uva 10763 Foreign Exchange

题目:
　　A到B地，要有B到A地才行得通
分析：
　　初始化 a[i] = i，然後讀入x,y,交換a[x],a[y]。最後判斷是否a[i]==i

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>


using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)

/******** 程序部份 ********************/

const int MAXN = 500005;

int a[MAXN],n;

int main(){

#ifndef ONLINE_JUDGE
	freopen("sum.in","r",stdin);
	//freopen("sum.out","w",stdout);
#endif

    while(RD(n),n){
        rep1(i,500000)
            a[i] = i;
        int x,y;
        rep1(i,n){
            RD2(x,y);
            swap(a[x],a[y]);
        }
        bool ok = true;
        rep1(i,500000)
            if(a[i]!=i){
                ok = false;
                break;
            }
        ok?puts("YES"):puts("NO");
    }

	return 0;
}