hdu5821(2016多校第8场，暴力)

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 883    Accepted Submission(s): 290

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

5
4 1
0 0 1 1
0 1 1 1
1 4
4 1
0 0 1 1
0 0 2 2
1 4
4 2
1 0 0 0
0 0 0 1
1 3
3 4
4 2
1 0 0 0
0 0 0 1
3 4
1 3
5 2
1 1 2 2 0
2 2 1 1 0
1 3
2 4

 

Sample Output

No
No
Yes
No
Yes

思路：分别判断出每个球对应的位置，用nex数组存起来，然后针对每个球判断经过所有操作之后能否到达，如果能到达则是yes，否则no。

#include <iostream>
#include<string.h>
#include<vector>
#include<queue>
#include<algorithm>
#include<stdio.h>
#include<math.h>
#include<map>
#include<stdlib.h>
#include<time.h>
using namespace std;
typedef long long ll;
int l[1010],r[1010];
int jl[1010],nex[1010];
vector<int>a[1010],b[1010];
int A[1010],B[1010];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(jl,0,sizeof(jl));
        memset(nex,0,sizeof(nex));
        for(int i=0;i<=1005;i++)
        {a[i].clear();b[i].clear();}
        int n,m,flag=1;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&A[i]),jl[A[i]]++;
            a[A[i]].push_back(i);
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&B[i]),jl[B[i]]--;
            b[B[i]].push_back(i);
        }
        for(int i=0; i<=1005; i++)
            if(jl[i]!=0)
            {
                flag=0;
                break;
            }
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&l[i],&r[i]);
            if(l[i]>r[i])swap(l[i],r[i]);
        }
        if(!flag)puts("No");
        else
        {
            bool vis[1010]= {0};
            for(int i=1; i<=n; i++)
                if(!vis[A[i]])
                {
                    vis[A[i]]=1;
                    for(int j=0,l=a[A[i]].size(); j<l; j++)
                    {
                        nex[a[A[i]][j]]=b[A[i]][j];
                    }
                }
            int ff=1;
            for(int i=1; i<=n; i++)
            {
                ff=0;
                int now=i,to=nex[i],ml=i,mr=i;
                if(now==to)
                {
                    ff=1;
                    continue;
                }
                for(int j=0; j<m; j++)
                {
                    if(l[j]<=ml&&r[j]>=mr)
                        ml=l[j],mr=r[j];
                    else if(l[j]>=ml&&l[j]<=mr&&r[j]>mr)
                        mr=r[j];
                    else if(r[j]<=mr&&r[j]>=ml&&l[j]<ml)
                        ml=l[j];
                    if(to>=ml&&to<=mr)
                    {
                        ff=1;
                        break;
                    }
                }
                //cout<<i<<":"<<ml<<"->"<<mr<<endl;
                if(!ff)
                    break;
            }
            //cout<<endl;
            if(!ff)puts("No");
            else puts("Yes");
        }
    }
    return 0;
}