Problem B. 即时战略 ———2019.10.12

题目：

 

代码~：感谢土蛋

 #include <iostream>
 #include <cstring>
 #include <cmath>
 #include <cstdio>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <map>
 #include <complex>

 #define lc k << 1
 #define rc k << 1 | 1

 #define inf 0x3f3f3f3f

 using namespace std;
 typedef long long ll;

 int a[][];
 int n, m, p;

 ll mn = 1e13, mx = ;

 inline int dis(int x1, int y1, int x2, int y2){
     return min(abs(x2 - x1) + abs(y2 - y1), p);
 }

 void solve(int x){
     int y = ;
     ll lres = , rres = ;
     ll lsum = , rsum = ;
     for(int i = x - p; i <= x + p; i ++){
         for(int j = y - p; j <= y; j ++){
             if(dis(i, j, x, y) != p) lsum += a[i + ][j + ]; lres += a[i + ][j + ] * (p - dis(i, j, x, y));
         }
     }
     for(int i = x - p; i <= x + p; i ++){
         for(int j = y + ; j <= y + p; j ++){
             if(dis(i, j, x, y) != p) rsum += a[i + ][j + ]; rres += a[i + ][j + ] * (p - dis(i, j, x, y));
         }
     }
     //printf("%d %d %lld %lld %lld %lld\n", x, y, lres, rres, lsum, rsum);
     mn = min(lres + rres, mn);
     mx = max(mx, lres + rres);
     while(y < m){
         lres -= lsum;
         for(int i = y - p + ; i <= y; i ++){
             lsum -= a[x + (p -  - y + i) + ][i + ];
             if(p -  - y + i != ) lsum -= a[x - (p -  - y + i) + ][i + ];
         }
         y ++;
         for(int i = y; i <= y + p - ; i ++){
             rsum += a[x + (p -  - i + y) + ][i + ];
             if(p -  - i + y != ) rsum += a[x - (p -  - i + y) + ][i + ];
         }
         rres += rsum;
         for(int i = x - p + ; i <= x + p - ; i ++){
             rsum -= a[i + ][y + ]; rres -= a[i + ][y + ] * (p - dis(i, y, x, y));
             lsum += a[i + ][y + ]; lres += a[i + ][y + ] * (p - dis(i, y, x, y));
         }
     //    printf("%d %d %lld %lld %lld %lld\n", x, y, lres, rres, lsum, rsum);
         mn = min(lres + rres, mn);
         mx = max(mx, lres + rres);
     }
 }

 int main(){
     freopen("rts.in", "r", stdin);
     freopen("rts.out", "w", stdout);

     scanf("%d%d%d", &n, &m, &p);
     memset(a, , sizeof(a));
     for(int i = ; i <= n; i ++){
         for(int j = ; j <= m; j ++){
             scanf("%d", &a[i + ][j + ]);
             //a[i + 200][j + 200] = 1;
         }
     }
     for(int i = ; i <= n; i ++){
         solve(i);
     }
     printf("%lld %lld\n", mn, mx);
     return ;
 }