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[805][805];
int n, m, p;

ll mn = 1e13, mx = 0;

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 = 1;
    ll lres = 0, rres = 0;
    ll lsum = 0, rsum = 0;
    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 + 200][j + 200]; lres += a[i + 200][j + 200] * (p - dis(i, j, x, y));
        }
    }
    for(int i = x - p; i <= x + p; i ++){
        for(int j = y + 1; j <= y + p; j ++){
            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));
        }
    }
    //printf("%d %d %lld %lld %lld %lld
", 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 + 1; i <= y; i ++){
            lsum -= a[x + (p - 1 - y + i) + 200][i + 200];
            if(p - 1 - y + i != 0) lsum -= a[x - (p - 1 - y + i) + 200][i + 200];
        }
        y ++;
        for(int i = y; i <= y + p - 1; i ++){
            rsum += a[x + (p - 1 - i + y) + 200][i + 200];
            if(p - 1 - i + y != 0) rsum += a[x - (p - 1 - i + y) + 200][i + 200];
        }
        rres += rsum;
        for(int i = x - p + 1; i <= x + p - 1; i ++){
            rsum -= a[i + 200][y + 200]; rres -= a[i + 200][y + 200] * (p - dis(i, y, x, y));
            lsum += a[i + 200][y + 200]; lres += a[i + 200][y + 200] * (p - dis(i, y, x, y));
        }
    //    printf("%d %d %lld %lld %lld %lld
", 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, 0, sizeof(a));
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= m; j ++){
            scanf("%d", &a[i + 200][j + 200]);
            //a[i + 200][j + 200] = 1;
        }
    }
    for(int i = 1; i <= n; i ++){
        solve(i);
    }
    printf("%lld %lld
", mn, mx);
    return 0;
}