hdu6853 Jogging

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6853

提示：头文件一定要多...不然编译不过doge

#include<iostream>
#include<fstream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<set>
#include<cstdio>
#include<sstream>
#include<cassert>
#include<bitset>
#include<unordered_map>
#include<climits>

#define ll long long
using namespace std;

int X[] = {0, -1, -1, 0, 1, 1, 1, 0, -1};
int Y[] = {0, 0, 1, 1, 1, 0, -1, -1, -1};
ll res1, res2;

void BFS(ll x0, ll y0)
{
    res1 = res2 = 0;
    set<pair<ll, ll> > vis;
    queue<pair<ll, ll> > q;

    q.emplace(x0, y0);
    vis.emplace(x0, y0);

    while(!q.empty()){
        ll nowx, nowy;
        tie(nowx, nowy) = q.front();//系起队首の元素
        res2++;//初始点已在其中，因而最后不必再加一
        q.pop();

        if(nowx == nowy){
            res1 = 0;
            res2 = 1;
            break;
        }
        for(int i = 1 ; i <= 8 ; i++){
            ll xi = nowx + X[i], yi = nowy + Y[i];

            if(__gcd(xi, yi) > 1){
                res2++;//总
                if(vis.count(make_pair(xi, yi)))   continue;
                vis.emplace(xi, yi);
                q.emplace(xi, yi);
            }
        }
        if(res1 == 0)   res1 = res2;//
    }

    ll gcd = __gcd(res1, res2);
    res1 /= gcd;
    res2 /= gcd;
    here:
    printf("%lld/%lld
",res1,res2);
    //起点の度数+1 / 总の度数+k //k表示连通块的个数（本题就一个连通块叭）
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll a, b;
        scanf("%lld %lld",&a,&b);
        BFS(a, b);
    }

    return 0;
}