题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5821点击打开链接
Ball
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1378 Accepted Submission(s): 781
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
将b数组看做排序后的结果 然后每次操作就是对区间排序
如此一来便可以将a数组通过下标转化成乱序的数组 对每次操作进行桶排序 最后检验是否一致即可
注意构造下标时因为不会出现交叉情况 因此重复数字对其没有影响
这里因为数的范围时0~n 会少了离散操作
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <set>
using namespace std;
int a[2222];
int b[2222];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
vector<int > pos[n+1];
int point[n+1];
int num[n+1];
int nmid=n+1;
for(int i=0;i<=n;i++)
point[i]=0;
for(int i=0;i<n;i++)
{
pos[b[i]].push_back(i);
}
for(int i=0;i<=n;i++)
num[i]=pos[i].size();
for(int i=0;i<n;i++)
{
if(point[a[i]]<num[a[i]])
{
int mid=a[i];
a[i]=pos[a[i]][point[a[i]]];
point[mid]++;
}
else
a[i]=nmid++;
}
for(int i=0;i<m;i++)
{
int l,r;
scanf("%d%d",&l,&r);
int tong[2222];
for(int i=0;i<2222;i++)
tong[i]=0;
for(int i=l-1;i<=r-1;i++)
{
tong[a[i]]++;
}
int cnt=0;
while(!tong[cnt])
cnt++;
for(int i=l-1;i<=r-1;i++)
{
a[i]=cnt;
tong[cnt]--;
while(!tong[cnt])
cnt++;
}
}
int flag=0;
for(int i=0;i<n;i++)
if(a[i]!=i)
{
flag=1;
}
if(flag)
cout << "No" << endl;
else
cout << "Yes" << endl;
}
}
