[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

  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 }

 

posted @ 2017-10-27 16:31  Z-Y-Y-S  阅读(546)  评论(0)  编辑  收藏  举报