2016多校8 hdu5821 Ball 智商题
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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. For1≤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.
You are given the initial configuration of the balls. For
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.
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
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;}
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