Hdu-5821 Ball(贪心)

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

这题开始错误贪心卡了我好久，最后还是队友A了过去，智商不够。

官方题解：假设有4个红球，初始时从左到右标为1,2,3,4。那么肯定存在一种方案，使得最后结束时红球的顺序没有改变，也是1,2,3,4。 那么就可以把同色球都写成若干个不同色球了。所以现在共有n个颜色互异的球。按照最终情况标上1,2，。。，n的序号，那么贪心的来每次操作就是把一个区间排序就行了。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MAXN 1005
using namespace std;
int t,n,m,a[MAXN],b[MAXN],pre[MAXN];
vector<int> color[MAXN];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		memset(pre,0,sizeof(pre));
		scanf("%d%d",&n,&m);
		for(int i = 0;i <= n;i++) color[i].clear();
		for(int i = 1;i <= n;i++) scanf("%d",&a[i]);
		for(int i = 1;i <= n;i++) 
		{
			scanf("%d",&b[i]);
			color[b[i]].push_back(i);
		}
		for(int i = 1;i <= n;i++)
         a[i] = color[a[i]][pre[a[i]]++];
        for(int i = 1;i <= m;i++)
        {
        	int l,r;
        	scanf("%d%d",&l,&r);
        	sort(a+l,a+1+r);
		}
		bool flag = true;
		for(int i = 1;i <= n;i++)
		 if(a[i] != i) 
		 {
		 	flag = false;
		 	break;
		 }
		if(flag) cout<<"Yes"<<endl;
		else cout<<"No"<<endl; 
	}
 }