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
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define ull unsigned long long 5 #define mst(a,b) memset((a),(b),sizeof(a)) 6 #define mp(a,b) make_pair(a,b) 7 #define pi acos(-1) 8 #define pii pair<int,int> 9 #define pb push_back 10 const int INF = 0x3f3f3f3f; 11 const double eps = 1e-6; 12 const int MAXN = 15e4 + 10; 13 const int MAXM = 2e6 + 10; 14 15 int n,k; 16 int dx[4] = {-1,0,0,1}; 17 int dy[4] = {0,-1,1,0}; 18 map<pii,bool>ma; 19 map<pii,int>sub; 20 21 bool judge(int x,int y) { 22 if(x < 0 || x >= n || y < 0 || y >= n) return false; 23 return true; 24 } 25 26 int check(int x,int y) { 27 if(x == 0 || x == n - 1) { 28 if(y == 0 || y == n - 1) return 3; 29 else return 4; 30 } else { 31 if(y == 0 || y == n - 1) return 4; 32 else return 5; 33 } 34 } 35 36 int main() { 37 #ifdef local 38 freopen("data.txt", "r", stdin); 39 #endif 40 int cas = 1; 41 int t; 42 scanf("%d",&t); 43 while(t--) { 44 ma.clear(); 45 sub.clear(); 46 scanf("%d%d",&n,&k); 47 if(n == 1) { 48 printf("Case #%d: 1/1 ",cas++); 49 continue; 50 } 51 ll ans1 = 16ll * (n - 2) + 5ll * (n - 2) * (n - 2) + 12; 52 ll ans2 = 5ll * (1ll * n * (n + 1) / 2 - 2ll * (n - 2) - 3) + 8ll * (n - 2) + 9; 53 while(k--) { 54 int x,y; 55 scanf("%d%d",&x,&y); 56 if(ma[mp(x,y)]) { 57 ans1 -= sub[mp(x,y)]; 58 if(x + y >= n - 1) ans2 -= sub[mp(x,y)]; 59 sub[mp(x,y)] = 0; 60 } else { 61 ma[mp(x,y)] = true; 62 ans1 -= check(x,y); 63 if(x + y >= n - 1) ans2 -= check(x,y); 64 sub[mp(x,y)] = 0; 65 } 66 for(int i = 0; i < 4; i++) { 67 int nx = x + dx[i], ny = y + dy[i]; 68 if(!judge(nx,ny)) continue; 69 if(ma[mp(nx,ny)]) { 70 if(sub[mp(nx,ny)]) { 71 ans1--; 72 if(nx + ny >= n - 1) ans2--; 73 sub[mp(nx,ny)]--; 74 } 75 } else { 76 ma[mp(nx,ny)] = true; 77 sub[mp(nx,ny)] = check(nx,ny) - 1; 78 ans1--; 79 if(nx + ny >= n - 1) ans2--; 80 } 81 } 82 } 83 ll gcd = __gcd(ans1,ans2); 84 ans1 /= gcd, ans2 /= gcd; 85 printf("Case #%d: %lld/%lld ",cas++,ans2,ans1); 86 } 87 return 0; 88 }