HDU 5821 Ball (贪心)
Ball
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1079 Accepted Submission(s): 648
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.
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.
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
题意：
有N个盒子，每个盒子最多装一个球. 球的颜色不一定相同.
现在要进行m次区间操作：
每次操作 [l, r] 后可以随意将区间内的球重新分配回去.
问经过上述操作后是否有可能达到给定的状态.
题解：
贪心.
为每个球标记它在最终结果中的序号. 对于颜色相同的球：左边的尽量分配小的序号.
对于m次区间操作，就将区间[l,r]中的球按最终序号排序.
每次排序都相当于让区间中的球向它们的最终位置更近一步.
最终再比较是否每个球都到位即可.
#include<stdio.h> #include<iostream> #include<string.h> #include<cmath> #include<algorithm> using namespace std; const int maxn=1111; int t; int n,m; int flag; struct node { int next; int color; }a[maxn]; int cmp(node a,node b) { return a.next<b.next; } int l,r; int b[maxn]; int main() { scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=0;i<n;i++) { scanf("%d",&a[i].color); a[i].next=-1; } for(int i=0;i<n;i++) { scanf("%d",&b[i]); for(int j=0;j<n;j++) { if(b[i]==a[j].color&&a[j].next==-1) { a[j].next=i; break; } } } for(int i=0;i<m;i++) { scanf("%d%d",&l,&r); sort(a+l-1,a+r,cmp); } flag=1; for(int i=0;i<n;i++) { if(a[i].color!=b[i]) { flag=0; break; } } if(flag)printf("Yes\n"); else printf("No\n"); } return 0; }
