Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 775    Accepted Submission(s): 463 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 思路：在b中寻找a对应的第一个没有被标记过的位置，然后在区间(l,r)上根据位置从小到大排序。M次排序之后如果与b一致则Yes，否则No。

#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int MAX = 1005; struct ss { int num, id; ss() { id = 10010; } }a[MAX]; bool cmp(ss a, ss b) { return a.id < b.id; } int b[MAX], vis[MAX]; int T, N, M, l, r; int main() { scanf("%d", &T); while(T--) { memset(vis, 0, sizeof(vis)); scanf("%d %d", &N, &M); for(int i = 1;i <= N; ++i) scanf("%d", &a[i].num); for(int i = 1;i <= N; ++i) scanf("%d", &b[i]); for(int i = 1;i <= N; ++i) for(int j = 1;j <= N; ++j) { if(a[i].num == b[j] && !vis[j]) { a[i].id = j; vis[j] = 1; break; } } for(int i = 0;i < M; ++i) { scanf("%d %d", &l, &r); sort(a+l, a+r+1, cmp); } bool key = true; for(int i = 1;i <= N; ++i) { if(a[i].num != b[i]) { key = false; cout <<"No"<< endl; break; } } if(key) cout <<"Yes"<< endl; } return 0; }