浙江财经大学
信息管理与工程学院

HDU 5821 Ball

本文由 Ocrosoft 于 2016-08-15 21:53:49 发表

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 854    Accepted Submission(s): 502

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
 

Author
学军中学
 

Solution
题意:有n个球,给定一种初始位置和一种最终位置,给m次操作,每次操作给出L,R,可以对[L,R]的球进行任意排序。问经过m种操作后,能不能到达最终状态。

思路:读取B数组的时候同时对a数组进行处理,a数组每个数记录下最终要到达的下标,每次操作对a数组按照要到达的下标从小到大排序,这样可以使每一个球都朝着目标前进。如果操作结束后,还是不能到达目标状态,那么就是不能达到。

官方解释:假设有4个红球,初始时从左到右标为1,2,3,4。那么肯定存在一种方案,使得最后结束时红球的顺序没有改变,也是1,2,3,4。 那么就可以把同色球都写成若干个不同色球了。所以现在共有n个颜色互异的球。按照最终情况标上1,2,。。,n的序号,那么贪心的来每次操作就是把一个区间排序就行了。(真暴力)

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
typedef long long ll;
const int INF = INT_MAX;
const int MAXN = 100000 + 10;
using namespace std;
struct node
{
	int num, dir;
	bool operator < (const node b)const
	{
		return dir < b.dir;
	}
}t;
int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		vector<node> v;
		vector<int> b;
		int n, m;
		cin >> n >> m;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &t.num); t.dir = -1;
			v.push_back(t);
		}
		for (int i = 0; i < n; i++)
		{
			int t; scanf("%d", &t);
			b.push_back(t);
			for (int j = 0; j < n; j++)
			{
				if (b[i] == v[j].num&&v[j].dir == -1)
				{
					v[j].dir = i; break;
				}
			}
		}
		for (int i = 0; i < m; i++)
		{
			int l, r; scanf("%d%d", &l, &r);
			l--;//sort的参数2是要加1的,所以r就不减了。
			sort(v.begin() + l, v.begin() + r);
		}
		bool f = 1;
		for (int i = 0; i < n; i++)
			if (v[i].num != b[i])
			{
				f = 0; break;
			}
		if (f)printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}
 

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