Ball

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.

 

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.

 

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

 

 

Author

学军中学

Source

2016 Multi-University Training Contest 8

 

题意：

有N个盒子，每个盒子最多装一个球. 球的颜色不一定相同.
现在要进行m次区间操作：
每次操作 [l, r] 后可以随意将区间内的球重新分配回去.
问经过上述操作后是否有可能达到给定的状态.

题解：

贪心.
为每个球标记它在最终结果中的序号. 对于颜色相同的球：左边的尽量分配小的序号.
对于m次区间操作，就将区间[l,r]中的球按最终序号排序.
每次排序都相当于让区间中的球向它们的最终位置更近一步.
最终再比较是否每个球都到位即可.

 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1111;
int t;
int n,m;
int flag;
struct node
{
    int next;
    int color;
}a[maxn];
int cmp(node a,node b)
{
    return a.next<b.next;
}
int l,r;
int b[maxn];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i].color);
            a[i].next=-1;
        }
        for(int i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
            for(int j=0;j<n;j++)
            {
                if(b[i]==a[j].color&&a[j].next==-1)
                {
                    a[j].next=i;
                    break;
                }
            }
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&l,&r);
            sort(a+l-1,a+r,cmp);
        }
        flag=1;
        for(int i=0;i<n;i++)
        {
            if(a[i].color!=b[i])
            {
                flag=0;
                break;
            }
        }
        if(flag)printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}