专题测试四 图论 A - pSort

1. 题目
   

   One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = d_i, where d_i is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.

   The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.

   Input

   The first line contains positive integer n (1 ≤ n ≤ 100) — the number of cells in the array. The second line contains n distinct integers from 1 to n — permutation. The last line contains n integers from 1 to n — favourite numbers of the cells.

   Output

   If the given state is reachable in the described game, output YES, otherwise NO.

   Sample 1

   Input	Output
   5 5 4 3 2 1 1 1 1 1 1	YES

   Sample 2

   Input	Output
   7 4 3 5 1 2 7 6 4 6 6 1 6 6 1	NO

   Sample 3

   Input	Output
   7 4 2 5 1 3 7 6 4 6 6 1 6 6 1	YES

2. 思路
   写之前先吐槽下，这个星期的题都是些什么高达
   元素可以换到自己最喜欢的距离上，意味着这个元素可以与自己距离d的元素任意交换
   对于任意两个元素a和b，如果a能和c交换，b也能和c交换，那么经由c进行中转，a、b也能交换，而且可以不改变c处元素的值
   所以任意两个可以交换的元素可以视为在一个连通块中，在一个连通块中的元素可以任意交换而不产生其他影响
   用并查集判断每个元素当前位置和目标位置在不在一个连通块中，都满足则YES，不满足则NO
3. 代码
   

   #include<cstdio>
   #include<cstring>
   #include<algorithm>
   #include<cmath>
   
   int fa[105],p[105],love[105];
   
   int find(int x)
   {
   	if(fa[x]!=x)
   		fa[x] = find(fa[x]);
   	return fa[x];
   }
   
   void build(int x,int y)
   {
   	int a=find(x);
   	int b=find(y);
   	fa[a] = b;
   }
   
   int main()
   {
   	int n,q;
   	long long ans;
   	scanf("%d",&n);
   	for(int i=1;i<=n;i++)
   		fa[i] = i;
   	for(int i=1;i<=n;i++)
   	{
   		scanf("%d",&q);
   		p[q]=i;
   	}
   	for(int i=1;i<=n;i++)
   	{
   		scanf("%d",&love[i]);
   	}
   	for(int i=1;i<=n;i++)
   	{
   		if(i-love[i]>=1)
   		build(i,i-love[i]);
   		if(i+love[i]<=n)
   		build(i,i+love[i]);
   	}
   	int f=1;
       for(int i=1;i<=n;i++)
       {
   		if(find(p[i])!=find(i))
   		{
   			f=0;
   			break;
   		}
       }
   	if(f==1)
   	{
   		printf("YES\n");
   	}
   	else
   	{
   		printf("NO\n");
   	}
   	return 0;
   }