[BALTIC 2008] Grid

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=1169

[算法]

首先DFS枚举出横着切的

然后二分 + 贪心即可

时间复杂度 : O(2 ^ N * N ^ 2logN)

[代码]

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define N 20
const int inf = 2e9;

int n , m , r , s , cnt;
ll ans;
ll a[N][N] , sum[N][N] , dp[N][N] , b[N];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
inline int calc(int x)
{
        int ret = ;
        for (int i = x; i; i -= i & (-i))
                ++ret;
        return ret;
}
inline ll calc_sum(int X1 , int Y1 , int X2 , int Y2) {
        return sum[X2][Y2] - sum[X1 - ][Y2] - sum[X2][Y1 - ] + sum[X1 - ][Y1 - ];
}
inline bool check(ll limit) {
        int pre =  , cut = ;
        for (int i = ; i <= m; ++i) {
                for (int j = ; j <= cnt; ++j) {
                        if (calc_sum(b[j - ] +  , i , b[j] , i) > limit)
                                return false;
                }
        }
        for (int i = ; i <= m; ++i) {
                ll value = ;
                for (int j = ; j <= cnt; ++j) {
                        if (calc_sum(b[j - ] +  , pre , b[j] , i) <= limit)
                                continue;
                        else {
                                pre = i;
                                ++cut;
                        }
                }
        }
        return cut <= s;
}
inline ll getans(ll S) {
        cnt = ;
        for (int i = ; i < n; ++i)
                if (S & ( << i)) b[++cnt] = i + ;
        b[++cnt] = n;
        ll l =  , r = ans , ret = inf;
        while (l <= r) {
                int mid = (l + r) >> ;
                if (check(mid))
                {
                        ret = mid;
                        r = mid - ;
                } else l = mid + ;
        }
        return ret;
}

int main() {

        read(n); read(m); read(r); read(s);
        for (int i = ; i <= n; ++i) {
                for (int j = ; j <= m; ++j) {
                        read(a[i][j]);
                        sum[i][j] = sum[i - ][j] + sum[i][j - ] - sum[i - ][j - ] + a[i][j];
                }
        }
        ans = sum[n][m];
        for (int i = ; i < ( << n); ++i)
        {
                if (calc(i) == r)
                        chkmin(ans , getans(i));
        }
        printf("%lld\n" , ans);

        return ;

}