hdu 5821 Ball 思维题

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 461    Accepted Submission(s): 273

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

 

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 题意：两个数组，其中一个数组是当前的，另外一个是需要变换到的，你有m个按次序的操作，每次操作可以把[l,r]范围内的所有数任意排列，判断是否可以通过依次进行这些操作将当前数组变换到目标数组。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int  inf =0x7f7f7f7f;
const double pi=acos(-1);
const int N=1005;

struct node{
  int x,y;
}a[N];

bool cmp(node a,node b)
{
    return a.y<b.y;
}

int b[N];
int main()
{
    int cas,n,m;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) {scanf("%d",&a[i].x);a[i].y=-1;}
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&b[j]);
            for(int i=1;i<=n;i++)
                if(a[i].x==b[j]&&a[i].y==-1)
                   {
                       a[i].y=j;
                       break;
                   }
        }
        while(m--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            sort(a+l,a+r+1,cmp);
        }
        bool flag=true;
        for(int i=1;i<=n;i++)
            if(a[i].y!=i) {flag=false;break;}
        if(flag) printf("Yes
");
        else printf("No
");
    }
    return 0;
}

　　分析：比赛时感觉这道题跟以前做过的猪圈的那道最大流题目很像，，，后来发现还是不能那样做。

其实对于当前数组中的每种数值，找到目标数组中与之相同的数值，然后按着相对位置，给当前数组标上

目标数组中相对位置一样的数所在的实际位置就好了，，是一种贪心吧