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
一開始以為是個數學題，沒想到是個圖論。。
對于這種各個元素之間有聯系的，一定要往圖論上想想，哪怕他不是。。
我們對于一開始可以移動的點，把它們放到一個集合中。處理完之后，判斷前后兩個位置上的點是否在一個集合中。
代碼如下：

#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)/~

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