Problem B. 即时战略 ———2019.10.12
题目：
代码~：感谢土蛋
1 #include <iostream> 2 #include <cstring> 3 #include <cmath> 4 #include <cstdio> 5 #include <algorithm> 6 #include <queue> 7 #include <vector> 8 #include <map> 9 #include <complex> 10 11 #define lc k << 1 12 #define rc k << 1 | 1 13 14 #define inf 0x3f3f3f3f 15 16 using namespace std; 17 typedef long long ll; 18 19 int a[805][805]; 20 int n, m, p; 21 22 ll mn = 1e13, mx = 0; 23 24 inline int dis(int x1, int y1, int x2, int y2){ 25 return min(abs(x2 - x1) + abs(y2 - y1), p); 26 } 27 28 void solve(int x){ 29 int y = 1; 30 ll lres = 0, rres = 0; 31 ll lsum = 0, rsum = 0; 32 for(int i = x - p; i <= x + p; i ++){ 33 for(int j = y - p; j <= y; j ++){ 34 if(dis(i, j, x, y) != p) lsum += a[i + 200][j + 200]; lres += a[i + 200][j + 200] * (p - dis(i, j, x, y)); 35 } 36 } 37 for(int i = x - p; i <= x + p; i ++){ 38 for(int j = y + 1; j <= y + p; j ++){ 39 if(dis(i, j, x, y) != p) rsum += a[i + 200][j + 200]; rres += a[i + 200][j + 200] * (p - dis(i, j, x, y)); 40 } 41 } 42 //printf("%d %d %lld %lld %lld %lld\n", x, y, lres, rres, lsum, rsum); 43 mn = min(lres + rres, mn); 44 mx = max(mx, lres + rres); 45 while(y < m){ 46 lres -= lsum; 47 for(int i = y - p + 1; i <= y; i ++){ 48 lsum -= a[x + (p - 1 - y + i) + 200][i + 200]; 49 if(p - 1 - y + i != 0) lsum -= a[x - (p - 1 - y + i) + 200][i + 200]; 50 } 51 y ++; 52 for(int i = y; i <= y + p - 1; i ++){ 53 rsum += a[x + (p - 1 - i + y) + 200][i + 200]; 54 if(p - 1 - i + y != 0) rsum += a[x - (p - 1 - i + y) + 200][i + 200]; 55 } 56 rres += rsum; 57 for(int i = x - p + 1; i <= x + p - 1; i ++){ 58 rsum -= a[i + 200][y + 200]; rres -= a[i + 200][y + 200] * (p - dis(i, y, x, y)); 59 lsum += a[i + 200][y + 200]; lres += a[i + 200][y + 200] * (p - dis(i, y, x, y)); 60 } 61 // printf("%d %d %lld %lld %lld %lld\n", x, y, lres, rres, lsum, rsum); 62 mn = min(lres + rres, mn); 63 mx = max(mx, lres + rres); 64 } 65 } 66 67 int main(){ 68 freopen("rts.in", "r", stdin); 69 freopen("rts.out", "w", stdout); 70 71 scanf("%d%d%d", &n, &m, &p); 72 memset(a, 0, sizeof(a)); 73 for(int i = 1; i <= n; i ++){ 74 for(int j = 1; j <= m; j ++){ 75 scanf("%d", &a[i + 200][j + 200]); 76 //a[i + 200][j + 200] = 1; 77 } 78 } 79 for(int i = 1; i <= n; i ++){ 80 solve(i); 81 } 82 printf("%lld %lld\n", mn, mx); 83 return 0; 84 }
