ACM-ICPC 2017 沈阳赛区现场赛 M. Wandering Robots && HDU 6229（思维+期望）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6229

参考题解：https://blog.csdn.net/lifelikes/article/details/78452558

　　　　   https://www.cnblogs.com/cxhscst2/p/8215717.html

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define ull unsigned long long
 #define mst(a,b) memset((a),(b),sizeof(a))
 #define mp(a,b) make_pair(a,b)
 #define pi acos(-1)
 #define pii pair<int,int>
 #define pb push_back
 const int INF = 0x3f3f3f3f;
 const double eps = 1e-;
 const int MAXN = 15e4 + ;
 const int MAXM = 2e6 + ;
 
 int n,k;
 int dx[] = {-,,,};
 int dy[] = {,-,,};
 map<pii,bool>ma;
 map<pii,int>sub;
 
 bool judge(int x,int y) {
     if(x <  || x >= n || y <  || y >= n) return false;
     return true;
 }
 
 int check(int x,int y) {
     if(x ==  || x == n - ) {
         if(y ==  || y == n - ) return ;
         else return ;
     } else {
         if(y ==  || y == n - ) return ;
         else return ;
     }
 }
 
 int main() {
 #ifdef local
     freopen("data.txt", "r", stdin);
 #endif
     int cas = ;
     int t;
     scanf("%d",&t);
     while(t--) {
         ma.clear();
         sub.clear();
         scanf("%d%d",&n,&k);
         if(n == ) {
             printf("Case #%d: 1/1\n",cas++);
             continue;
         }
         ll ans1 = 16ll * (n - ) + 5ll * (n - ) * (n - ) + ;
         ll ans2 = 5ll * (1ll * n * (n + ) /  - 2ll * (n - ) - ) + 8ll * (n - ) + ;
         while(k--) {
             int x,y;
             scanf("%d%d",&x,&y);
             if(ma[mp(x,y)]) {
                 ans1 -= sub[mp(x,y)];
                 if(x + y >= n - ) ans2 -= sub[mp(x,y)];
                 sub[mp(x,y)] = ;
             } else {
                 ma[mp(x,y)] = true;
                 ans1 -= check(x,y);
                 if(x + y >= n - ) ans2 -= check(x,y);
                 sub[mp(x,y)] = ;
             }
             for(int i = ; i < ; i++) {
                 int nx = x + dx[i], ny = y + dy[i];
                 if(!judge(nx,ny)) continue;
                 if(ma[mp(nx,ny)]) {
                     if(sub[mp(nx,ny)]) {
                         ans1--;
                         if(nx + ny >= n - ) ans2--;
                         sub[mp(nx,ny)]--;
                     }
                 } else {
                     ma[mp(nx,ny)] = true;
                     sub[mp(nx,ny)] = check(nx,ny) - ;
                     ans1--;
                     if(nx + ny >= n - ) ans2--;
                 }
             }
         }
         ll gcd = __gcd(ans1,ans2);
         ans1 /= gcd, ans2 /= gcd;
         printf("Case #%d: %lld/%lld\n",cas++,ans2,ans1);
     }
     return ;
 }