HDU 5821 Ball（贪心）

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 138    Accepted Submission(s): 77

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

 

Author

学军中学

 

Source

2016 Multi-University Training Contest 8

 

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题目大意：
    有N个箱子，每个箱子有一个颜色为Ai的球，或没有球。给出M个区间，每次可以把这个区间的球重新排列，问经过这些排列后能不能形成序列B。

解题思路：
    比赛时一眼看过去感觉像网络流的最大流，然后霸占着电脑建了两个小时的图，最后自己证明了这题用网络流不可做(╯°□°）╯︵ ┻━┻
    正解是用贪心。我们可以观察发现，对于同一种颜色的球如果可以通过交换得到目标序列，那么第一个对应第一个，第二个对应第二个……一定是可行的。所以我们对B的每个球重新编号，让A中和它颜色一样的第一个球和它编号相同。这样就变成了N个不同的球，而且最终要求得到的是一个0～N－1的序列。所以，我们只要贪心地对每个区间排序，这样得到的一定是最接近目标序列的。最后我们判断，排序后的区间和目标序列是否相同，如果相同则输出“Yes”，否则输出“No”。

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <queue>using namespace std;const int maxn=1000+3;int N,M,A[maxn],B[maxn];bool vis[maxn];int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(vis,0,sizeof vis);        bool flag=true;        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]);            int j;            for(j=0;j<N;++j)                if(!vis[j]&&A[j]==B[i])                {                    A[j]=i;                    vis[j]=true;                    break;                }            if(j==N)//A中没有能够与其匹配的球                flag=false;            B[i]=i;        }        for(int i=0;i<M;++i)        {            int l,r;//区间左右端点            scanf("%d%d",&l,&r);            sort(A+l-1,A+r);        }        for(int i=0;i<N;++i)            if(A[i]!=B[i])//A经过交换不能变成B            {                flag=false;                break;            }        if(flag)            puts("Yes");        else puts("No");    }        return 0;}