HDU5821 Ball ( 贪心）

http://acm.hdu.edu.cn/showproblem.php?pid=5821

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

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题意： 归纳概括就是，一个序列，随机取一个区间[l,r]。 然后将这个区间的数随机再排列组合。 问是否能得到第二个序列。

解题思路： 构造一个序列，每个数放的是第一个序列在第二个序列中的位置。如果有相同的数，那默认从左到右增加。对于每个区间操作，直接对该构造的序列排序，这样贪心的就离结果更近一步。最后判断该序列是不是1..n。

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#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 1100; int arr[N], brr[N], crr[N]; bool vis[N]; int main() { int T; scanf("%d",&T); while(T--) { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d",&arr[i]); for(int i=1;i<=n;i++) scanf("%d",&brr[i]); memset(vis,false,sizeof(vis)); memset(crr,0,sizeof(crr)); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(vis[j]) continue; if(brr[j]!=arr[i]) continue; crr[i] = j; vis[j] = true; break; } } while(m--) { int l,r; scanf("%d%d",&l,&r); sort(crr+l,crr+r+1); } bool flag = true; for(int i=1;i<=n;i++) { if(crr[i]!=i) { flag = false; break; } } puts(flag?"Yes":"No"); } return 0; }