HDU 5821 Ball【贪心】

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1083    Accepted Submission(s): 651

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
  

 

Author

学军中学

 

Source

2016 Multi-University Training Contest 8

 

贪心：
为每个球标记它在最终结果中的序号. 对于颜色相同的球：左边的尽量分配小的序号.
对于m次区间操作，就将区间[l,r]中的球按最终序号排序.
每次排序都相当于让区间中的球向它们的最终位置更近一步.
最终再比较是否每个球都到位。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b)  memset(a,b,sizeof(a))
const int M=1e6+10;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int i,j,k,n,m;
int a[M];
int b[M];
int flag[M];
bool vis[M];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)scanf("%d",&b[i]);
        ms(vis,0);
        ms(flag,0);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(!vis[j]){
                    if(a[i]==b[j]){
                        vis[j]=1;
                        flag[i]=j;
                        break;
                    }
                }
            }
        }
        while(m--){
            int l,r;
            scanf("%d%d",&l,&r);
            sort(flag+l,flag+r+1);
        }
        bool ans=0;
        for(int i=1;i<=n;i++){
            if(flag[i]!=i){
                ans=1;
                break;
            }
        }
        if(ans)printf("No\n");
        else printf("Yes\n");
    }
    return 0;
}