HDU 5821 Ball 思路题

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Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1093    Accepted Submission(s): 655

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

题意：有两行数字，现在想把第一行数字变成第二行数字，现在有m个区间，第一行所给的区间内的任意两个数字可以交换位置，可以交换很多次，问经过m个区间交换之后，第一行数字是否能变成第二行数字？

思路：把数字都重新编号，先将第二行的数字编号为0，1，2，...  ，n-1；然后根据已知的两行数字，将第一行数字也进行编号，

以最后一组样例为例子：

第一行：1  1  2  2  0

编号：    2  3  0  1  4

第二行： 2  2  1  1  0（这里第一个2应该对应离他最近的2（贪心思想），也就是第一行第3个2了，有编号的不能再次编号）

编号：    0  1  2  3  4

然后进行m次操作，只需要sort排序（相对应的区间）即可

最后对比两行数字是否相同。

这个思路我没想到，看了其他人博客里的一句话（这题有很多数字是一样的，为了区别，需要把一样的数字变成不同的）

不好做啊，多校的题就是难想啊（明天第一场多校就要开始了，我还这么菜，怎么办，啊....）

#include<iostream>#include <cstdio>#include<algorithm>#include<cmath>#define N 1005using namespace std;int a[N],aa[N];int b[N],bb[N];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]);        for(int i=0; i<n; i++)            bb[i]=i;        int flag=0;        int j;        for(int i=0; i<n; i++)        {            for(j=0; j<n; j++)                if(b[i]==a[j])                {                    aa[j]=i;                    a[j]=-1;                    break;                }            if(j==n)  {flag=1;break;}        }        while(m--)        {            int l,r;            scanf("%d%d",&l,&r);            if(!flag)                sort(aa+l-1,aa+r);        }        if(!flag)            for(int i=0; i<n; i++)                if(aa[i]!=bb[i])                { flag=1; break;}        if(flag)  printf("No\n");        else  printf("Yes\n");    }}