2016多校8 hdu5821 Ball 智商题

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 735    Accepted Submission(s): 432

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

这种题怎么分类。。。完全是看人脑子转不转得过来啊。。。

在在b［］中寻找a［］中的球的位置，将这个位置信息标志给a［］。

比如a｛1，0，1，0，1｝，b｛1，1，1，0，0｝这样的，经过处理后应该变为｛1，4，2，5，3｝

每次题中给的区间其实就是对这个区间排序，若要想让a变成b的样子，那么经过题中排序后a数组中应该是顺序排列的。。。

奶奶的，谁想出来的

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;
const int maxn = 1005;
int a[maxn];
int b[maxn];
int main(){
    int t,n,m,i,j;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        for(i=1;i<=n;i++){
            scanf("%d",&b[i]);
        }
        bool flag=true;
        for(i=1;i<=n;i++){
            bool find=false;
            for(j=1;j<=n;j++){
                if(a[i]==b[j]){
                    a[i]=j;
                    b[j]=-1;
                    find=true;
                    break;
                }
            }
            if(!find){
                flag=false;
                break;
            }
        }

        int l,r;
        for(i=1;i<=m;i++){
            scanf("%d%d",&l,&r);
            sort(a+min(l,r), a+max(r,l)+1);
        }
        if(!flag){
            printf("No\n");
            continue;
        }
        flag=true;
        for(i=1;i<=n;i++){
            if(a[i]!=i){
                flag=false;
                break;
            }
        }
        if(flag){
            printf("Yes\n");
        }else{
            printf("No\n");
        }
    }
    return 0;
}