Problem Description

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.

Input

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

Output

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.

Sample Input

20 4

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

Sample Output

3

題意：給一段長度為n，每個位置上的數都不同的序列a[1..n]和問答(xi, yi, ri)代表RMQ(a, x, y)=r，求給出最早的有矛盾的那個問答的編號。

思路：

有矛盾的情況有兩種，一是之前已更新區間最小值為x，又要更新此區間或他的子區間的最小值為更小的數；二是兩段區間的最小值相同，但沒有交集。

先用并查集合并再查找有矛盾的情況，由于要求最小的序號，采用二分查找，從1枚舉到m。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 10000001 #define MOD 123 #define E 1e-6 using namespace std; struct Node {int x;int y;int minn; }a[N],b[N]; int father[N]; int ans; int n,m; int read() {int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;for(;isdigit(ch);ch=getchar())x=(x*2)+(x*8)+ch-'0';return x*f; } int Find(int x) {if(father[x]==x)return x;elsereturn father[x]=Find(father[x]); } bool cmp(Node a,Node b) {return a.minn>b.minn; } bool check(int pos) {for(int i=1;i<=n+1;i++)father[i]=i;for(int i=1;i<=pos;i++)b[i]=a[i];sort(b+1,b+pos+1,cmp);int lmin=b[1].x,lmax=b[1].x,rmin=b[1].y,rmax=b[1].y;for(int i=2;i<=pos;i++){if(b[i].minn<b[i-1].minn)//情況一{if(Find(lmax)>rmin)return true;for(int j=lmin;j<=rmax;j++){j=Find(j);father[j]=Find(rmax+1);}lmax=lmin=b[i].x;rmax=rmin=b[i].y;}else if(b[i].minn==b[i-1].minn)//情況二{lmax=max(lmax,b[i].x);lmin=min(lmin,b[i].x);rmax=max(rmax,b[i].y);rmin=min(rmin,b[i].y);if(lmax>rmin)return true;}}if(Find(lmax)>rmin)return true;elsereturn false; } int main() {n=read();m=read();for(int i=1;i<=m;i++){a[i].x=read();a[i].y=read();a[i].minn=read();}int l=1,r=m,mid;while(r-l>1){int mid=(l+r)/2;if(check(mid)){ans=mid;r=mid;}elsel=mid;}printf("%d\n",ans);return 0; }

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