, y) = r, 要你给出最早的有矛盾的那个问答的编号。

输入输出格式

输入格式：

• Line 1: Two space-separated integers: N and Q

• Lines 2..Q+1: Each line contains three space-separated integers that represent a single query and its reply: Ql, Qh, and A

输出格式：

• Line 1: Print the single integer 0 if there are no inconsistencies
  among the replies (i.e., if there exists a valid realization of the hay
  stacks that agrees with all Q queries). Otherwise, print the index from
  1..Q of the earliest query whose answer is inconsistent with the answers
  to the queries before it.

输入输出样例

输入样例#1：
复制

20 4
1 10 7
5 19 7
3 12 8
11 15 12

输出样例#1： 复制

3

以下题解摘自洛谷题解，非常清楚

出现矛盾的区间符合两个条件之一：

1.题目中的两个干草堆没有任何数量是一样的，所以如果两个区间没有交集并且它们的最小值相同，则这两个区间产生矛盾

2.如果一个区间包含另一个区间，被包含的区间的最小值大于另一个区间，则两个区间产生矛盾

考虑对原先问答的顺序进行二分答案，对于一个二分出的mid作如下处理：

为了方便处理矛盾2，将从1到mid的每个区间的值按照从大到小进行排序

对于值相同的区间，求出并集和交集的范围，如果不存在并集，则mid不可行

维护一颗线段树，将交集的区间覆盖为1

查询并集的区间是否被覆盖为1，如果是，则mid不可行

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 struct Ask
 {
   int l,r,x;
 }a[],b[];
 int c[],n,q;
 bool cmp(Ask a,Ask b)
 {
   return a.x>b.x;
 }
 void build(int rt,int l,int r)
 {
   if (l==r)
     {
       c[rt]=;
       return;
     }
   int mid=(l+r)/;
   build(rt*,l,mid);
   build(rt*+,mid+,r);
   c[rt]=c[rt*]&c[rt*+];
 }
 void pushdown(int rt)
 {
   if (c[rt])
     {
       c[rt*]=c[rt];
       c[rt*+]=c[rt];
     }
 }
 void update(int rt,int l,int r,int L,int R)
 {
   if (l>=L&&r<=R)
     {
       c[rt]=;
       return;
     }
   int mid=(l+r)/;
   pushdown(rt);
   if (L<=mid) update(rt*,l,mid,L,R);
   if (R>mid) update(rt*+,mid+,r,L,R);
   c[rt]=c[rt*]&c[rt*+];
 }
 int query(int rt,int l,int r,int L,int R)
 {
   if (c[rt]) return ;
   if (l>=L&&r<=R)
     {
       return c[rt];
     }
   int mid=(l+r)/;
   int ll=,rr=;
   if (L<=mid) ll=query(rt*,l,mid,L,R);
   if (R>mid) rr=query(rt*+,mid+,r,L,R);
   c[rt]=c[rt*]&c[rt*+];
   return ll&rr;
 }
 bool check(int mid)
 {int i,j,l1,l2,r1,r2,k;
   for (i=;i<=mid;i++)
     b[i]=a[i];
   build(,,n);
   sort(b+,b+mid+,cmp);
   for (i=;i<=mid;i=j)
     {
       j=i;
       while (j<=mid&&b[j].x==b[i].x) j++;
       l1=2e9;r2=2e9;l2=-;r1=-;
       for (k=i;k<j;k++)
     {
       l1=min(l1,b[k].l);
       r1=max(r1,b[k].r);
       l2=max(l2,b[k].l);
       r2=min(r2,b[k].r);
     }
       if (l2>r2) return ;
       if (query(,,n,l2,r2)) return ;
       update(,,n,l1,r1);
     }
   return ;
 }
 int main()
 {int i;
   cin>>n>>q;
   for (i=;i<=q;i++)
     {
       scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].x);
     }
   int l=,r=q,ans=;
   while (l<=r)
     {
       int mid=(l+r)/;
       if (check(mid))
     {
       ans=mid;
       l=mid+;
     }
       else r=mid-;
     }
   cout<<(ans+)%(q+);
 }