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
- 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
20 4 1 10 7 5 19 7 3 12 8 11 15 12
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
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 struct Ask 8 { 9 int l,r,x; 10 }a[25001],b[25001]; 11 int c[4000001],n,q; 12 bool cmp(Ask a,Ask b) 13 { 14 return a.x>b.x; 15 } 16 void build(int rt,int l,int r) 17 { 18 if (l==r) 19 { 20 c[rt]=0; 21 return; 22 } 23 int mid=(l+r)/2; 24 build(rt*2,l,mid); 25 build(rt*2+1,mid+1,r); 26 c[rt]=c[rt*2]&c[rt*2+1]; 27 } 28 void pushdown(int rt) 29 { 30 if (c[rt]) 31 { 32 c[rt*2]=c[rt]; 33 c[rt*2+1]=c[rt]; 34 } 35 } 36 void update(int rt,int l,int r,int L,int R) 37 { 38 if (l>=L&&r<=R) 39 { 40 c[rt]=1; 41 return; 42 } 43 int mid=(l+r)/2; 44 pushdown(rt); 45 if (L<=mid) update(rt*2,l,mid,L,R); 46 if (R>mid) update(rt*2+1,mid+1,r,L,R); 47 c[rt]=c[rt*2]&c[rt*2+1]; 48 } 49 int query(int rt,int l,int r,int L,int R) 50 { 51 if (c[rt]) return 1; 52 if (l>=L&&r<=R) 53 { 54 return c[rt]; 55 } 56 int mid=(l+r)/2; 57 int ll=1,rr=1; 58 if (L<=mid) ll=query(rt*2,l,mid,L,R); 59 if (R>mid) rr=query(rt*2+1,mid+1,r,L,R); 60 c[rt]=c[rt*2]&c[rt*2+1]; 61 return ll&rr; 62 } 63 bool check(int mid) 64 {int i,j,l1,l2,r1,r2,k; 65 for (i=1;i<=mid;i++) 66 b[i]=a[i]; 67 build(1,1,n); 68 sort(b+1,b+mid+1,cmp); 69 for (i=1;i<=mid;i=j) 70 { 71 j=i; 72 while (j<=mid&&b[j].x==b[i].x) j++; 73 l1=2e9;r2=2e9;l2=-1;r1=-1; 74 for (k=i;k<j;k++) 75 { 76 l1=min(l1,b[k].l); 77 r1=max(r1,b[k].r); 78 l2=max(l2,b[k].l); 79 r2=min(r2,b[k].r); 80 } 81 if (l2>r2) return 0; 82 if (query(1,1,n,l2,r2)) return 0; 83 update(1,1,n,l1,r1); 84 } 85 return 1; 86 } 87 int main() 88 {int i; 89 cin>>n>>q; 90 for (i=1;i<=q;i++) 91 { 92 scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].x); 93 } 94 int l=1,r=q,ans=0; 95 while (l<=r) 96 { 97 int mid=(l+r)/2; 98 if (check(mid)) 99 { 100 ans=mid; 101 l=mid+1; 102 } 103 else r=mid-1; 104 } 105 cout<<(ans+1)%(q+1); 106 }