HDU 5821 Ball 贪心+映射

题目链接：HDU5821

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 713    Accepted Submission(s): 424

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n个盒子，每个盒子只能放一个球，球的种类用1～n表示，0为没有球。给出m次操作，每次把l到r的所有球取出来再随机放回去，问有没有可能转化为给定的状态。

题目分析：将2个状态a和b建立映射的关系，每个球建立与之最近的相同的球关系，注意关系必须1对1，我们用mp数组来记录这个关系，比如数组ab分别为：1 2 0 2 1 和1 0 1 2 2，则mp数组为0，3，1，5，2。然后再根据每次操作的l和r把mp数组的l r范围内排序，最后排序得到的结果mp[i]==i都成立则可以转换，否则不可以。

贪心要素在于每次操作都将mp数组排序，仔细想一下也没什么问题，原因在于每次都将各种元素尽可能的使数组有序，有序即为每个数放到其对应的位置。

////  main.cpp//  Ball////  Created by teddywang on 2016/8/12.//  Copyright © 2016年 teddywang. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[1005],b[1005];int n,m,vis[1005];int mp[1005];struct node {    int l,r;}q[1006];int main(){    int T;    scanf("%d\n",&T);    while(T--)    {        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<m;i++)        {            scanf("%d%d",&q[i].l,&q[i].r);        }        memset(vis,0,sizeof(vis));        memset(mp,-1,sizeof(mp));        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                if(a[i]==b[j]&&!vis[j])                {                    mp[i]=j;                    vis[j]=1;                    break;                }            }        }        for(int i=0;i<m;i++)        {            sort(mp+q[i].l-1,mp+q[i].r);        }        int flag=0;        for(int i=0;i<n;i++)        {            if(mp[i]!=i)            {                flag=1;                printf("No\n");                break;            }        }        if(flag==0) printf("Yes\n");    }}