HDU - 6229
题意：
在一个n*n的地图中，有一个初始在（0，0）位子的机器人，每次等概率的向相邻的格子移动或者留在原地。问最后留在格子（x，y）（x+y>=n-1)的地方的概率。
思路：
这道题由于每个格子的贡献是不同的，在四个角格子的贡献是3分（留下来，两个边来的），中间的5分，有一条边与边相连的4分。如果这个点是障碍物，则把这个点的贡献抹为0，再把其四周的格子贡献-1.
由于开不下数组，可以用map
// #pragma GCC optimize(3) // #pragma comment(linker, "/STACK:102400000,102400000") //c++ // #pragma GCC diagnostic error "-std=c++11" // #pragma comment(linker, "/stack:200000000") // #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> #include <map>using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queuetypedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3;//priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n'#define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que;const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x7f7f7f7f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e8+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601;template<typename T> inline T read(T&x){x=0;int f=0;char ch=getchar();while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x=f?-x:x; }/*-----------------------showtime----------------------*/map<pii,int>mp;int nx[4][4] = {{1,0}, {0,1} ,{-1,0},{0,-1}};int T,n,k;int get(int x,int y){if(x>0&&x<n-1 && y>0 && y<n-1)return 5;if(x==0&&y==0)return 3;if(x==n-1&&y==0)return 3;if(x==0&&y==n-1)return 3;if(x==n-1&&y==n-1)return 3;return 4;}ll gcd(ll a, ll b){if(b == 0)return a;return gcd(b, a % b);} int main(){scanf("%d", &T);for(int tt=1; tt<=T; tt++){mp.clear();scanf("%d%d", &n, &k);if(n == 1){printf("Case #%d: 1/1\n", tt);continue;}ll sum = 4*3+(n-2)*4*4+(n-2)*(n-2)*5;ll dec = 0;ll up = 3*3+(n-2)*2*4+(n-1)*(n-2)/2*5;for(int i=1; i<=k; i++){int x,y;scanf("%d%d", &x, &y);for(int i=0; i<4; i++){int tx = x + nx[i][0];int ty = y + nx[i][1];if(tx <0 || tx >= n||ty<0||ty>=n)continue;if(mp.count(pii(tx,ty))){if(mp[pii(tx,ty)] == 0)continue;mp[pii(tx,ty)] --;if(tx + ty >= n-1)up--;sum--;}else {mp[pii(tx,ty)] = get(tx,ty) - 1;if(tx + ty >= n-1)up--;sum--;}}if(mp.count(pii(x,y))) {if(mp[pii(x,y)] == 0)continue;if(x + y >= n-1)up -= mp[pii(x,y)];sum -= mp[pii(x,y)];mp[pii(x,y)] = 0;}else{mp[pii(x,y)] = 0;if(x + y >= n-1)up -= get(x,y);sum -= get(x,y);}}ll gg = gcd(up, sum);printf("Case #%d: %lld/%lld\n", tt, up/gg, sum/gg);// for(int i=0; i<n; i++){// for(int j=0; j<n; j++){// if(mp.count(pii(i,j)))// cout<<i<<" , "<<j<<"="<<mp[pii(i,j)]<<endl;// }// cout<<endl;// } }return 0; }
