pSort CodeForces - 28B（并查集）

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| = di, where di 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.

Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
一开始以为是个数学题，没想到是个图论。。
对于这种各个元素之间有联系的，一定要往图论上想想，哪怕他不是。。
我们对于一开始可以移动的点，把它们放到一个集合中。处理完之后，判断前后两个位置上的点是否在一个集合中。
代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;const int maxx=1e2+100;
int f[maxx];
int a[maxx];
int n;inline int getf(int u)
{if(u==f[u]) return u;else return f[u]=getf(f[u]);
}
inline void merge(int u,int v)
{int t1=getf(u);int t2=getf(v);if(t1!=t2){f[t1]=t2;}
}
int main()
{scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]),f[i]=i;int x;for(int i=1;i<=n;i++){scanf("%d",&x);if(i-x>=1) merge(i,i-x);if(i+x<=n) merge(i,i+x);}int flag=0;for(int i=1;i<=n;i++){if(getf(a[i])!=getf(i)) {flag=1;break;}}if(flag) puts("NO");else puts("YES");
}

努力加油a啊，(^o)/~

pSort CodeForces - 28B（并查集）相关推荐

A - Cthulhu CodeForces - 103B (并查集)
-Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for th ...
Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) C. String Reconstruction 并查集
C. String Reconstruction 题目连接: http://codeforces.com/contest/828/problem/C Description Ivan had stri ...
codeforces 400D Dima and Bacteria 并查集+floyd
题目链接:http://codeforces.com/problemset/problem/400/D 题目大意: 给定n个集合,m步操作,k个种类的细菌, 第二行给出k个数表示连续的xi个数属于i集 ...
Codeforces Round #375 (Div. 2) D. Lakes in Berland 并查集
http://codeforces.com/contest/723/problem/D 这题是只能把小河填了,题目那里有写,其实如果读懂题这题是挺简单的,预处理出每一块的大小,排好序,从小到大填就行了 ...
Secret Passwords CodeForces - 1263D（并查集）
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems fro ...
Codeforces Global Round 14 F. Phoenix and Earthquake 思维 + 并查集
传送门文章目录题意: 思路: 题意: 给你nnn个点,mmm条边,限制xxx,每个点都有沥青aia_iai,定义合并两个点即两点之间有边且au+av≥xa_u+a_v\ge xau+av≥x ...
Codeforces 699D Fix a Tree 并查集
原题:http://codeforces.com/contest/699/problem/D 题目中所描述的从属关系,可以看作是一个一个块,可以用并查集来维护这个森林.这些从属关系中会有两种环,第一种 ...
Codeforces Round #548 (Div. 2) C. Edgy Trees（dfs || 并查集)
题目链接:https://codeforces.com/contest/1139/problem/C 题意:给了一棵树,n个点,m条边.让从中选k个点,使得从a1到a2,a2到a3,ak-1到ak的路 ...
C. Edgy Trees---（思维题+并查集的运用）---Codeforces Round #548 (Div. 2)
Edgy Trees time limit per test 2 seconds memory limit per test 256 megabytes 题目链接http://codeforces.c ...

pSort CodeForces - 28B（并查集）

pSort CodeForces - 28B（并查集）相关推荐

最新文章

热门文章