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;

}