HDU-5821 多校训练第8场-1001（巧妙模拟）

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 306    Accepted Submission(s): 176

Problem Description

ZZX has a sequence of boxes numbered 1,2,...,n. Each box can contain at most one ball.

You are given the initial configuration of the balls. For 1≤i≤n, if the i-th box is empty then a[i]=0, otherwise the i-th box contains exactly one ball, the color of which is a[i], a positive integer. Balls with the same color cannot be distinguished.

He will perform m operations in order. At the i-th operation, he collects all the balls from boxes l[i],l[i]+1,...,r[i]-1,r[i], and then arbitrarily put them back to these boxes. (Note that each box should always contain at most one ball)

He wants to change the configuration of the balls from a[1..n] to b[1..n] (given in the same format as a[1..n]), using these operations. Please tell him whether it is possible to achieve his goal.

 

Input

First line contains an integer t. Then t testcases follow.
In each testcase: First line contains two integers n and m. Second line contains a[1],a[2],...,a[n]. Third line contains b[1],b[2],...,b[n]. Each of the next m lines contains two integers l[i],r[i].

1<=n<=1000,0<=m<=1000, sum of n over all testcases <=2000, sum of m over all testcases <=2000.

0<=a[i],b[i]<=n.

1<=l[i]<=r[i]<=n.

 

Output

For each testcase, print "Yes" or "No" in a line.

 

Sample Input

54 10 0 1 10 1 1 11 44 10 0 1 10 0 2 21 44 21 0 0 00 0 0 11 33 44 21 0 0 00 0 0 13 41 35 21 1 2 2 02 2 1 1 01 32 4

 

Sample Output

NoNoYesNoYes

 

题解：

编号 1  2  3  4  5                1   2   3    4   5

A       1  1  2  2  0     -- >B   2   2    1   1   0     把 A  变成 B序列嘛，就要把A1  送至 B3， A2送至B4 ，A3送至 B1 ，A4送至B2 ，A5送至 B5;

那么我们用数组  mp  表示这些" 送 "的关系  ，mp[1]=3   mp[2]=4   mp[3]=1   mp[4]=2  mp[5]=5  ,假如A B完全相同的话，那么mp[1]应该等于1，即mp[i]=i，所以我们进行[L - R ] 的操作时，就把mp[ L--R ]按照升序排序就好了，因为要使mp[ i ]=i ，那么他就是一个有序的数组啊

m次操作后，判断mp[i]是否等于 i ，然后就得出结果了，当然如果排序后颜色都不匹配肯定是No 啊

#include<stdio.h>#include<queue>#include<string.h>#include<math.h>#include<limits.h>#include <iostream>  #include <stack>  #include <string>  #include<algorithm>using namespace std;#define maxn 20005int ans[maxn];//ans数组表示一一对应关系 struct node{int x,y;//x表示颜色，y表示相应颜色小球的初始位置 }a[maxn],b[maxn]; bool comp(node p,node q){if(p.x==q.x)   return p.y<q.y;return p.x<q.x;}int  main(){int T,i,j,n,m,x,y;scanf("%d",&T);while(T--){int flag=0;scanf("%d%d",&n,&m);for(i=1;i<=n;i++){scanf("%d",&a[i].x);a[i].y=i; }  sort(a+1,a+n+1,comp);for(i=1;i<=n;i++){scanf("%d",&b[i].x);b[i].y=i;} sort(b+1,b+n+1,comp);for(i=1;i<=n;i++){if(a[i].x==b[i].x)  ans[a[i].y]=b[i].y;elseflag=1;}for(i=1;i<=m;i++){scanf("%d%d",&x,&y);sort(ans+x,ans+y+1);}for(i=1;i<=n;i++){if(ans[i]!=i){flag=1;break;}}if(flag)  printf("No\n");else  printf("Yes\n");}}